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Thermodynamics, PV Diagrams, Internal Energy, Heat, Work, Isothermal, Adiabatic, Isobaric, Physics

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in this video we're going to focus on

solving thermodynamic problems

particularly problems associated with

internal energy heat

work isothermal

isocoric adiabatic processes and even

isobar processes but first let's go over

the equations that we need to

know let's organize it into a table so

on the left side the First Column we're

talk about the different types of

processes that you need to

know and then here we're going to have

work Q which represents the heat flow

also Delta U the change in internal

energy so the first process that we're

process now for an isocoric process what

you need to know is that the change in

volume is equal to zero the volume is

constant and because the volume is

constant the work done for an isocoric

process will always be equal to

zero now the change in internal energy

W so therefore since W is zero the

change in internal energy will simply be

equal to

Q to calculate

Q you can use this equation it's equal

to n * CV * Delta

C now if you don't know what these

variables mean don't worry just sit back

relax take notes throughout the video

I'm going to explain how to use these

equations and also when to use them I'm

going to give you plenty of practice

problems for you to work

on but you need to know that n

represents the number of moles CV is

basically the molar heat

capacity when the volume is constant of

an idal gas and delta T is the change in

temperature delta T can be in Kelvin or

Celsius anytime you see just T in the

equation without the triangle the

temperature has to be in kelvin if you

see the triangle associated with it it

can be in Celsius or kelvin the change

in temperature in celsius and Kelvin is the

the

same but when you multiply or divide by

kelvin now the next thing that we need

in an isobaric process the change in

pressure is equal to zero the pressure is

constant under constant pressure

conditions you can calculate the work

using this

equation it's equal to the pressure

volume by the way you can also calculate

the work using this equation NR * delta T

we'll talk more about that throughout this

this

video now what about Q for an isobaric

process what equation will be useful to

us so like the equation above Q is going

to be equal to NC delta T but instead of

CV it's going to be

CP because we're dealing with constant

pressure CP is the m heat capacity of a

gas measured at constant pressure and

temperature now Delta U you can always

use this

equation it's the heat minus the amount of

work now another equation that you can

use to calculate Delta U for any process

equation it's ncv delta T keep in mind

Delta U is equal to Q for an isocoric

process but Delta is always equal to n

T now the next thing that we need to

process under this process the

temperature is

now because the change in temperature is

equal to

zero notice that Delta U will be equal

to zero so under isothermal conditions

the change in internal energy is always

Zer if that's the case then Q minus W is

equal to Z which means that Q is equal

so how do we calculate the work or the

heat exchange in an isothermal process

let's see if I can fit this equation in

here so one way to do it is to use this

log the final

another equation that you can use in

terms of

pressure keep in mind if you want to

find Q it equals this as well but you

can also use this

equation it's equal to

pressure divided by the final

pressure pressure and volume are

inversely related by the way make sure

you know the gas laws because with this

with these types of problems you can be

dealing with gases for example in an

isocoric process because the volume is

constant if the temperature increases

the pressure will increase

therefore you can use this

T1 is equal to P2 / T2 for an isocoric

process now what about an isobaric

process where the pressure is

constant whenever you increase the

temperature at constant pressure the

volume has to

increase if you've taken chemistry and

you've studied the gas laws this is

known as Charles's

Law whenever you heat a gas the gas will

expand now the equation that you need

for this type of

situation is

T2 now there's a lot more equations that

we need

under isothermal conditions if you

increase the

pressure the volume will decrease and

likewise if you increase the

volume the pressure will

decrease volume and pressure are

inversely related this is known as a

boils law so the equation that is

associated with Bo's law is P1 V1 is

now the last process that we need to go

over is the adiabatic

adiabatic

process under this

process Q is equal to zero so what this

means is that no heat flows into or out

of the system now this can occur if

let's say the process occurs quickly or

if it's well insulated because heat

takes time to flow from one place to

another a process that happens quickly

will allow a very small amount of heat

to travel through so basically you could

zero H I should have put

column now because Q is equal to zero

that means that the change in internal

energy is equal to negative

W so let's write that over here so Delta

U is equal to

W and Delta U is always equal to ncv

delta T which means that the work is

equal to

negative n the number of

moles times

CV the molar heat capacity at constant

volume times the change in

temperature now there are some other

equations associated with this

process and I'm going to add that

throughout this video but something that

you need to know is that CV for a

monoatomic gas is equal to 3 over2 *

R and that's true for any of these

processes a monoatomic gas is simply a

gas that contains one atom for a

diatomic gas we'll talk about what to do

in those circumstances

later make sure you know that CP

is equal to 5/2 * R for a monoatomic ideal

gas so those are the main equations that

you need to know for now there are more

equations but we'll go over that

video so what exactly is

thermodynamics thermodynamics is

basically it's associated with the study

of heat mechanical work and internal

energy and how these three correlates to

each other the first law of

thermodynamics is basically uh this

equation Delta U is equal to Q minus

w u represents the internal energy of a

system and it's proportional to

temperature Q represents the amount of

system there's basically two ways of

transferring energy

and that is through heat and mechanical

work if you apply a force on a box you

can increase its kinetic energy or you

add heat to a system the temperature

will go up and the kinetic energy of the

molecules will

increase whenever heat flows into the

system Q is

positive and Q is negative when heat

a positive Q

value increases the internal energy of a

system because heat is entering into the

system a negative Q value will decrease

the internal energy of the

system the next thing you need to know

is the signs for

work when work is done by the

system W is

positive now let's think about that

imagine if you're doing work if you're

running if you're exercising you're

burning energy as you do work your

internal energy will go

down so whenever the system does work

its internal energy decreases so work is

system W is negative if work is done on a

system that is if the surroundings

perform work on a system a good example

is if you compress a

gas and let's say the system is the gas

work is being done on the system the

internal energy will increase now if the gas

gas

expands then it's doing positive work

decreases now let's work on this

particular practice problem for each of

these questions make sure you pause the

video and work on these problems

yourself that's the best way you're

going to learn

thermodynamics calculate the change in internal

internal

energy the system absorbs 300 Jew of

heat and performs 500 Jews of work so

system now 300 jewles of energy enters the

the

system that means the internal energy

should go up by 300 and also the system

performs 500 jewles of work now because

heat is flowing into the system we know

Q is positive

300 if work is done by the system that means

means

positive so using the equation Delta U

is equal to Q - W this is going to be

300 - 500

so the change in internal energy is200

Jew now let's make sense of this heat

flows from the surroundings into the

system and 300 Jews of heat flows into

it and the system performs work to do

work it has to expend energy so it's to

losing 500 Jews of

energy so therefore we can see that the

net change is 200 it took in 300 jewles

as heat and it lost 500 Jews by means of

surroundings now what about Part B the

system absorbs 720 Jew of heat energy

and the surroundings perform 300 Jew on a

system so is cute positive or negative

in this particular example since heat is

absorbed Q is

positive now what about W is it positive or

negative well the surroundings perform

300 Jews of work on a

system so work is done on a system and

therefore W is negative whenever that's the

case so using the

equation Delta Al to Q minus W it's

going to be

720 minus -300 which is the same as 720

plus 300 and that's equal to 1,20

C what is w in this problem and what's

Q 400 Jew of work is done all on a

system so is W positive or

negative just like before anytime work

is done on a system W is negative now

the system loses 225 jewles of heat

energy so Q is negative because heat

went from the system to the surroundings

anytime the system loses heat or

releases heat into the surroundings Q is

negative so now using this

equation it's going to be -225

minus -400 which is the same as

-225 +

400 which is positive

175 so the internal energy of the system

went up because more work was done on a

system compared to the amount of heat

energy lost by the

system so the system gained a net of 175

jewles of

energy now what is the difference

between an open system a closed system

and an isolated

system in an open

system matter can enter into and out of

the system so let's say like oxygen gas

out in addition energy can flow into or

out of the system so heat can flow into

system matter cannot enter or leave in a closed

closed

system so oxygen gas O2 will not be able

to enter however heat can flow into or

out of a closed system so in an open

system matter and energy can flow into

and out of it but in a closed system

energy can flow into or out of it but

not matter

system no energy or matter can flow into

system so because heat can't flow into

it Q is equal to

zero and no type of energy can enter

into the system so that means that there

can be no work uh done on a system or by

the system so w is zero which means that

the internal energy is constant so the

change in internal energy of an isolated

system is always zero so keep that in

mind the equation Delta U is equal to Q

minus W typically applies to a closed

system where energy can enter into or

out of it for an open system you have to

take into account the amount of matter

that that enters or leave the

system but this equation usually applies

system let's try this problem how much

work is required to compress a

monoatomic ideal gas at a pressure of

2.5 * 10 5

pascals from an initial volume of 015

Cub M to a final volume of 01b

M so is this an isocoric process

isobaric adiabatic or isothermal process

notice that the pressure is constant as

a result this is an isobaric

process and what exactly is an ideal

gas some gases behave almost ideally

others do not basically if a gas behaves

in a way that fits this equation nicely

it behaves it's almost ideally this is

the isog gas law equation pressure time

volume is equal to the moles of the gas

temperature now what about a monoatomic

gas what is that

exactly the word or the prefix

mono means one a monoatomic gas is a gas

that contains molecules that have a

single atom so noble gases such as

helium helium argon

gases contain molecules with two atoms

the word d is associated with two so

oxygen gas nitrogen gas hydrogen gas

these are datomic

datomic

poly means many polyatomic gases have

molecules with many atoms sulfur dioxide

has a total of three atoms 1 plus 2 is

three sulfur trioxide is a polyatomic

gas carbon dioxide too these are gases

atoms now here's another question for

you the gas is being

compressed do you predict the sign of w

to be positive or negative in this

case anytime a gas expands the work done

compression the work is done on the gas

by the surroundings so therefore the

work is negative so our final answer for

the first part of this problem should be negative

so what equation can we use to calculate

the work required to compress this gas

now ear in a video I mentioned that for

isobaric processes the work is equal to

the pressure times the change in volume

but let's derive that

if we have a gas

and let's say this

to compress the

gas the Piston is going to move a certain

D now the work done to apply a force to

move any object is simply the force

and anytime you apply a force over a

certain area you're applying a pressure

pressure

pressure is equal to force / area so if

we multiply both sides by a we can see

that the pressure times the area is

equal to the

force so therefore we can replace f with

p *

a so the work is therefore equal to the

pressure times the

area times the

displacement or the distance that the

gas now for a rectangular solid volume

is equal to the length time the width

time the height and we know that the

area of a rectangle is length time width

so we can view the volume as being the

area times the height and you can think

of the height as being the

change in the displacement of the Piston

so we can say V is equal to a times Delta

D so therefore we can replace this term

V so that's how we can come up with the

V now let's draw a PV diagram for this

isobaric process this is a graph that

contains pressure and volume we're going

to put pressure on the Y AIS volume on

the x-

pascals and we have two volumes the

lower value is 01 and the higher value is

now because the gas is being compressed

we're going to start at the higher value

and travel towards the lower value let's

call this position a and position

B so we're going to travel from A to

B it turns out that the area of the PB

diagram is equal to the work done on a

gas or by the

now we know that the area is equal to

the length times the

width the length of this rectangle is

pressure and the

width of the rectangle is the change in

volume It's the final volume minus the initial

initial

volume so we can say the width is equal

to Delta V the change volume so

therefore another way to get the

equation for work for this particular PV

diagram is equal to the pressure which

is L times the width or the change in

volume so that's another way in which

you can derive this equation from the PV

diagram so sometimes you need to use the

fact that the work is equal to the

change in area it's going to help you to

calculate the work um for problems that

but now let's finish this

problem so w is equal to P Delta V for

an isobaric

pascals the change in volume is the

final volume which is

01 minus the initial volume of 015 and

units for that is cubic

meters now you need to pay attention to

the units 1 Pascal * 1 cubic m is equal

to one

Jewel sometimes you might see the units

liters and

ATMs 1 lit * 1

Jew so you don't need to convert ATM

into pascals and liters into cubic

meters you can simply multiply these two

and convert it once into Jews if you

that now for those of you who want to

convert ATM into pascals and liters into

cubic meters here are the conversion

factors that you need to

know 1 atm one atmospheric pressure is

1.013 * 10 5 pascals

and 1 cubic meter is equal to 1,000

lit so you may or may not need those

conversions but you have them now just

in case if they ever become important in

the future but now let's go back to this

problem so if you type this in exactly

the way you see it in your

1250w and we could see why it's negative

because the gas is compressed by an outside

outside

force so now that we have the work done

on the

gas what is the change in the internal

system now we have the heat and heat is

being released by the system so Q is- 350

so we can use the equation Delta U is

equal to Q minus W so that's

-350 minus

1250 so

therefore the change in internal

Jews let's try this problem how much

work does a gas perform at a constant

pressure of 8.4 ATM as it expands from a

volume of 2 L to 4.5

l so under constant pressure conditions

we can use this equation once

more work is equal to pressure times the

change in

volume so the pressure is 8.4

atmospheres and the volume changes from

final minus initial 4.5 minus

minus 2.0

l so this is equal to

21 L *

ATM now typically you want your answer

to be in Jews so let's convert

it as we mentioned before there's 101.3

atm notice that because the gas expanded

the final answer is 2,127

jewles 4.5 moles of an ideal gas is

heated at constant pressure from 30 C to 150

150

C calculate the work performed by this

gas draw a PB diagram for the process as

well so the equation that we can use is

this equation work is equal to n * R * delta

T now keep in mind PV is equal to nrt

this is the idog gas law equation so

therefore the pressure times the change

in volume must be equal to n r * the

change in

t so so if work is equal to P Delta V

T N is the number of moles we have 4.5

moles of an Ido

gas R is the energy constant for a gas it's

8.3145 jewles per mole per Kelvin

now the temperature difference in

Celsius is equal to the temperature

difference in kelvin anytime you have a

delta T value it can be in kelvin or

Celsius the change in temperature 150

minus 30 is

Jews so let's multiply these three numbers

numers so the final answer for this

problem is

jewles now the work is positive because

as you heat the gas it will expand

according to Charles's Law whenever you

increase the temperature of a gas at

constant pressure the volume of that gas must

diagram now that we know the volume is

increasing we know which direction in

the PV diagram we're moving

towards so here's the pressure here's the

the

volume and the current pressure is about

we don't know what the current pressure

now we don't know what V1 or V2 is equal

to but we can just call it V1 V2 we do

increasing so if we call this point a

and point B at Point a the temperature

is 30° C and at point B it's

150 to increase the volume you have to

increase the temperature or at least

that's one way in which you can increase the

the

volume so imagine if you have a a

piston and then you have a gas

inside which is at 30°

C if you add heat to this

gas particularly at a slow rate so that

the pressure is

constant the temperature of the gas will

increase and as that happens the vol

volume will

increase the gas will exert a

force that will push up the Piston

pressure and as the volume increases the

gas is performing work against the

surroundings so anytime you heat up a

gas at constant pressure the volume will increase

the pressure of a gas inside a rigid 2.4

L cylinder is decreased from 4.8 to 3.4

ATM so how can we accomplish this well

let's say if we have a metal rigid

cylinder and there's a gas inside let's

say the temperature of this gas is 300

Kelvin and at this temperature let's say

that the pressure is 4.8 ATM

now the volume is constant because we

have a rigid cylinder the cylinder has a

fixed volume of 2.4 l so therefore this

is an isocoric

process under constant volume the only

way to really decrease the

temperature if the air of the

surroundings let's say if it's less than

300 Kelvin let's

put it at 200 Kelvin heat is going to

naturally flow from hot to cold so heat

energy is going to leave the gas into

the colder surroundings and as that

happens the pressure will

decrease the relationship between

pressure and temperature under constant volume

volume

conditions is this this is uh believe

it's called gayus sax

law temperature and pressure are

directly related

so we can use this equation to calculate

the new pressure or the new temperature

uh whichever we need but we don't need

to do all of that for this problem

whenever you have an isocoric

process you need to know that the work

is always equal to zero whenever the

constant for one thing if you use the

equation work equals P Delta

V the change in volume is zero so the

work has to be zero that's one way you

4.8 and it's decreasing into 3.4 at a

l so we're going from position a to position

position

B notice that this graph has no

area because the area is

zero that is the area between this curve

and the x axis since there's no area the

work is therefore equal to zero so for

any isocoric process W is always equal to

zero how much heat energy is required to

increase the temperature of 7 moles

of each of the following gases by 50

kin so first we have helium at a

constant volume of 5.4

L perhaps in an earlier chapter you've

learned that Q is equal to M cap where m

is the mass C is the specific heat

capacity and delta T is the

temperature Q can also be used in terms

of moles it's equal to n c * Delta V

where C is the molar heat capacity

at constant volume you want to use CV

the molar heat capacity measured at a

constant volume

process and under constant pressure

conditions you want to use

pressure now what

exactly is CV for

helium we see we have an isocoric

process the volume is constant but what

is is CV for H how can we find it Well

the first thing you should do is look it

up in your

12.47 Jews per mole per

Kelvin now if you don't have this value

you can estimate it using this equation

CV is approximately equal to 32 * R for

gas now let's see how close the answer

is going to be or the CV value rather so

8.3145 and let's multiply that by 3 and

divid by

two this is equal to

12.47 2 which is very close to this

value so for most monoatomic gases

cval 32 R is a good approximation of the actual

actual

value so chances are for a question like

this you probably won't be given the CV

value you just have to know that CV is

equal to 3 over2 * R for monatomic

gas so now let's use the

equation Q is equal to ncv delta

T so we have seven moles of gas CV let's use

use

12.47 and delta T the change in

Kelvin so

Q is equal to positive

positive

43645 juw

a positive Q value is typically

associated with an increase in

temperature so the change in temperature

positive now what about Part B at a

constant pressure of 9.2

ATM by the way if you have the ch in

temperature you really don't need to use

the value of the volume or the value of the

the

9.2 it's only important if you don't

know the change in

temperature so for Part B Q is going to

be equal to n * CP * delt T so what is

CP for a monoatomic gas argon is a monoatomic

monoatomic

gas it's a gas that is composed of

particles that have only one atom N2 is

molecule now we said that CV is 3 over2

R for a monoatomic gas but CP is going

to be 5 over 2 * R the actual value for

20.78 and if we take 5 over2 and

8.3145 this will give us

20.78 which is still pretty close to this

number so for monoatomic gases make sure

that you know that CP is equal to 5

over2 * R because if you don't have

access to a table or if this value is

not provided you have to use uh this

fact to calculate

CP so Q is going to be stimes

Kelvin so this is positive

now what about part C where we have a

diatomic molecule or part D where we

have a molecule with three atoms what

can we do so let's

review so CV is equal to 3 over2 * R for

a monoatomic gas that is a gas basically

with one atom in a molecule or in a

particle for diatomic gases it's

approxim equal to 5 2 * R and nitrogen

fits that

description now for polyatomic gases

with three molecules it turns out that

it's approximately 7/2 * R however at

this point it begins to deviate so let's

use an approximately equal 2

symbol so CO2 would fit this example it

has three atoms per

gas it's 7 over2 * R for a diatomic gas like

approximately 9 /2 *

triatomic so now let's go back to

N2 if you look up in your textbook the

values for CP and CV for N2 you'll see

that CV is 20.7

CP is 29.7

for N2 at least that's according to the

textbook that I

have so for diatomic gas who said it's 5

over 2 * R so if you take 8.3145

multiply by 5 and divide by two this is

going to give you 20. 786 which we

calculated that earlier and as you can

see it's not exactly the same but it's

still pretty close so for monoatomic

gases and diatomic gases

you can use these equations to get a good

good

approximation for CP it's 7/2 *

R so 8.3145 * 7 / 2 is about

29.1 which is still a good approximation for

for

CP once you start having molecules with

three atoms then it begins to

deviate but let's finish part C so Q is

going to be n times not CP but we're

going to use

CV since we want to find out how much

heat is required to increase the

volume so it's going to be 7 *

50 and this is

now for Part D CO2 let's talk about the

values of CV and

2846 and

CP is equal to 36.9

4 so now if we approximate CV using the

equation 7/2

R this is going to be 8.31 4 5 * 7 / 2

and that's

29.1 as you can see there's a slight

difference uh between these two

values you might be maybe 2% off but

it's still pretty close so if you don't

have the actual value of CV for CO2 go

ahead and use this approximation but if

you have it preferably use this one now

delta

T so it's 7 *

Jew the temperature of a sample of ideal

gas increase from 300 Kelvin to 500

kin calculate the change of the internal

energy if there were five moles of a monoatomic

gas anytime you want to find a

change in the Eternal energy of the

system you can use this equation Delta U

is equal to n CV delta

T now keep in mind for a monoatomic ideal

ideal

gas the CV value is 32 *

R so Delta U is equal to n or simply 3

over2 NR delta T for monoatomic idog

gas so it's going to be 3 over2 * 5

moles R is

8.3145 and the change in

temperature final minus the initial

value that's 500 - 300 which is a change

12,472 Jews round it to the nearest whole

number now what about Part B for Part B

we can use the same equation

but for a datomic gas

CV is equal to 5/ 2 *

R so therefore for datomic gas we have

T and this part n is going to

be 8 R is

so this is equal to

how much work is required to compress

7.5 moles of an idal gas at 400 Kelvin

from 5.2 L to 2.1

l so what type of process do we

diagram notice that the temperature is

constant under constant temperature

conditions we have an isothermal

process and during an isothermal

process the change in internal energy is

zero so that's the answer for the second

part of the

problem keep in mind Delta U is always

equal to n CV delta T for any

process so for an isothermal process

delta T is zero which means that that

zero now just because the temperature is

constant doesn't mean q is zero heat can

still flow into the system or out of the

system it just happens in such a way

constant now during an ISO thermal

process as the volume of the gas

increases the pressure will decrease

now it's not going to decrease like this

according to B's law it's going to

decrease more like

that it's not a straight line it's actually

curve but how can we calculate the work required

to compress this particular gas so if

the gas is being compressed let's say if

we're going from position a to position B

we're going this way as you travel

towards the left of a PV diagram the

work done is going to be negative

anytime you're compressing a

gas W is negative anytime the gas

expands W is

positive now the work done is going to

be equal to the area under the curve so

how can we calculate this particular

area it doesn't look like a rectangle so

we can't use the area equation which is

length time

width to get the formula that you need

calculus the work required which is the

area under the curve is simply the

anti-derivative of the pressure

so it's the anti-derivative of P

DV now according to the ideal gas law

nrt let's get p as a function of v so if

we divide both sides by

v p is equal to

V so there therefore we have this

equation we can now replace p with nrt over

V now since the differential is based on

the variable

V we're going to treat nrt as constants

so we can move it in front of the

integral so the work is going to be

nrt times the

anti-derivative of 1 / V DV

now for those of you who have taken

calculus you know that the

anti-derivative of 1 /x is equal to Ln X

for those who haven't taken calculus

yet just sit back relax and then you'll

see the equation that you need in order

to solve this problem I'm just going to

equation so if the anti-derivative of 1

/x is L and X the anti-derivative

log of V evaluated from position a to

position B which position a was the

initial volume and position B was the final

volume so it's going to be

nrt we need to plug in these two values

the final value minus the initial

value so it's nrt Ln V

final minus nrt

nrt

initial now if you recall from your

algebra class or pre-calculus

class the natural log of a minus the

natural log of B can be Rewritten as Ln

a / B so therefore

we can combine these two expressions as

a single log so the equation that we

need for this problem is

nrt natural

now let's finish this problem n is the

number of moles which is

7.5 R is the energy constant for gases

8.3145 T is the temperature 400

Kelvin and then this is going to be Ln V

final which is

22617

Jew and we can see why it's negative

because the gas is being compressed from

a high volume to a low volume so that's

how you can calculate the work for an isothermal

isothermal

process now what about the heat we know

that the internal energy the change and

the internal energy is zero but how can

we calculate the heat transferred during this

this process well one equation that you can

process well one equation that you can use is Delta U is equal to Q minus W if

use is Delta U is equal to Q minus W if we add W to both sides we can see that Q

we add W to both sides we can see that Q is Delta u+ W and since Delta U is zero

is Delta u+ W and since Delta U is zero Q is equal to W so Q is also 22,6

Q is equal to W so Q is also 22,6 17 that's how much heat is transferred

17 that's how much heat is transferred so basically let's understand what's

so basically let's understand what's happening in this

situation consider a cylinder with a movable

cylinder with a movable piston and let's say that we have a gas

inside now as work is being done on the gas to compress it so work is

gas to compress it so work is negative we can see that the volume is

negative we can see that the volume is decreasing the gas is releasing heat

decreasing the gas is releasing heat into the environment now this is

into the environment now this is occurring at such a rate that the

occurring at such a rate that the temperature remains

temperature remains constant so the energy that's going into

constant so the energy that's going into the system to compress the gas is

the system to compress the gas is leaving the system as

leaving the system as heat so 22,00 ,

heat so 22,00 , 66 Jew of

66 Jew of work goes into the system to compress

work goes into the system to compress the

gas so that action increases the internal energy by that amount however

internal energy by that amount however heat

heat leaves from the gas and about

leaves from the gas and about 22,616 Jew

22,616 Jew leave the

leave the gas at the same time so that decrease es

gas at the same time so that decrease es the internal energy so therefore the

the internal energy so therefore the change in internal energy is zero the

change in internal energy is zero the amount of energy that is being

amount of energy that is being transferred into the gas by means of the

transferred into the gas by means of the mechanical work is equal to the amount

mechanical work is equal to the amount of energy that leaves a gas um by means

of energy that leaves a gas um by means of heat so the gas gains no internal

of heat so the gas gains no internal energy the internal energy remains

energy the internal energy remains constant in this particular

problem 3.2 moles of an idal gas expands from a pressure of 5 ATM to 1.4 ATM at a

from a pressure of 5 ATM to 1.4 ATM at a constant temperature of

constant temperature of 127° is work done by the gas or on the

127° is work done by the gas or on the gas anytime a gas expands the work is

gas anytime a gas expands the work is positive and whenever the work is

positive and whenever the work is positive work is done by the gas on the

surroundings so now how can we calculate the work for this

the work for this problem once again the temperature is

problem once again the temperature is constant so we have an isothermal

constant so we have an isothermal process with a PV

process with a PV diagram that looks like

diagram that looks like this so this time we're traveling

this so this time we're traveling towards the right anytime you travel

towards the right anytime you travel towards the right of a PV diagram the

towards the right of a PV diagram the work done is

positive now this problem is slightly different from the last one because in

different from the last one because in the last problem we had two values for

the last problem we had two values for volume but here we have two values of

volume but here we have two values of pressure so what can we do in this

pressure so what can we do in this problem well let's start with this

problem well let's start with this equation W is equal to

equation W is equal to nrt natural

nrt natural log V final / V

log V final / V initial for an isothermal process since

initial for an isothermal process since temperature is constant whenever you

temperature is constant whenever you increase the volume the pressure will

increase the volume the pressure will decrease according to Bo's law P1 V1 is

decrease according to Bo's law P1 V1 is Al to P2 V2 so we could say p initial *

Al to P2 V2 so we could say p initial * V initial is equal to P final * V

V initial is equal to P final * V final now let's divide both sides by V

final now let's divide both sides by V initial and also by

initial and also by pfal if we do that notice what happens V

pfal if we do that notice what happens V initial cancels on the left side pfal

initial cancels on the left side pfal cancels on the right side so therefore P

cancels on the right side so therefore P initial / P

initial / P final is equal to V final / V initial so

final is equal to V final / V initial so we can replace V final over V initial

we can replace V final over V initial with P initial P

with P initial P final so therefore this is the equation

final so therefore this is the equation that you want to use for this

that you want to use for this problem so during an

problem so during an isothermal expansion or compression you

isothermal expansion or compression you can use this equation W is equal to

can use this equation W is equal to nrt

nrt Ln P initial / P final

so now let's finish the problem n in this problem is 3.2 moles R

problem n in this problem is 3.2 moles R is still

8.3145 now the temperature has to be in kelvin if you have any equation where

kelvin if you have any equation where you just see T like in pval nrt or this

you just see T like in pval nrt or this equation it has to be in

equation it has to be in kelvin but for equations where you would

kelvin but for equations where you would see the term delta T it can be in kelvin

see the term delta T it can be in kelvin or Celsius just keep that in

or Celsius just keep that in mind so to get the Kelvin temperature we

mind so to get the Kelvin temperature we need to add

need to add 273 to

127 which is about 400 Kelvin let's multiply that by the

Kelvin let's multiply that by the natural log of the initial pressure

natural log of the initial pressure which is 5 ATMs divided by the final

which is 5 ATMs divided by the final pressure of 1.4

pressure of 1.4 for so the work done should be

for so the work done should be positive and it's done by the

gas so the answer to this problem is positive

positive 13, 548

jewles now as you mentioned before for an iso thermal process Delta U is equal

an iso thermal process Delta U is equal to zero and Q is equal to

to zero and Q is equal to W so therefore the system absorbed

W so therefore the system absorbed 13548 jewles of heat energy and at the

13548 jewles of heat energy and at the same time it used that energy to perform

same time it used that energy to perform 13548 jewles of

13548 jewles of work so the energy that entered the

work so the energy that entered the system as heat left it by means of

system as heat left it by means of mechanical work

an ideal monoatomic gas expands from 2.5 L at 300 kin to 8 L at a constant

L at 300 kin to 8 L at a constant pressure of 3

pressure of 3 ATM what is the temperature when the

ATM what is the temperature when the volume is 8

l so let's draw a PB diagram now the first thing you want to do is identify

first thing you want to do is identify what type of process we have and it

what type of process we have and it turns out that this is an isobaric

turns out that this is an isobaric process since the pressure is constant

process since the pressure is constant the volume changes from 2.5

the volume changes from 2.5 L to 8

ATM so the work has to be positive since we're traveling towards the right

since we're traveling towards the right of the PV

of the PV diagram so we're going from position a

diagram so we're going from position a to position B at position a the

to position B at position a the temperature is 300

temperature is 300 Kelvin we need to calculate the

Kelvin we need to calculate the temperature at position

temperature at position B so how can we do

B so how can we do that

that well we can use this equation V1 / T1 is

well we can use this equation V1 / T1 is equal to V2 / T2 under constant pressure

equal to V2 / T2 under constant pressure conditions this is known as Charles's

conditions this is known as Charles's Law V1 is 2.5 liters T1 the temperature

Law V1 is 2.5 liters T1 the temperature that corresponds to it is 300 Kelvin so

that corresponds to it is 300 Kelvin so we want to find the new

we want to find the new temperature at a volume of 8 l so let's

temperature at a volume of 8 l so let's cross multiply 300 * 8

cross multiply 300 * 8 is400 and this is going to be equal to

is400 and this is going to be equal to 2.5 *

T2 so 2400 divided by 2.5 will give us a

2400 divided by 2.5 will give us a temperature of 960

K now what about Part B how many moles of gas are

present let's use the equation PV is equal to nrt

equal to nrt the pressure in this

the pressure in this problem is 3

problem is 3 ATM now we can use point a or point B

ATM now we can use point a or point B let's focus on point a at Point a the

let's focus on point a at Point a the volume is 2.5 L we're looking for

volume is 2.5 L we're looking for n r is going to be

8206 now the R value that you want to use in this equation is dependent on the

use in this equation is dependent on the units of p and v there's two R values

units of p and v there's two R values that you need to be aware

that you need to be aware of you've seen this one

of you've seen this one 8.3145 jewles per mole per

8.3145 jewles per mole per Kelvin if you're going to use this r

Kelvin if you're going to use this r value the pressure has to be in pascals

value the pressure has to be in pascals and the volume has to be in cubic meters

and the volume has to be in cubic meters because 1 Pascal * 1 Cub met is equal to

because 1 Pascal * 1 Cub met is equal to a

Jew now the other r value which is the one that we're

one that we're using

using 8206 it has the units liters time

8206 it has the units liters time ATM / moles time Kelvin and since the

ATM / moles time Kelvin and since the pressure is in ATM and the volume is in

pressure is in ATM and the volume is in liters this is the r value that we want

liters this is the r value that we want to use in this

to use in this equation T at Point a is 300

equation T at Point a is 300 Kelvin if you use the values at Point V

Kelvin if you use the values at Point V I mean not V but point B you should get

I mean not V but point B you should get the same answer so let's calculate n

the same answer so let's calculate n it's going to be 3 *

it's going to be 3 * 2.5 ID 300 and then take that result ID

2.5 ID 300 and then take that result ID 08

206 so you should get

get 304 66

moles now let's calculate the work performed by this

performed by this gas so we can use the equation W is

gas so we can use the equation W is equal to NR

equal to NR delta

delta T so we have n which

T so we have n which is

30466 now for R I'm going to use 8.3145 because I want my answer to be in

8.3145 because I want my answer to be in Jews and delta T the change in

Jews and delta T the change in temperature the final minus the initial

temperature the final minus the initial temperature that's 660

Kelvin so multiplying these three numbers

numbers will give

will give us a value of

us a value of 1672 jewles so that's how much work is

1672 jewles so that's how much work is performed by this

performed by this gas now keep in mind you can also

gas now keep in mind you can also use this

use this equation W is also equal to P Delta

equation W is also equal to P Delta V under constant pressure

V under constant pressure conditions so the pressure is three the

conditions so the pressure is three the change in volume the final minus the

change in volume the final minus the initial volume that's 8 minus

initial volume that's 8 minus 2.5 which is uh

2.5 which is uh 5.5 * 3 that's equal to

5.5 * 3 that's equal to 16.5 L time

16.5 L time ATM now you need to convert this to

ATM now you need to convert this to jewels so we need to multiply by

jewels so we need to multiply by 101.3 jewles per liter per

ATM and if you do that you should get approximately around 1671

approximately around 1671 jewles which is pretty close to this

jewles which is pretty close to this answer by the way keep in mind this

answer by the way keep in mind this answer is

rounded now how can we calculate the change in the internal

change in the internal energy of the

energy of the system to do that the equation that we

system to do that the equation that we need is Delta U is equal to n CV delta

need is Delta U is equal to n CV delta T this is true for any process so make

T this is true for any process so make sure you keep this equation in mind so n

sure you keep this equation in mind so n is still

30466 now we have a monoatomic gas so CV is 3 over2 * R that's 3 over2

gas so CV is 3 over2 * R that's 3 over2 * I mean time

8.3145 we want Delta U to be in Jews and the change in temperature is 6 60

kin so this gives us a value of 2500 and8 Jew so that's the change in

2500 and8 Jew so that's the change in the internal energy of the

system so now how can we answer Part D how much heat energy was

how much heat energy was transferred well there's different ways

transferred well there's different ways in which we can do

in which we can do that so first let's make some

space one equation that we can use is the equation that is associated

use is the equation that is associated with the first law of thermodynamics

with the first law of thermodynamics Delta U is equal to Q minus W therefore

Delta U is equal to Q minus W therefore Q is Delta u+

Q is Delta u+ W since we have those two values Delta

W since we have those two values Delta to U is

to U is 258 and W is

1672 so therefore the amount of heat transferred is about

transferred is about 4,180 Jews now this is not the only way

4,180 Jews now this is not the only way in which we can get this

in which we can get this answer we can also get it using other

answer we can also get it using other equations as well

equations as well you can use this equation Q is equal to

you can use this equation Q is equal to n CP delta

n CP delta T So n is still

so this will also give you 4,180 Jew rounded to the nearest whole

4,180 Jew rounded to the nearest whole number so that's how you can find Q if

number so that's how you can find Q if you have uh the change in internal

you have uh the change in internal energy and if you have the work or you

energy and if you have the work or you can also find it in terms of temperature

can also find it in terms of temperature or you can find it in terms of

or you can find it in terms of pressure and volume so let me give you a

pressure and volume so let me give you a third equation that might be useful if

third equation that might be useful if you don't have the

you don't have the temperature and it's uh Q is equal to P

temperature and it's uh Q is equal to P *

* CP Delta V / R by the way this equation

CP Delta V / R by the way this equation only applies

only applies to processes dealing with constant

to processes dealing with constant pressure that is isobaric

pressure that is isobaric processes and this equation also so

processes and this equation also so applies to isobaric

applies to isobaric processes so if you have that chart that

processes so if you have that chart that I gave you earlier it might be good to

I gave you earlier it might be good to expand that chart and add these

expand that chart and add these equations to the isobaric uh

row I couldn't fit all of the equations that I had on the screen so I'm going

that I had on the screen so I'm going over it

over it now so let's plug in the values that we

now so let's plug in the values that we have so P which is constant is 3 ATM

have so P which is constant is 3 ATM and

and CP that's going to be 5 over 2 *

R since we have a monoatomic gas at constant

constant pressure and Delta

pressure and Delta V is 8 L minus 2.5 so that's a change of

V is 8 L minus 2.5 so that's a change of positive

5.5 let me uh get rid of something first since this is in the

way now I really don't have to plug in the r value because as you can see R

the r value because as you can see R will

cancel so therefore it's going to be 3 * 2.5 which is 5 over 2 *

5.5 now this answer is in liters times ATM because pressure was in ATM and the

ATM because pressure was in ATM and the volume was in liters so we need to

volume was in liters so we need to convert

convert it so as of now you should get

41.25 lers time ATM so we need to multiply by 101.3

jewles and if you do that you should get about 4,179

as you can see these answers are very very close so that's how you can

very close so that's how you can calculate the heat

calculate the heat transferred in terms of pressure and

transferred in terms of pressure and volume instead of using

temperature so you have many different ways in which you can calculate q but

ways in which you can calculate q but now let's talk about how we can derive

now let's talk about how we can derive that

equation so it comes from this equation Q is equal to n CP delta T

Q is equal to n CP delta T now we know that PV is equal to nrt the

now we know that PV is equal to nrt the ideal gas law

ideal gas law equation and in this problem the

equation and in this problem the pressure is constant but the volume is

pressure is constant but the volume is not constant the volume changes so

not constant the volume changes so therefore we could say that P Delta V is

therefore we could say that P Delta V is equal to NR delta T since pressure is

equal to NR delta T since pressure is constant only volume and temperature is

constant only volume and temperature is changing so I added a triangle to only

changing so I added a triangle to only that portion of the

equation now we need to solve for delta T So P Delta V / NR is equal to delta T

T So P Delta V / NR is equal to delta T so now let's replace Delta t with this

so now let's replace Delta t with this value so Q is equal to n * CP * P Delta

value so Q is equal to n * CP * P Delta V /

V / NR so we can cancel n so therefore Q is

NR so we can cancel n so therefore Q is equal to P the pressure time CP the

equal to P the pressure time CP the molar heat capacity at constant pressure

molar heat capacity at constant pressure times the change in volume /

times the change in volume / R so keep in mind CP is equal to 5/2 R

R so keep in mind CP is equal to 5/2 R for a monoatomic gas for a diatomic gas

for a monoatomic gas for a diatomic gas it's 7 over2 R and if you have a gas

it's 7 over2 R and if you have a gas with three atoms it's going to be 9/ 2 R

with three atoms it's going to be 9/ 2 R approximately so you can use that

approximately so you can use that estimation if you don't have the exact

estimation if you don't have the exact value but if you can look up the value

value but if you can look up the value in your textbook for that particular gas

in your textbook for that particular gas molecule make sure you use that value

molecule make sure you use that value because it's more accurate

700 jewles of work was done on an igil gas in imp perfectly insulated

gas in imp perfectly insulated container how much heat energy was

container how much heat energy was transferred so

transferred so first is the work positive or is it

first is the work positive or is it negative since work is done on the gas

negative since work is done on the gas the work is negative 700 Jew Anytime

the work is negative 700 Jew Anytime work is done on a gas the gas under go

work is done on a gas the gas under go compress

ression now what type of process do we have is this an isobaric process

have is this an isobaric process isochoric adiabatic or

isochoric adiabatic or isothermal the key expression here is

isothermal the key expression here is the gas is contained in a perfectly

the gas is contained in a perfectly insulated

insulated container so if it has good insulation

container so if it has good insulation it's going to be difficult for heat to

it's going to be difficult for heat to flow into or out of the

flow into or out of the system so therefore for Q is equal to

system so therefore for Q is equal to zero if heat energy cannot flow into or

zero if heat energy cannot flow into or out of the system Q is zero which means

out of the system Q is zero which means that we have uh an adiabatic

process so the answer to part A is zero now what about Part

zero now what about Part B well Delta U is equal to Q minus W

B well Delta U is equal to Q minus W since Q is zero for an adiabatic process

since Q is zero for an adiabatic process Delta U is negative w

Delta U is negative w and W

and W is700 which means Delta U is equal to

is700 which means Delta U is equal to positive

700 so in an adiabatic process the work performed on a

process the work performed on a gas goes directly to change its Eternal

gas goes directly to change its Eternal energy so 700 jewles of work was done on

energy so 700 jewles of work was done on a gas as a result the internal energy

a gas as a result the internal energy went up by 700

went up by 700 Jew now will the temperature increase or

Jew now will the temperature increase or decrease well Delta U is proportional to

decrease well Delta U is proportional to delta

delta T so if Delta U is

T so if Delta U is positive delta T will be positive which

positive delta T will be positive which means that the temperature will

means that the temperature will increase so anytime you have an

increase so anytime you have an adiabatic compression the temperature

adiabatic compression the temperature will go up and during an adiabatic

will go up and during an adiabatic expansion the temperature will go down

so as we compressed the gas adiabatically the net effect was to

adiabatically the net effect was to increase its internal energy and

increase its internal energy and therefore the temperature went

up let's work on this one 3 moles of Neon gas expands adiabatically against a

Neon gas expands adiabatically against a piston and the temperature drops from

piston and the temperature drops from 800 Kelvin to 500 Kelvin what is the

800 Kelvin to 500 Kelvin what is the change in the internal energy of the

change in the internal energy of the system or of the gas

to find the change in the internal energy we can use this equation ncv

energy we can use this equation ncv delta T which applies to any

delta T which applies to any process so n is 3

process so n is 3 moles now neon is a monoatomic gas so CV

moles now neon is a monoatomic gas so CV is equal to 3 over2 * R so that's 3

is equal to 3 over2 * R so that's 3 over2

over2 *

8.3145 and delta T the change in temperature 500 minus 800 so it drops by

temperature 500 minus 800 so it drops by 300

300 kin so delta T is

Jews Part B the amount of heat transferred during any adiabatic process

transferred during any adiabatic process is always zero

is always zero so in this case Delta U is equal

so in this case Delta U is equal to W which means that W is equal to

to W which means that W is equal to negative Delta U and if Delta U is

negative Delta U and if Delta U is 11,225 that means the work done by the

11,225 that means the work done by the gas is positive

gas is positive 11,225

11,225 Jew anytime a gas expands the work done

Jew anytime a gas expands the work done by that gas is equal to a positive value

let's try this problem we have .5 moles of a datomic gas particularly nitrogen

of a datomic gas particularly nitrogen gas and it's placed inside a 5.0 rigid

gas and it's placed inside a 5.0 rigid container at a pressure of 2 ATM this

container at a pressure of 2 ATM this gas is heated until it reaches a

gas is heated until it reaches a pressure of 8 ATM so let's begin by

pressure of 8 ATM so let's begin by drawing a PV diagram

so the volume is 5.0 lit the initial pressure is two

lit the initial pressure is two atmospheres and it increases to eight

atmospheres notice that this is an isocoric process the volume is constant

isocoric process the volume is constant since the gas is contained in a fixed

since the gas is contained in a fixed rigid 5.0 lit

container for an isocoric process to work done by the gas is zero so no work

work done by the gas is zero so no work is done in this

is done in this process so how can we calculate how much

process so how can we calculate how much heat energy will be absorbed by the

gas particularly in terms of pressure and volume an equation that you can use

and volume an equation that you can use is this one Q is equal to V * CV * Delta

is this one Q is equal to V * CV * Delta P /

R the volume is 5 L CV is

L CV is 20.7 the change in

20.7 the change in pressure 8 minus 2 so that's an increase

pressure 8 minus 2 so that's an increase of 6 ATM and R is 8.3145

so this gives us a value of 74.9 but V has the units liters P has

74.9 but V has the units liters P has unit

unit ATM so we need to multiply this number

ATM so we need to multiply this number by

by 101.3 to convert it to

88w so that's how much heat energy is absorbed in the

absorbed in the process and we know that Delta U is Q

process and we know that Delta U is Q minus W and since W is zero Delta U is

minus W and since W is zero Delta U is equal to Q in this

problem so as a result Delta U is also 7,500 and

88 now we know that Delta U is always equal to this equation

is always equal to this equation ncv delta T so you can also find Q using

ncv delta T so you can also find Q using that

equation so n is 0.5 CV is 20.6 the only thing we're missing is

20.6 the only thing we're missing is delta

delta T which we can find that as we work on

T which we can find that as we work on Part B

this so what is the temperature when the pressure is 2 ATM

temperature when the pressure is 2 ATM and when the volume is

and when the volume is five so how can we find it well let's go

five so how can we find it well let's go back to the ideal gas law equation PV is

back to the ideal gas law equation PV is equal to

nrt so at point a the pressure is 2 atmospheres the volume is 5 l n

atmospheres the volume is 5 l n is5 r is

8206 and we could solve for T so it's going to be 2 * 5 which is 10 ID 0.5

going to be 2 * 5 which is 10 ID 0.5 which is 20 ID 8

which is 20 ID 8 206 so at Point a the temperature is

B this time the pressure is going to be eight instead of five I mean instead of

two so basically if the pressure is increased by a factor of

pressure is increased by a factor of four the temperature should be increased

four the temperature should be increased by a factor of four so if we multiply 2

by a factor of four so if we multiply 2 43.7 * 4 it should be about

9748 but that answer is round an answer because I the this

because I the this answer so let's use the equation 8 * 5 /

answer so let's use the equation 8 * 5 / .5 that's 80 ID

.5 that's 80 ID 8206 so that's

now we can find the heat transferred using the other

using the other equation so delta T is going to be the

equation so delta T is going to be the final temperature which is

final temperature which is 9749 Kelvin minus the initial

9749 Kelvin minus the initial temperature of 243.166

in this particular process is positive

process is positive 7,000

7,000 590 Jew which is very very close to the

590 Jew which is very very close to the other answer of

other answer of 7,588

7,588 Jew so as you can

Jew so as you can see you can find

see you can find Q if you have the change of temperature

Q if you have the change of temperature using this equation for a constant

using this equation for a constant volume process

volume process or if you have the volume and

or if you have the volume and pressure you can use this equation Q is

pressure you can use this equation Q is equal to V CV Delta P / R but now let's

equal to V CV Delta P / R but now let's talk about how we can derive that

talk about how we can derive that equation because it really comes from

equation because it really comes from this

this equation n CV Delta

equation n CV Delta C Now using the ideal gas law equation

C Now using the ideal gas law equation pval

pval nrt we need to replace Delta t with

nrt we need to replace Delta t with something now in this problem the volume

something now in this problem the volume is constant so the pressure is changing

is constant so the pressure is changing so Delta P * V is equal to NR delta T So

so Delta P * V is equal to NR delta T So solving for delta T it's Delta P * V

solving for delta T it's Delta P * V over

NR so let's replace Delta t with this value so it's

replace Delta t with this value so it's going to be n CV Delta P * V over NR

so Q is equal to the volume times the molar heat capacity

volume times the molar heat capacity measured at constant volume times the

measured at constant volume times the change in pressure / R so that's how you

change in pressure / R so that's how you can find Q if you have an isocoric

can find Q if you have an isocoric process and if you're given the volume

process and if you're given the volume and the two different pressure

and the two different pressure values if you have the change in

values if you have the change in temperature then it's easier to use this

temperature then it's easier to use this equation

and also keep in mind that Q is equal to Delta U for any isocor process since W

Delta U for any isocor process since W is equal to0 so make sure you write

is equal to0 so make sure you write these down um in your

these down um in your chart so you can organize all of the

chart so you can organize all of the equations you need for the different

equations you need for the different types of processes that you'll see on

types of processes that you'll see on your

exam an ideal monoatomic gas is compressed from 12 L to 3 l at a

compressed from 12 L to 3 l at a constant pressure of 2.58

constant pressure of 2.58 heat is added to increase the pressure

heat is added to increase the pressure at constant volume until the temperature

at constant volume until the temperature reaches its original value calculate the

reaches its original value calculate the total work done by the gas and the total

total work done by the gas and the total heat flow into the gas and also draw a

heat flow into the gas and also draw a PV diagram so let's do that

first so we're starting at high volume a volume of 12

volume of 12 L and the volume is decreased into 3 l

L and the volume is decreased into 3 l at a pressure of 2.5

ATM so we're going this way let's call this position a position

way let's call this position a position B so that's an isobaric

B so that's an isobaric process now heat is added if heat is

process now heat is added if heat is added the temperature will increase and

added the temperature will increase and the pressure will increase as as well

the pressure will increase as as well and this occurs at constant volume so

and this occurs at constant volume so that's an isocoric process so we need to

that's an isocoric process so we need to go up in the PV

diagram and then the temperature will reach its original

reach its original value calculate the total work done by

value calculate the total work done by the gas and the total heat flow into the

the gas and the total heat flow into the gas let's call this position

C so what can we do in this problem how can we figure out the

answer well we need to calculate the

well we need to calculate the work done as we travel from position a

work done as we travel from position a to position C so let's find it from A to

to position C so let's find it from A to B so that's an isobaric process so we

B so that's an isobaric process so we can use the equation P Delta

can use the equation P Delta V the pressure is 2.5

V the pressure is 2.5 ATM and the change in volume the final

ATM and the change in volume the final minus initial that's 3 - 12 and then we

minus initial that's 3 - 12 and then we need to multiply it by

need to multiply it by 101.3 to convert it to

101.3 to convert it to Jews so this is 2.5 * 9 *

101.3 so the work done from A to B is 2279

2279 JW now what about the work done from B

JW now what about the work done from B to C from B to C that's an isocoric

to C from B to C that's an isocoric process volume is constant and the work

process volume is constant and the work is always zero therefore the work done

is always zero therefore the work done from a to

from a to c is the sum total of these two values

c is the sum total of these two values which is just - 2279

Jew now what about going from a to c notice that this is an isothermal

notice that this is an isothermal path the key expression is the

path the key expression is the temperature the temperature remains at

temperature the temperature remains at the same

the same value it reaches its original value so

value it reaches its original value so whatever the temperature was at Point a

whatever the temperature was at Point a is the same as Point C and for an

is the same as Point C and for an isothermal uh graph or process Delta U

isothermal uh graph or process Delta U is equal to

is equal to zero so from a to c the fact that Delta

zero so from a to c the fact that Delta U is equal to zero which means

U is equal to zero which means that Q is equal to

that Q is equal to w q is w+

w q is w+ Delta

Delta u based on this equation if you

u based on this equation if you rearrange it so if Delta U is z q is

rearrange it so if Delta U is z q is equal to W which is -

equal to W which is - 2279 now let's see if there's another

2279 now let's see if there's another way we can get the same answer by

way we can get the same answer by calculating Q through the individual

calculating Q through the individual paths so we need to get -

paths so we need to get - 2279 uh during this other technique or

2279 uh during this other technique or other method

so first let's assign the temperature value to point

value to point a let's say the temperature is 1,000

a let's say the temperature is 1,000 Kelvin now notice the volume decreased

Kelvin now notice the volume decreased from 12 to 3 l whenever the volume

from 12 to 3 l whenever the volume decreases the temperature decreases as

decreases the temperature decreases as well if the pressure is held constant

well if the pressure is held constant the volume decreases by a factor of

the volume decreases by a factor of four if you divide 12 by 4 you should

four if you divide 12 by 4 you should get three likewise if we divide 1,000

get three likewise if we divide 1,000 Kelvin by 4 it should give us a

Kelvin by 4 it should give us a temperature of 250

temperature of 250 Kelvin so that's going to be the

Kelvin so that's going to be the relative temperature at point B if point

relative temperature at point B if point a is 1,000 Kelvin now at Point C the

a is 1,000 Kelvin now at Point C the temperature is 1,000 Kelvin because it's

temperature is 1,000 Kelvin because it's back to its original value but going

back to its original value but going from B to C the temperature increases by

from B to C the temperature increases by factor four which means the pressure

factor four which means the pressure must increase by factor four at constant

must increase by factor four at constant volume if you raise the temperature the

volume if you raise the temperature the pressure will

pressure will increase so 2.5 * 4 is 10

ATM now how can we calculate Q in the isobaric process from A to B there's two

isobaric process from A to B there's two equations that you can use you can use

equations that you can use you can use this one Q is equal to n CP * delta T

this one Q is equal to n CP * delta T during a constant uh pressure or an

during a constant uh pressure or an isobaric

isobaric process or we can use the other equation

process or we can use the other equation Q is equal

to P CP Delta V /

CP Delta V / R ideally we want to use this equation

R ideally we want to use this equation because we have the pressure and the

because we have the pressure and the change in volume we don't have the moles

change in volume we don't have the moles we do have the temperature but if we are

we do have the temperature but if we are going to use the first one we need to

going to use the first one we need to find n which we could do using equation

find n which we could do using equation PV equals nrt we could find it at any

PV equals nrt we could find it at any points uh point a b or

points uh point a b or c but we don't have to do that so let's

c but we don't have to do that so let's use this

equation so Q is going to equal p which is

is 2.58m time

2.58m time CP and this is a monoatomic

CP and this is a monoatomic gas so

gas so therefore CP is going to be 5 over 2 * R

therefore CP is going to be 5 over 2 * R CV is 3 over2 * r

CV is 3 over2 * r Delta V the change in volume final minus

Delta V the change in volume final minus initial is 3 - 12 and then divid

R and don't forget you need to multiply this value by 101.3 to convert it to

this value by 101.3 to convert it to Jews so it's 2.5 * 5 / 2 *

Jews so it's 2.5 * 5 / 2 * 9 you should get

9 you should get 56.25 and then multiply that by

101.3 so Q from A to B is equal to 5,698

5,698 Jew now what about from B to C how can

Jew now what about from B to C how can we calculate

Q so from B to C you can use this equation Q is equal to n CV Delta C

equation Q is equal to n CV Delta C since the volume is solid constant or we

since the volume is solid constant or we can use this equation V

can use this equation V CV Delta

CV Delta P /

P / R let's go ahead and use that

R let's go ahead and use that one so

one so V is 3

V is 3 l from position B to C it's a constant 3

l from position B to C it's a constant 3 l and CV that's 3/ 2 * R and Delta P the

l and CV that's 3/ 2 * R and Delta P the change in pressure final minus the

change in pressure final minus the initial value 10 - 2.5 that's a change

initial value 10 - 2.5 that's a change of

of 7.5 /

7.5 / R and then multiplied

R and then multiplied by

by 101.3 to convert it back to

101.3 to convert it back to jewels so it's 3 * 3 / 2 *

jewels so it's 3 * 3 / 2 * 7.5 that's

7.5 that's 33.75 *

33.75 * 101.3 so this is equal to 3,400

18.9 now to calculate the total heat which we said was - 2279 Jew we could

which we said was - 2279 Jew we could simply add the heat absorbed or released

simply add the heat absorbed or released by processes A to B and B to

by processes A to B and B to C so if we take

C so if we take 5,698 and add it to 341

5,698 and add it to 341 18.9 this will give you -22 79

18.9 this will give you -22 79 Jew which is the same answer that we had

Jew which is the same answer that we had before but now you know how to

before but now you know how to calculate the heat energy absorbed or

calculate the heat energy absorbed or released during an isobaric process and

released during an isobaric process and an isocoric

process how much work is performed by the gas as the system moves in a cyclic

the gas as the system moves in a cyclic path A to B to C to D to

path A to B to C to D to a well a simple way to find the work is

a well a simple way to find the work is to calculate the

to calculate the area that is the area of the Shaded

area that is the area of the Shaded region so we have the shape of a

region so we have the shape of a rectangle and the area is simply the

rectangle and the area is simply the length times the width the length of the

length times the width the length of the rectangle as you can see is the change

rectangle as you can see is the change in

in pressure and the width of the

pressure and the width of the rectangle is the change in volume the

rectangle is the change in volume the change in pressure is the difference

change in pressure is the difference between 1 and 8 atmospheres so that's

between 1 and 8 atmospheres so that's seven and the change in volume 6 - 2 so

seven and the change in volume 6 - 2 so that's

that's 4 and we need to convert it to Jews so

4 and we need to convert it to Jews so we're going to multiplied by 101.3 Jew

we're going to multiplied by 101.3 Jew per liter per

atmosphere so the work perform is 2,8 36.4 Jew

2,8 36.4 Jew now this answer is it positive or is it

now this answer is it positive or is it negative what would you

negative what would you say it turns out that if you're

say it turns out that if you're traveling in the clockwise

traveling in the clockwise Direction the work done is

Direction the work done is positive now if you travel in let's say

positive now if you travel in let's say the counterclockwise

the counterclockwise Direction the work done is going to be

Direction the work done is going to be negative now let's prove

negative now let's prove that first we need to prove this answer

to do that let's calculate the work perform in each step so let's start with

perform in each step so let's start with the first step that is from A to

B so that's an isobaric process so we can use the equation work is equal to P

can use the equation work is equal to P Delta

Delta V and anytime you go to the right the

V and anytime you go to the right the work done is positive so the pressure is

work done is positive so the pressure is 8 the final volume is 6us 2 so that's uh

8 the final volume is 6us 2 so that's uh 8 * 4 which is 32 *

8 * 4 which is 32 * 101.3 so the work done from A to B is

101.3 so the work done from A to B is positive

positive 3,2

41.6 now the work done going from B to C that's an isocoric process and as a

that's an isocoric process and as a result the work is

zero so from B to C and D to a the work is

is zero so now we need to calculate the

zero so now we need to calculate the work going from C to

D so the pressure is now One D is the final position C is the

One D is the final position C is the initial position so it's going to be 2 -

initial position so it's going to be 2 - 6 so that's -4 *

6 so that's -4 * 101.3 and that's equal to -

101.3 and that's equal to - 45.2 Jew

45.2 Jew so if we add these two

so if we add these two values 32

values 32 41.6 + -

45.2 this is going to be equal to 28 36.4 and if you go in the reverse

36.4 and if you go in the reverse Direction the numbers will be reversed

Direction the numbers will be reversed instead of this value being 2 - 6 it's

instead of this value being 2 - 6 it's going to be 6 - 2 so this is going to be

going to be 6 - 2 so this is going to be positive 405 and this will be - 32 41.6

positive 405 and this will be - 32 41.6 you can try but going in the

you can try but going in the counterclockwise Direction the work will

counterclockwise Direction the work will be negative 28 36.4 that's the answer to

be negative 28 36.4 that's the answer to the second part of the problem but for

the second part of the problem but for the first part of the problem traveling

the first part of the problem traveling in the clockwise Direction the work done

in the clockwise Direction the work done is

positive now if you want to understand why that's the case keep in mind anytime

why that's the case keep in mind anytime you travel towards the right the work is

you travel towards the right the work is positive and if you travel towards the

positive and if you travel towards the left in the PV diagram the work is

left in the PV diagram the work is negative as you travel towards the right

negative as you travel towards the right notice that the pressure is higher and

notice that the pressure is higher and therefore the work done by the gas is

therefore the work done by the gas is going to be higher as you travel towards

going to be higher as you travel towards the left the pressure is lower it's one

the left the pressure is lower it's one and so the negative work performed by

and so the negative work performed by the gas is going to be lower so the net

the gas is going to be lower so the net work done is

work done is positive because the higher side is

positive because the higher side is moving towards the

moving towards the right if you mooving towards the

right if you mooving towards the left at the higher pressure then the

left at the higher pressure then the overall work is going to be negative the

overall work is going to be negative the work is determined by the direction

work is determined by the direction of the system at high pressure because

of the system at high pressure because at high pressure that's where most of

at high pressure that's where most of the work will be

accomplished go ahead take a minute pause the video try this problem and

pause the video try this problem and then unpause it when you want to see the

then unpause it when you want to see the solution so let's begin by drawing a PV

diagram now the volume of a gas increases from 3 l to 9 L

and it occurs at a constant pressure of 4 ATM which is lower than

10 so we're traveling in this direction let's call this position a and position

b or state a and state B next heat is added to increase the

B next heat is added to increase the pressure to 10 ATM

so that's an isocore process finally heat is removed to

process finally heat is removed to decrease the pressure and volume back to

decrease the pressure and volume back to its original

its original values that is at three and

values that is at three and four and it's going to travel back to

four and it's going to travel back to the original um state in a straight line

the original um state in a straight line path so calculate the total work done in

path so calculate the total work done in this process

this process to find the answer we got to calculate

to find the answer we got to calculate the area under the

curve and this is the area of a triangle so it's going to be 1 12 base time

so it's going to be 1 12 base time height the base of the triangle is the

height the base of the triangle is the change in

change in volume and the height of the triangle is

volume and the height of the triangle is the change in pressure

so the change in volume is going to be equal to 6 L and the change in

equal to 6 L and the change in pressure is 10- 4 so that's a difference

pressure is 10- 4 so that's a difference of

of six * 101.3 to convert it to

six * 101.3 to convert it to Jews so 6 * 6 is 36 half of that is

Jews so 6 * 6 is 36 half of that is 18 so the work done is going to be

18 so the work done is going to be 1,823 point4 jewles now here's a

1,823 point4 jewles now here's a question for you is the work done

question for you is the work done positive or negative

so notice the direction that we're traveling in that is the system is

traveling in that is the system is rotating in

rotating in the counterclockwise Direction which

the counterclockwise Direction which means that work in this process is

means that work in this process is negative if it's moving in a cyclic

negative if it's moving in a cyclic clockwise Direction the work should be

clockwise Direction the work should be positive but now let's see if we can get

positive but now let's see if we can get this

this answer by calculating the work done in

answer by calculating the work done in each individual

step so let's start from A to B that's an isobaric process so we can use the

an isobaric process so we can use the equation P Delta

equation P Delta V the pressure is four at this point the

V the pressure is four at this point the change in volume final minus initial 9 -

change in volume final minus initial 9 - 3 that's 6 *

101.3 so 24 * 101.3 gives us a value of 2431

101.3 gives us a value of 2431 .2 and the work done is

.2 and the work done is positive now between B and C that's an

positive now between B and C that's an isocor

isocor process so the work done for that region

process so the work done for that region is zero let's see if I can fit it

is zero let's see if I can fit it here now what about in this section

here now what about in this section going from C to a how much work is

going from C to a how much work is performed by the gas and taking that

performed by the gas and taking that route it turns out that you can use this

route it turns out that you can use this equation to get the answer but you have

equation to get the answer but you have to use it in a certain way notice that

to use it in a certain way notice that the pressure is not constant it changes

the pressure is not constant it changes from 10 to 4 so what value of P should

from 10 to 4 so what value of P should you plug in to this equation to make it

you plug in to this equation to make it work it turns out that if the pressure

work it turns out that if the pressure increases at a constant linear rate you

increases at a constant linear rate you can use the average

can use the average pressure so the

pressure so the equation for the path to to calculate

equation for the path to to calculate the work from C to a it's going to be

the work from C to a it's going to be 12 P initial plus P final * Delta V to

12 P initial plus P final * Delta V to find the average of these two numbers

find the average of these two numbers you need to add them up and divide by

you need to add them up and divide by two so the average pressure is the

two so the average pressure is the initial and the final pressure is the

initial and the final pressure is the sum of the two values ID two or half of

sum of the two values ID two or half of the

the sum so using that equation 10 + 4 is 14

sum so using that equation 10 + 4 is 14 half of that is seven so the work is

half of that is seven so the work is going to be the average pressure of

going to be the average pressure of seven time the change in

seven time the change in volume final minus initial so 3 - 9

volume final minus initial so 3 - 9 which is -6 *

which is -6 * 101.3 so -42 *

101.3 so -42 * 101.3 gives us a value of

101.3 gives us a value of negative

42546 now the reason why it's negative is because we're traveling towards the

is because we're traveling towards the left of the PV diagram here it's

left of the PV diagram here it's positive because we're going towards the

positive because we're going towards the right of the PV

right of the PV diagram now in terms of the magnitude of

diagram now in terms of the magnitude of these two values the one at the top is

these two values the one at the top is greater because the average pressure is

greater because the average pressure is higher than the pressure at the bottom

higher than the pressure at the bottom which is

which is four so keep in mind the direction of

four so keep in mind the direction of the system at high pressure determines

the system at high pressure determines the overall sign of the work done for

the overall sign of the work done for this entire cyclic process that's why

this entire cyclic process that's why the overall work done is

the overall work done is negative because at the top is moving

negative because at the top is moving towards

towards left now let's add these two values so

left now let's add these two values so -42 54.6 + 24

-42 54.6 + 24 31.2 gives you this answer 8

23.4 which of the following is not a state

state function pressure internal energy work

function pressure internal energy work temperature volume which one

it turns out that work is not a state function pressure internal energy

function pressure internal energy everything else is a state

everything else is a state function let's say if we have a PB

diagram and let's say we have three points point A B and

C the work that's required to go from a to c by means of B it's going to be

to c by means of B it's going to be different than the work required to go

different than the work required to go from a to c it's not going to be the

from a to c it's not going to be the same let's say a to c is an isothermal

same let's say a to c is an isothermal path however the change in internal

path however the change in internal energy going from ABC or AC

energy going from ABC or AC directly is going to be the

directly is going to be the same it may not be zero but it's going

same it may not be zero but it's going to be the same

the change in internal energy does not depend on the path it depends on the

depend on the path it depends on the state as long as you go from a to c

state as long as you go from a to c regardless if you pass through this path

regardless if you pass through this path or this path the change in internal

or this path the change in internal energy is the same which means it's a

energy is the same which means it's a state function the work will not be the

state function the work will not be the same the work depends on the path that

same the work depends on the path that you take if you go from a to c by taking

you take if you go from a to c by taking this route the work is different than if

this route the work is different than if you take this route because the area on

you take this route because the area on the curve is different also heat is not

the curve is different also heat is not a state

a state function now pressure temperature and

function now pressure temperature and volume they will be the same regardless

volume they will be the same regardless of the path

taken if I take the path A to B to C the pressure at C will still be the same

pressure at C will still be the same let's say it's two

atmospheres if I go from a to c at C the pressure would still be two

pressure would still be two atmospheres so as you can see the

atmospheres so as you can see the pressure is a state function it depends

pressure is a state function it depends on the initial and final States just

on the initial and final States just like internal energy it doesn't depend

like internal energy it doesn't depend on the path taken whereas work and heat

on the path taken whereas work and heat which are forms of energy transfer they

which are forms of energy transfer they depend on the path taken and so work and

depend on the path taken and so work and heat are not State functions

let's try this problem so let's focus on path a to d to

problem so let's focus on path a to d to C which is this

C which is this path and we're given the amount of heat

path and we're given the amount of heat that flows into the

system so because heat is absorbed by the system Q is

the system Q is positive and we're told that 45 Jew of

positive and we're told that 45 Jew of work is done by the system whenever the

work is done by the system whenever the system does work or performs

system does work or performs work the work done is

BC the work performed by the gas is 30

Jew how much heat energy is absorbed in path

path ABC to answer this question it's

ABC to answer this question it's important to realize

important to realize that internal energy is a state function

that internal energy is a state function the change in the internal energy and

the change in the internal energy and going from ABC or ADC is the

going from ABC or ADC is the same so if we could find Delta U for

same so if we could find Delta U for this path we can use it to find Q in

this path we can use it to find Q in path

ABC Delta U is equal to Q - W so that's 125 - 45

125 - 45 which is

which is 80 so we can use this value for path

80 so we can use this value for path ABC solving for Q Q is Delta U plus W so

ABC solving for Q Q is Delta U plus W so that's 30 +

that's 30 + 80 so 110 Jew of heat

80 so 110 Jew of heat energy was absorbed in path

ABC now what about the next part when a system returns from C to a

part when a system returns from C to a along the Curve Earth path that's the

along the Curve Earth path that's the path in the

path in the middle 38 Jew of energy is transferred

middle 38 Jew of energy is transferred by

work now this energy transfer is it positive or is it

negative anytime a gas is compressed or if you travel towards the left of a PV

if you travel towards the left of a PV diagram the work done is negative as you

diagram the work done is negative as you travel towards the right the work done

travel towards the right the work done will be positive so this is4

5 now in order to find q that is the amount of heat energy

q that is the amount of heat energy transferred during this process we need

transferred during this process we need to find Delta

to find Delta U so how can we do

U so how can we do it well we know that going from A to B

it well we know that going from A to B to C Delta U is positive

to C Delta U is positive 80 going back from C to B to a it's

80 going back from C to B to a it's going to be negative 80 because the

going to be negative 80 because the internal energy is a state function it

internal energy is a state function it really doesn't matter the path that you

really doesn't matter the path that you take take as long as you get from a to c

take take as long as you get from a to c it's POS 80 so going back to C to a it's

it's POS 80 so going back to C to a it's 80 it's also 80 if we take the curve

80 it's also 80 if we take the curve path the change in the internal energy

path the change in the internal energy of a system is independent of the path

of a system is independent of the path taken so Delta U going from C to a is

taken so Delta U going from C to a is going to be 80 instead of positive 80 so

going to be 80 instead of positive 80 so now we could find Q which is Delta U + W

now we could find Q which is Delta U + W so that's

so that's uh80 Plus

45 actually that's not supposed to be 45 it's supposed to be

45 it's supposed to be 38 I guess I was looking at this number

38 I guess I was looking at this number but going from C to a 38 Jews of energy

but going from C to a 38 Jews of energy is transferred by work so this is

38 so if we add these two numbers this is going to be

-118 so that's how much heat energy is transferred

transferred now because Q is negative that means

now because Q is negative that means that heat was released during this

that heat was released during this process as a system went from C to

process as a system went from C to a now what about the next part if UA is

a now what about the next part if UA is equal to Z and U is equal to 20 how much

equal to Z and U is equal to 20 how much heat energy is absorbed in paths ab and

BC well first let's find the internal energy at path c as you mentioned before

energy at path c as you mentioned before going going from A to B to C the change

going going from A to B to C the change in the internal energy is 80 so if a is

in the internal energy is 80 so if a is zero C is 80 Jew higher than a so at C

zero C is 80 Jew higher than a so at C the internal energy is

the internal energy is 80 so with that information we can

80 so with that information we can calculate

calculate Q now the work done for any isocoric

Q now the work done for any isocoric process which is B to C or a to d is

process which is B to C or a to d is zero so let's uh make some space

so w is equal to zero and it's equal to zero on this side

zero and it's equal to zero on this side as

well now the work going from ad DC is

45 so that's here to here so if a to D is already

is already Zero D to C must be 45

5 the work done from a to d plus the work done from D to C is equal to work

work done from D to C is equal to work done from a to d to C so this is zero

done from a to d to C so this is zero this is

this is 45 this other value must be

45 now what about the work done from A to

to B we know the work done from A to B plus

B we know the work done from A to B plus that for B to C is equal to work done

that for B to C is equal to work done for A to B to C now this is zero so

for A to B to C now this is zero so these two must be equal to each other so

therefore the work done from A to B is 30 so now that we have these

values we can calculate Q so focusing on path

Q so focusing on path a q is equal to Delta u+ w

a q is equal to Delta u+ w Delta U is the difference between these

Delta U is the difference between these two values UA is zero U is 20 so the

two values UA is zero U is 20 so the change going from a 2B is positive 20

change going from a 2B is positive 20 going from B to a must be negative - 20

going from B to a must be negative - 20 and W is 30 so that means Q

and W is 30 so that means Q AB is equal to postive 50

Jew now what about from or path B to C and path B to C W is zero but going from

and path B to C W is zero but going from B to C the change in the internal

B to C the change in the internal energy is positive 60 it changes from 20

energy is positive 60 it changes from 20 to 80 so final minus initial 80us 20 is

to 80 so final minus initial 80us 20 is 60 so Q is 60 therefore

60 so Q is 60 therefore QBC is equal to positive

qabc which is 50 + 60 which is 110 and as we can see these two are equal to

as we can see these two are equal to each

other now what about the last part of the

problem if UD is equal to 30 how much heat is transferred in paths a and

heat is transferred in paths a and DC so let's put 30

DC so let's put 30 here now let's focus on

here now let's focus on ad so Q is Delta U + w

ad so Q is Delta U + w so if we add Z Plus the change which is

so if we add Z Plus the change which is 30 we can see that q a

30 we can see that q a d is equal to POS

30 now what about Q

Q DC what's the answer for

DC what's the answer for that well notice that the change in the

that well notice that the change in the internal energy going from D to C it

internal energy going from D to C it increases by 50 80 minus 30 is 50

and the work done going from D to C is positive

positive 45 so Q is Delta u+ W so 50 + 45 that's

45 so Q is Delta u+ W so 50 + 45 that's equal to positive

95 now if we add 30 and 95 it should give us the total of 125 which it does

give us the total of 125 which it does so everything is balanced so to speak

so everything is balanced so to speak the PV diagram shown below applies to

the PV diagram shown below applies to 2.1 moles of an ideal diatomic gas

2.1 moles of an ideal diatomic gas calculate the temperature at all

points now we need to use the equation PV is equal to nrt at least for the

PV is equal to nrt at least for the first point and then we can easily find

first point and then we can easily find the other points now notice that the

the other points now notice that the volume is in cubic met 1 Cub met is

volume is in cubic met 1 Cub met is equal to 1,000

equal to 1,000 l so to convert these values into lers

l so to convert these values into lers multiply by th000 so this is basically

multiply by th000 so this is basically 10 L and this is equivalent to 50

L now let's calculate the temperature at Point a since that temperature is going

Point a since that temperature is going to be the lowest temperature the

to be the lowest temperature the temperature increases whenever you move

temperature increases whenever you move in this general

in this general direction the temperature increases as

direction the temperature increases as you go towards the right in the P

you go towards the right in the P diagram and it increases as you go up

diagram and it increases as you go up in a PV

in a PV diagram so using this equation at Point

diagram so using this equation at Point a the pressure is five atmospheres the

a the pressure is five atmospheres the volume is 10 L we have uh 2.1 moles R is

volume is 10 L we have uh 2.1 moles R is 8206 and we're looking for

8206 and we're looking for T So 5 * 10 is 50 / 2.1 and then take

T So 5 * 10 is 50 / 2.1 and then take that result divided by

that result divided by 8206 so the temperature at Point

8206 so the temperature at Point a is 290

Kelvin now what about the temperature at Point D notice that the volume increases

Point D notice that the volume increases from 10 L to 50

from 10 L to 50 L if you increase the temperature at

L if you increase the temperature at constant pressure the volume will

constant pressure the volume will increase proportionally so the volume

increase proportionally so the volume increased by a factor of five which

increased by a factor of five which means the temperature should also

means the temperature should also increase by a factor of 5 290 * 5 is

increase by a factor of 5 290 * 5 is equal to 1450 Kelvin now if you want to

equal to 1450 Kelvin now if you want to you can use the PV equals nrt equation

you can use the PV equals nrt equation to get that

to get that answer if you plug in a pressure of 5 a

answer if you plug in a pressure of 5 a volume of

volume of 50 and 2.1 for n and the same r

value 5 * 50 is 250 ID 2.1 that's 19 19 divide that by

2.1 that's 19 19 divide that by 8206 and you get about 1450

Kelvin now it's really close to 1451 it's

it's 14

14 .7 because this answer was

rounded so I'm going to leave it as 1450 I'm going to round it to a nice whole

I'm going to round it to a nice whole number

so this answer is an estimate so to speak now what about the temperature at

speak now what about the temperature at point

point B notice that the pressure increases

B notice that the pressure increases from 5 to 40 going from A to B so 40 / 5

from 5 to 40 going from A to B so 40 / 5 is 8 so the pressure increases by a

is 8 so the pressure increases by a factor of eight at constant volume if

factor of eight at constant volume if you increase the temperature the

you increase the temperature the pressure will increase proportionately

pressure will increase proportionately that means that we have to increase this

that means that we have to increase this value by factor of a two 90 *

value by factor of a two 90 * 8 is equal to 2320

Kelvin now going from D to C the pressure increases by a factor of eight

pressure increases by a factor of eight so we can multiply 1450 by 8 and that

so we can multiply 1450 by 8 and that will give us 11,600 Kelvin or we can go

will give us 11,600 Kelvin or we can go from uh B to C where the volume

from uh B to C where the volume increases by a factor of

11600 so now what about Part B how can we calculate the heat transferred and

we calculate the heat transferred and paths AB b c c d and

paths AB b c c d and da well let's start with

da well let's start with AB there's many different equations that

AB there's many different equations that we can use but the first thing that we

we can use but the first thing that we need to realize is this is an isocoric

need to realize is this is an isocoric process the volume is constant so we can

process the volume is constant so we can use this equation n CV delta

use this equation n CV delta T So n is

T So n is 2.1 CV for an ideal datomic gas is 5/2 *

R and the change in temperature going from A to B it's a

temperature going from A to B it's a final minus

final minus initial so that's 2320

initial so that's 2320 -

290 so this is going to be about

be about 88,500 round into three6

fix now what about Q BC here we have an isobaric

BC here we have an isobaric process so we can use the equation n CP

process so we can use the equation n CP delta

T now the CP value for a diatomic gas is going to be 7 /2 *

R and going from B to C the final temperature is

temperature is 11,600 minus the initial temperature of

2320 so this is going to be about

about 567,000

so let me take a minute and just clear away a few things

in the meantime go ahead and calculate the other two Q

CD this time let's use a different equation we have another isocoric

equation we have another isocoric process but instead of using the ncv

process but instead of using the ncv delta T

delta T equation let's

equation let's use V CV Delta P /

use V CV Delta P / R the volume is 50

R the volume is 50 L CV for datomic gas is 52 *

L CV for datomic gas is 52 * R the change in

R the change in pressure is -35

and we can cancel R and let's not forget to multiply by

101.3 so this is going to be negative

negative 443,000

fix now what about the last part Q

part Q da so going from D to a how can we

da so going from D to a how can we calculate it so let's use this equation

calculate it so let's use this equation since we have a isobaric process the

since we have a isobaric process the pressure is constant so let's use the

pressure is constant so let's use the equation P

equation P CP Delta V /

CP Delta V / R so the pressure is 5 ATM CP is 7/2 * R

R so the pressure is 5 ATM CP is 7/2 * R and the change in volume final minus

and the change in volume final minus initial 10 - 50 is

initial 10 - 50 is -40 so we can cancel

R so as you can see whenever you go up and to the right Q is

and to the right Q is positive when you go down and towards

positive when you go down and towards the left Q is

negative and you can tell by the change in

in temperature to increase the temperature

temperature to increase the temperature typically you have to put heat into the

typically you have to put heat into the system and to decrease the temperature

system and to decrease the temperature from 11,000 to

from 11,000 to 1450 you have to remove heat from the

system so now let's calculate this value so five * 3.5 * -4 *

so five * 3.5 * -4 * 101.3 this is about - 70,000 and 900

Jew so now that we have the Q values for each

each step let's calculate the net Q value for

step let's calculate the net Q value for the cyclic process so as we travel from

the cyclic process so as we travel from position A to B B to C C to d d to a

position A to B B to C C to d d to a let's find the net amount of heat energy

let's find the net amount of heat energy absorbed or released into the

system so we just got to add these four values

values 88600 + 567,000 - 443,000

88600 + 567,000 - 443,000 -

- 7,900 this is equal to

7,900 this is equal to positive

141,000 now what is the work done in a cyclic process we know the work done is

cyclic process we know the work done is equal to the area of the Shaded

equal to the area of the Shaded region and since we have a rectangle

region and since we have a rectangle it's basically the length times the

it's basically the length times the width the length is basically the change

width the length is basically the change in pressure the width is the change in

in pressure the width is the change in volume the change in pressure is

volume the change in pressure is 35 if you subtract 40 minus 5 and the

35 if you subtract 40 minus 5 and the change in volume 50 l - 10 L that's 40

change in volume 50 l - 10 L that's 40 L multiplied 101.3 to converted to

jewels and this will give you about

about 141,000

141,000 820 but let's round these values to

820 but let's round these values to approximately

approximately 142,000 it's about the same if we keep

142,000 it's about the same if we keep it to three

it to three sigfigs as we've been doing for each of

sigfigs as we've been doing for each of these

problems now BAS on rounding errors there's going to be some slight

there's going to be some slight differences but theoretically these two

differences but theoretically these two should be

should be equal which they

equal which they are now let's think about what this

are now let's think about what this means we see that q and W are the

means we see that q and W are the same the question is

same the question is why now we started at position a we went

why now we started at position a we went all around the rectangle and we ended

all around the rectangle and we ended back at position a so going from a to

back at position a so going from a to a the internal energy is is the same

a the internal energy is is the same there's no change therefore Delta U is

there's no change therefore Delta U is equal to zero for any cyclic

equal to zero for any cyclic process and if Delta U is equal to zero

process and if Delta U is equal to zero and we know that Q is Delta u+ W

and we know that Q is Delta u+ W therefore Q is equal to a W for any

therefore Q is equal to a W for any cyclic

cyclic process and that's why we can see

process and that's why we can see they're the

they're the same the area is not only equal to the

same the area is not only equal to the work done but also it's equal to Q

work done but also it's equal to Q because it's a cyclic process

because it's a cyclic process now let's talk about something else but

now let's talk about something else but first let's get rid of a few

things what is the total amount of heat energy that went into the

energy that went into the system let's say

system let's say if this block represents the

if this block represents the system if we add these two values we're

system if we add these two values we're going to get how much heat energy flowed

going to get how much heat energy flowed into the system

and that's about 65,66 Jew of energy went into the

qh now how much heat energy went out of the system that's basically these values

the system that's basically these values because they're

because they're negative if we add

them this is going to be negative or just simply

53,9 this is also known as a ql that's the amount of heat energy lost

ql that's the amount of heat energy lost to the environment qh is the amount of

to the environment qh is the amount of heat energy that enters into the

heat energy that enters into the system the work is the difference

system the work is the difference between these two

values 655,000 minus 53,9

655,000 minus 53,9 is equal to approximately 142,000 if you

is equal to approximately 142,000 if you round

it so that's how much energy that came from heat and that was transferred into

from heat and that was transferred into mechanical

work now let's calculate the efficiency of the

efficiency of the system the efficiency is basically the

system the efficiency is basically the output work divided by the total heat

output work divided by the total heat energy that went into the system that is

energy that went into the system that is qh time

100% so that's going to be 142,000 ided by the

so the efficiency is 21.7% so what this means is that this

21.7% so what this means is that this particular system can take

particular system can take 21.7% of the heat energy that is

21.7% of the heat energy that is transferred to it and convert that heat

transferred to it and convert that heat energy into useful mechanical

energy into useful mechanical work the other

work the other 78.3% is lost to the environment which

78.3% is lost to the environment which is a lot of energy that's lost so these

is a lot of energy that's lost so these um devices are not very efficient but

um devices are not very efficient but nevertheless you can still use it to

nevertheless you can still use it to perform useful mechanical

work ethane gas has a gam ratio of 1.22 what is the molar heat capacity of

1.22 what is the molar heat capacity of ethane at constant volume and not

ethane at constant volume and not constant

constant pressure so how can we uh find CP and

pressure so how can we uh find CP and CV given this gamma

CV given this gamma ratio

ratio gamma is CP /

gamma is CP / CV it's the ratio of these two different

CV it's the ratio of these two different M he

M he capacities and we know that

capacities and we know that CP is equal to 5 over 2 R that's

CP is equal to 5 over 2 R that's basically uh the same as

basically uh the same as 2.5r and CV is 3 over2 R for U monatomic

2.5r and CV is 3 over2 R for U monatomic gas that's 1.5

gas that's 1.5 R now

R now 2.5r is equal to 1.5 r + 1 R 1.5 + 1 is

2.5r is equal to 1.5 r + 1 R 1.5 + 1 is 2.5 so therefore we can say that CP is

2.5 so therefore we can say that CP is equal to CV plus

equal to CV plus r so using uh this

r so using uh this relationship which is very accurate for

relationship which is very accurate for ideal gases

we could say that gamma is equal to CV + r / CV so basically we replace

to CV + r / CV so basically we replace CP with CV plus

CP with CV plus r and we could separate this into two uh

r and we could separate this into two uh fractions we could divide CV by itself

fractions we could divide CV by itself and R by CV which means that uh gamma is

and R by CV which means that uh gamma is 1 + r divided by uh

CV so let's subtract one from both sides if we do that we're going to get

this and if we this is the same as gamma minus one

we this is the same as gamma minus one over one if we flip both sides this is 1

over one if we flip both sides this is 1 over gamma minus 1 is equal to CV /

over gamma minus 1 is equal to CV / R and now if we multiply both sides by R

R and now if we multiply both sides by R we can see that CV is equal to R over

we can see that CV is equal to R over gamma minus one so this is how you could

gamma minus one so this is how you could find

equation now there's one more equation that we can

that we can write so keep we said that gamma is CP /

write so keep we said that gamma is CP / CV so if we let's move

CV so if we let's move this this way if we multiply both sides

this this way if we multiply both sides by

CV CP is gamma time CV and CV is equal to

CP is gamma time CV and CV is equal to that equation so CP is R * gamma over

that equation so CP is R * gamma over gamma minus

gamma minus one so this is another useful

one so this is another useful equation so with these two equations we

equation so with these two equations we can find CP and CV if we're given the

can find CP and CV if we're given the gamma ratio for a

gamma ratio for a gas so let's calculate CV that's going

gas so let's calculate CV that's going to be uh

8.3145 divid by 1.22 -

1.22 - one 1.22 -1 is simply 22 on the bottom

one 1.22 -1 is simply 22 on the bottom so

8.3145 /22 gives us a CV

/22 gives us a CV value of

value of 37.8 so that's the first answer that's

37.8 so that's the first answer that's the molar heat capacity at constant

the molar heat capacity at constant volume now at constant

volume now at constant pressure we can uh take R which is

pressure we can uh take R which is 8.3145 multiply by gamma

/ gamma minus one so 8.3145 * 1.22

1.22 id22 is

id22 is 46.1 so that's the M capacity at

46.1 so that's the M capacity at constant pressure so that's how you can

constant pressure so that's how you can find it given the gamma

find it given the gamma ratio here's another problem that we

ratio here's another problem that we could try how much heat is required to

could try how much heat is required to raise the temperature of 6.3 moles of

raise the temperature of 6.3 moles of propane from 30 C to 50 c at a constant

propane from 30 C to 50 c at a constant pressure of 2 ATM and we're given the

pressure of 2 ATM and we're given the gamma

gamma ratio well to find Q at constant

ratio well to find Q at constant pressure we can use the

pressure we can use the equation Q is equal to n CP delta T so

equation Q is equal to n CP delta T so we got to find CP

we got to find CP first and we can use the equation that

first and we can use the equation that we got um in the last problem

we got um in the last problem which is

which is uh R * gamma over gamma minus one so

uh R * gamma over gamma minus one so that's

1.27 divid by 1.27

by 1.27 minus1 which on the bottom is

minus1 which on the bottom is 0.127 so 8.3145 * 1.27 /

0.127 so 8.3145 * 1.27 / .27 is about

73 78 so using

78 so using this we can now calculate Q so Q is

this we can now calculate Q so Q is going to be equal to n the number of

going to be equal to n the number of moles which is

moles which is 6.5 time CP which is

6.5 time CP which is 73.7 times the change in temperature

73.7 times the change in temperature final minus initial 50 minus 30 that's

final minus initial 50 minus 30 that's uh 20

uh 20 so 73.7 *

so 73.7 * 6.5 * 20 is equal

6.5 * 20 is equal to

to 9,5

9,5 91.4

91.4 jewles so that's the aners to the first

jewles so that's the aners to the first part of the

part of the problem now for the second part what is

problem now for the second part what is the change in the internal energy of the

the change in the internal energy of the gas to find the change in internal

gas to find the change in internal energy we can use Q minus W but we don't

energy we can use Q minus W but we don't know the work so we can also use uh n CV

know the work so we can also use uh n CV delta T which is true for any

delta T which is true for any process so we got to find CV um before

process so we got to find CV um before we can use that equation so let's uh go

we can use that equation so let's uh go ahead and do

ahead and do that and we know that CV is equal to R /

that and we know that CV is equal to R / gamma minus one so that's

gamma minus one so that's 8.3145

8.3145 /

/ 1.27 minus 1

1.27 minus 1 so 8.3145 divid

so 8.3145 divid 0.127 is around

6547 so Delta U is going to equal to the number of moles

number of moles 6.5 time

6.5 time CV time delta T which is still

CV time delta T which is still 20 so 6.5 *

20 so 6.5 * 6547 *

6547 * 20 is equal to 8,500

20 is equal to 8,500 11.1

11.1 Jew and that's it for this

problem now here's a question for you let's say if you have a PV

let's say if you have a PV diagram with three isothermal graphs

diagram with three isothermal graphs which looks like

which looks like this let's say the first one is graph a

this let's say the first one is graph a this is B and this is C and there's

this is B and this is C and there's three temperature values 300 Kelvin 400

three temperature values 300 Kelvin 400 Kelvin and 500

Kelvin and 500 Kelvin so with this information which

Kelvin so with this information which graph is which one which one occurs at

graph is which one which one occurs at which temperature let's say we have

which temperature let's say we have pressure on the y- AIS volume on the x-

pressure on the y- AIS volume on the x- axis if you ever see a question like

axis if you ever see a question like this you need to know that the

this you need to know that the temperature increases as you go up in

temperature increases as you go up in this general

this general direction whenever you go to the right

direction whenever you go to the right the temperature increases because the

the temperature increases because the volume increases at constant pressure

volume increases at constant pressure according to Charles law when you

according to Charles law when you increase the temperature the volume goes

increase the temperature the volume goes up and when you go

up and when you go up the increase in temperature causes an

up the increase in temperature causes an increase in pressure at constant volume

increase in pressure at constant volume at constant volume when you increase the

at constant volume when you increase the temperature the pressure goes up so

temperature the pressure goes up so therefore C is at the highest

therefore C is at the highest temperature that is at 500 Kelvin a is

temperature that is at 500 Kelvin a is at the lowest temperature of 300 K which

at the lowest temperature of 300 K which means B is in the middle at 400 kin so

means B is in the middle at 400 kin so anytime you go in this general direction

anytime you go in this general direction C the temperature

C the temperature increases make sure you understand that

increases make sure you understand that for PV

diagrams now let's spend some time talking about the graph of an adiabatic

talking about the graph of an adiabatic expansion and let's see how it differs

expansion and let's see how it differs from an isothermal

from an isothermal expansion so let's say this is an ISO

expansion so let's say this is an ISO thermal expansion from A to

thermal expansion from A to B keep in mind for isothermal

B keep in mind for isothermal graphs or any isothermal process delta T

graphs or any isothermal process delta T is equal to zero the change in

is equal to zero the change in temperature is zero so the temperature

temperature is zero so the temperature remains

constant now for an adiabatic expansion it looks very similar but notice that

it looks very similar but notice that the curve is lower so a to c is anotic

the curve is lower so a to c is anotic expansion and for that Q is equal to

expansion and for that Q is equal to zero no heat enters or leaves the

zero no heat enters or leaves the system but notice that the temperature

drops so this is let's call this a d arotic expansion this is the isothermal

arotic expansion this is the isothermal graph so for adiabatic

graph so for adiabatic expansion when the volume

expansion when the volume increases the pressure decreases but it

increases the pressure decreases but it happens in such a way that the

happens in such a way that the temperature drops for isothermal graph

temperature drops for isothermal graph when the volume goes up the pressure

when the volume goes up the pressure goes

goes down but because the temperature is

down but because the temperature is constant for an ismal graph you can use

constant for an ismal graph you can use the equation P1 V1 is equal to P2

the equation P1 V1 is equal to P2 V2 now because the temperature changes

V2 now because the temperature changes in the adiabatic expansion because it

in the adiabatic expansion because it cools

cools down you need to use the variant of this

down you need to use the variant of this equation the equation that you need to

equation the equation that you need to use is uh

use is uh P1 V1 raised to to the gamma

P1 V1 raised to to the gamma power is equal to P2 V2 raised to the

power is equal to P2 V2 raised to the gamma

gamma power so as you can see it looks like

power so as you can see it looks like that equation but slightly different and

that equation but slightly different and there's also another equation that you

there's also another equation that you need to know it's T1

need to know it's T1 V1 raised to the gamma minus

V1 raised to the gamma minus one which is equal to T2 V2 rais to

one which is equal to T2 V2 rais to gamma minus one we can talk about how to

gamma minus one we can talk about how to derive this equation later well let's

derive this equation later well let's work on a problem

work on a problem using those

using those equations a car engine takes in air at

equations a car engine takes in air at 25° C and 1 atm and compresses it

25° C and 1 atm and compresses it adiabatically to 0.1 times the original

adiabatically to 0.1 times the original volume the air has a gamma ratio of 1.4

volume the air has a gamma ratio of 1.4 what is the final temperature and

what is the final temperature and pressure so we know that the temperature

pressure so we know that the temperature decreases during adiabatic uh expansion

decreases during adiabatic uh expansion but during adiabatic compression

but during adiabatic compression increases and if you decrease the volume

increases and if you decrease the volume the pressure will go up and it turns out

the pressure will go up and it turns out the temperature will go up as

the temperature will go up as well so let's work out this

well so let's work out this problem let's begin by using equation P1

problem let's begin by using equation P1 V1 raised to the gamma value is equal to

V1 raised to the gamma value is equal to P2

P2 V2 raised to

V2 raised to gamma now let's make a list of the

gamma now let's make a list of the information that we have T1 is 25 cels

information that we have T1 is 25 cels and if you add 273 to it that's going to

and if you add 273 to it that's going to equal 298

equal 298 Kelvin now we don't know what T2 is

Kelvin now we don't know what T2 is equal

equal to we know that P1 is 1

to we know that P1 is 1 atm and we don't have the value for P2

atm and we don't have the value for P2 we need to find that as

we need to find that as well we don't know what V1 or V2 is

well we don't know what V1 or V2 is equal to however we do know the ratio

equal to however we do know the ratio it's 0.1 so let's say if V1 is one liter

it's 0.1 so let's say if V1 is one liter that means V2 is going to be 0.1 time

that means V2 is going to be 0.1 time that .1 L it really doesn't matter the

that .1 L it really doesn't matter the the actual values of V1 and V2 the

the actual values of V1 and V2 the relative values is what matters if V1

relative values is what matters if V1 was 10 L V2 would be 10 * .1 which is 1

was 10 L V2 would be 10 * .1 which is 1 so regardless if you use 1 in 0.1 or 10

so regardless if you use 1 in 0.1 or 10 and one the answer will still be the

and one the answer will still be the same now if you've taken General

same now if you've taken General chemistry perhaps you've seen the

chemistry perhaps you've seen the combined gas law equation

now if we had T2 or P2 we could find a missing value but here we have two

missing value but here we have two missing values so therefore the combined

missing values so therefore the combined gas law won't help

gas law won't help us we need to use this equation to find

us we need to use this equation to find uh P2 and then the other equation to

uh P2 and then the other equation to find T2 so let's begin P1 is 1 V1 is 1

find T2 so let's begin P1 is 1 V1 is 1 raised to the gamma which is

raised to the gamma which is 1.4 we're looking for P2 V2 2 is

1.4 we're looking for P2 V2 2 is .1 raised to the uh

.1 raised to the uh 1.4 so to solve for

1.4 so to solve for P2 it's going to be 1 raised to the 1.4

P2 it's going to be 1 raised to the 1.4 which is simply 1 / 0.1 raised to the

which is simply 1 / 0.1 raised to the 1.4 so

P2 is about 25.1

ATM now let's use the other equation T1

equation T1 V1 gamma

V1 gamma -1 is equal to T2

-1 is equal to T2 V2 gamma minus one so T1 is 298 Kelvin

V2 gamma minus one so T1 is 298 Kelvin V1 is 1 gamma minus 1 that's 1.4 minus 1

V1 is 1 gamma minus 1 that's 1.4 minus 1 which is

which is uh4 we're looking for

uh4 we're looking for T2 and V2 is uh 0.1 raised to the

T2 and V2 is uh 0.1 raised to the point4 so T2 is going to be

point4 so T2 is going to be 298 times 1 raised to anything is 1 so 1

298 times 1 raised to anything is 1 so 1 to the point4 is 1 so it's 298 / .1 ra

to the point4 is 1 so it's 298 / .1 ra to

to the4 and you should get

the4 and you should get 7485

Kelvin if you subtract that by 273 that's

now let's talk about this but first let's make some

space now the volume changed from 1 liter to .1

liter to .1 lit so the volume decreased by a factor

lit so the volume decreased by a factor of

of 10 when the volume goes down the

10 when the volume goes down the pressure goes

pressure goes up but notice that all we did was we

up but notice that all we did was we decreased the volume by a factor of 10

decreased the volume by a factor of 10 and the pressure increased more than a

and the pressure increased more than a factor of 10 how did that

factor of 10 how did that happen well it has to do with the change

happen well it has to do with the change in

in temperature by decrease in the volume by

temperature by decrease in the volume by a factor of 10 the pressure actually

a factor of 10 the pressure actually goes up by a factor of 10 they're

goes up by a factor of 10 they're proportional according to B's law but

proportional according to B's law but let's look at the temperature the

let's look at the temperature the temperature changed from 298

temperature changed from 298 kelv to

kelv to 7485

7485 Kelvin when you increase the temperature

Kelvin when you increase the temperature you also increase the pressure if we

you also increase the pressure if we divide

divide 7485 by

7485 by 298 it tells us that the the temperature

298 it tells us that the the temperature went up by

went up by 2.51 so therefore the pressure should go

2.51 so therefore the pressure should go up by

up by 2.51 if you multiply these two values

2.51 if you multiply these two values you can see the pressure went up by a

you can see the pressure went up by a combined value of

25.1 so it increased by a factor of 10 due to the decrease in volume and it

due to the decrease in volume and it increased by a factor of 2.51 due to the

increased by a factor of 2.51 due to the increase in

increase in temperature so as you can see the

temperature so as you can see the pressure and the temperature greatly

pressure and the temperature greatly increased in this

increased in this situation which means adiabatic

situation which means adiabatic compression is very useful for engines

compression is very useful for engines if you think about it

if you think about it as the gas compresses

as the gas compresses adiabatically the increase in pressure

adiabatically the increase in pressure and the increase in

and the increase in temperature greatly accelerates the

temperature greatly accelerates the reaction between the air and the

reaction between the air and the fuel at high pressures gas molecules

fuel at high pressures gas molecules Collide more frequently which increases

Collide more frequently which increases the rate of the reaction also if you

the rate of the reaction also if you increase the temperature they Collide

increase the temperature they Collide not only more frequently but also with

not only more frequently but also with more energy making the reaction more

more energy making the reaction more spontaneous and as a result

spontaneous and as a result when you compress the when you compress

when you compress the when you compress the air fuel mixture

the air fuel mixture adiabatically you really don't need a

adiabatically you really don't need a spark plug the high temperature and

spark plug the high temperature and pressure could ignite the mixture

pressure could ignite the mixture spontaneously which is very useful for

engines now let's talk about how to derive those equations for an adiabatic

derive those equations for an adiabatic process we know that there's no exchange

process we know that there's no exchange of

of heat the heat flow into or out of the

heat the heat flow into or out of the system is zero so so Q is z and

system is zero so so Q is z and according to the first law of

according to the first law of thermodynamics the change in internal

thermodynamics the change in internal energy is Q minus W so if Q is zero that

energy is Q minus W so if Q is zero that means Delta U is equal to

means Delta U is equal to W and we know that Delta U for any

W and we know that Delta U for any process is

process is ncv time delta

T and the work is typically pressure times a change in

times a change in volume and according to the ID gas law

volume and according to the ID gas law PV is equal to nrt so if we solve for p

PV is equal to nrt so if we solve for p it's nrt divid

it's nrt divid V so let's replace p with this

V so let's replace p with this expression so now we have the expression

expression so now we have the expression ncv delta T is equal to

ncv delta T is equal to nrt over V * Delta V so we can cancel

nrt over V * Delta V so we can cancel the number of moles which is n

the number of moles which is n and so we're left with uh CV delta T is

and so we're left with uh CV delta T is equal to

equal to RT Delta V divided by

RT Delta V divided by V so now what we're going to do is

V so now what we're going to do is divide both sides

divide both sides by T and CV so I'm going to divide the

by T and CV so I'm going to divide the left side by CV and T and the right side

left side by CV and T and the right side by CV and

by CV and T so on the left side T is going to

T so on the left side T is going to cancel and on the right side CV will

cancel and on the right side CV will cancel

so I now have the expression delta T / T is equal to RV

expression delta T / T is equal to RV over CV Delta V /

over CV Delta V / V now if you recall CV is equal to R /

V now if you recall CV is equal to R / gamma minus one if we multiply both

gamma minus one if we multiply both sides by gamma minus one or even if we

sides by gamma minus one or even if we cross

cross multiply CV * gamma -

multiply CV * gamma - 1 is equal to 1 * R which is

1 is equal to 1 * R which is R so if we divide both sides by CV gamma

R so if we divide both sides by CV gamma minus 1 is equal to R over

minus 1 is equal to R over CV so now we

CV so now we can

can replace R over CV with gamma minus one

replace R over CV with gamma minus one so we have delta T / T is equal

so we have delta T / T is equal to gamma - 1

to gamma - 1 Delta v

Delta v v now let's take this expression let's

v now let's take this expression let's move it to this

move it to this side so it's delta T / T is equal to

side so it's delta T / T is equal to positive well not equal to but plus

positive well not equal to but plus gamma minus1 time Delta V over V and

gamma minus1 time Delta V over V and this is equal to

zero now delta T if the change is very small we can say it's approxim equal to

small we can say it's approxim equal to DT and Delta V given a small change in V

DT and Delta V given a small change in V we could say that's equal to

we could say that's equal to DV so I'm going to rewrite the

DV so I'm going to rewrite the expression as 1 / T *

expression as 1 / T * DT

DT plus Gamma

plus Gamma minus1 1 / V *

minus1 1 / V * DV and this is equal to

DV and this is equal to zero so now we're going to find the

zero so now we're going to find the anti-derivative of both sides of the

anti-derivative of both sides of the equation so let's integrate 1 / T 1 V

equation so let's integrate 1 / T 1 V and Z gam minus one is a constant so we

and Z gam minus one is a constant so we can move it in front of the integral now

can move it in front of the integral now if you recall we said the

if you recall we said the anti-derivative of 1 /x is L and X so

anti-derivative of 1 /x is L and X so the anti-derivative of 1/ T is Ln T and

the anti-derivative of 1/ T is Ln T and the anti-derivative of 1/ V is simply

the anti-derivative of 1/ V is simply lnv times the constant in front of it

lnv times the constant in front of it which is gamma minus one the

which is gamma minus one the anti-derivative of zero is equal to a

anti-derivative of zero is equal to a constant

now a property of natural logs allows us to take the constant in the front and

to take the constant in the front and move it to the exponent position so 2 Ln

move it to the exponent position so 2 Ln a is equivalent to Ln a^

a is equivalent to Ln a^ 2 so using that property of natural logs

2 so using that property of natural logs we can take the value gamma minus one

we can take the value gamma minus one and move it onto the exponent of V so Ln

and move it onto the exponent of V so Ln t plus Ln

t plus Ln V raised to the gamma minus 1 is equal

V raised to the gamma minus 1 is equal to a

to a constant another property of logs that

constant another property of logs that we can employ is this property LNA a

we can employ is this property LNA a plus Ln B is equivalent to Ln a *

plus Ln B is equivalent to Ln a * B so we can take two natural logs and

B so we can take two natural logs and compress it into a single natural log so

compress it into a single natural log so Ln T * V raised to the gamma minus one

Ln T * V raised to the gamma minus one is equal to a

constant so Ln T1 V1 gam minus1 is equal to a constant which is equal to Ln T2 at

to a constant which is equal to Ln T2 at a different temperature and V2 at a

a different temperature and V2 at a different volume but raised to the gamma

different volume but raised to the gamma minus

minus one so if these two are equal to the

one so if these two are equal to the same constant they must be equal to each

same constant they must be equal to each other so we can say that T1

other so we can say that T1 V1 gamma minus 1 is equal to T2 V2 G

V1 gamma minus 1 is equal to T2 V2 G minus

one now at this point let's go back to the ideal gas law equation PV is equal

the ideal gas law equation PV is equal to nrt this time we're going to solve

to nrt this time we're going to solve for a t PV / NR is equal to T if you

for a t PV / NR is equal to T if you divide both sides by the NR value

if you do that D will cancel and you get this expression so let's replace t with

this expression so let's replace t with PV over NR so T1 is P1 V1 / NR R is

PV over NR so T1 is P1 V1 / NR R is constant and we're not going to change

constant and we're not going to change the moles of gas so n will be

constant so T2 is going to equal P2 V2 over NR * V2 gamma -

over NR * V2 gamma - so n r is the same we can cancel these

so n r is the same we can cancel these two now according to the rules of

two now according to the rules of algebra if you multiply let's say two

algebra if you multiply let's say two common

common variables you can add the exponents so

variables you can add the exponents so X2 * X Cub is x 5 2 + 3 is 5 therefore

X2 * X Cub is x 5 2 + 3 is 5 therefore if we multiply V1 raised to the first

if we multiply V1 raised to the first Power with V1 raised to the gamma minus

Power with V1 raised to the gamma minus one

one we can add 1 + gamma - 1 which is equal

we can add 1 + gamma - 1 which is equal to gamma so V1 * V1 gamma minus one is

to gamma so V1 * V1 gamma minus one is just V1 raised to the

gamma so these two combine will simply become V1 raised to the gamma power and

become V1 raised to the gamma power and the same is true for the other side it's

the same is true for the other side it's going to be V2 raised to the gamma so

going to be V2 raised to the gamma so that's how we we can get this

equation so now what we're going to do is go over all of the equations that we

is go over all of the equations that we went in this video we're just going to

went in this video we're just going to do a final summary and that's going to

do a final summary and that's going to be uh it for this

be uh it for this video now the first and most important

video now the first and most important equation that you need to know is the

equation that you need to know is the first law of

first law of thermodynamics which is basically the

thermodynamics which is basically the change in internal energy is equal to Q

change in internal energy is equal to Q minus W anytime heat is absorbed by the

minus W anytime heat is absorbed by the system Q is positive and when heat is

system Q is positive and when heat is released Q is

released Q is negative when work is done by the system

negative when work is done by the system work is positive and when work is done

work is positive and when work is done on the system work is negative so let me

on the system work is negative so let me just write that this is by the

just write that this is by the system

system and this is on the

system now the system performs work whenever the gas

whenever the gas expands now to do work on a system

expands now to do work on a system ideally on a gas you need to compress it

ideally on a gas you need to compress it so compress in a gas work is always

so compress in a gas work is always negative for that

process Now Delta U is always dependent on

on temperature Delta U is equal to n CV

temperature Delta U is equal to n CV delta T for any

delta T for any process so if the change in temperature

process so if the change in temperature is zero Delta U is zero if the

is zero Delta U is zero if the temperature

temperature increases then the internal energy will

increases then the internal energy will increase as well so this is the next

increase as well so this is the next equation that you

equation that you need now keep in mind for monoatomic

need now keep in mind for monoatomic gases like

gases like helium uh neon

helium uh neon argon CV is approximately equal to 3

argon CV is approximately equal to 3 over2 * R and CP is 5 over2 *

over2 * R and CP is 5 over2 * R now R has two values

for the most part R is usually 8.3145 jewles per mole per

8.3145 jewles per mole per Kelvin but if you're dealing with gases

Kelvin but if you're dealing with gases where p is in ATM and V is in liters

where p is in ATM and V is in liters sometimes you may use this

equation typically you may need to use this RV value for the ideal gas law

this RV value for the ideal gas law equation PV is equal to nrt

but going back to gases if we have a datomic gas like N2 or O2 or even

datomic gas like N2 or O2 or even H2 you can approximate CV using this

H2 you can approximate CV using this equation 5 over 2 * R and CP is 7 over2

equation 5 over 2 * R and CP is 7 over2 *

* R and if you have a gas with three atoms

R and if you have a gas with three atoms like CO2

like CO2 SO2 it's better to look up the CP and CV

SO2 it's better to look up the CP and CV values but if you don't have it if you

values but if you don't have it if you can't look it up the best thing you

can't look it up the best thing you could do is estimate the answer

could do is estimate the answer so CV is going to be equal to 7/2 R and

so CV is going to be equal to 7/2 R and CP is approximately equal to 92 R so

CP is approximately equal to 92 R so this is approximately equal

this is approximately equal to there will be some uh

to there will be some uh variations now the first process that we

variations now the first process that we need to talk about is an isocoric

process in a PV diagram it looks like this it's basically a straight line

this it's basically a straight line going up or down

for an isocoric process the change in volume is

volume is zero so therefore the work performed is

zero so therefore the work performed is zero since there's no area there's no

zero since there's no area there's no shade of

shade of region because work is equal to

region because work is equal to zero Q is equal to W I mean not W but

zero Q is equal to W I mean not W but Delta

Delta U so this equation still applies Delta U

U so this equation still applies Delta U is Q minus W and since W is zero Delta U

is Q minus W and since W is zero Delta U equal Q

equal Q so you can use this equation to find

so you can use this equation to find Delta u n CV delta

Delta u n CV delta T this will always

T this will always work and Q is also equal to ncv do T

work and Q is also equal to ncv do T because the volume is

constant so those equations are the same since they equal each

other now this more for an isocoric process where the change in volume is

process where the change in volume is constant I mean is equal to zero you can

constant I mean is equal to zero you can also use this equation Q is equal to V

also use this equation Q is equal to V CV Delta P /

CV Delta P / R so this equation

R so this equation applies and also since the volume is

applies and also since the volume is constant when the temperature goes up

constant when the temperature goes up the pressure will go up

the pressure will go up proportionally so you can also use this

proportionally so you can also use this equation P1 / T1

equation P1 / T1 is equal to P2 / T2 during an isocoric

is equal to P2 / T2 during an isocoric process so that's it for this process

process so that's it for this process now let's move on to the next one now

now let's move on to the next one now let's talk about the isomeric

let's talk about the isomeric process this is when the pressure is

process this is when the pressure is constant so Delta p is equal to zero and

constant so Delta p is equal to zero and the PV diagram looks like

this you can travel to the right or to the

the left now keep in mind whenever you

left now keep in mind whenever you travel towards the right the work

travel towards the right the work performed by the gas is

performed by the gas is positive and if you travel towards the

positive and if you travel towards the left the work performed by the gas is

left the work performed by the gas is negative and remember the work performed

negative and remember the work performed is equal to the

is equal to the area of the Shader

region so for an isobaric process the work is equal to P Delta V V this is the

work is equal to P Delta V V this is the equation that you want to use now you

equation that you want to use now you can also find the work in terms of

can also find the work in terms of temperature it's equal to NR * delta

temperature it's equal to NR * delta T since PV is equal to nrt P Delta V is

T since PV is equal to nrt P Delta V is equal to NR delta T so that's another

equal to NR delta T so that's another way you could find the work during an

way you could find the work during an isobaric

isobaric process now to calculate Q during an

process now to calculate Q during an isobaric process you can use the

isobaric process you can use the equation n CP delta T you can also use

equation n CP delta T you can also use the equation

the equation P

P CP Delta V since the volume changes ID

R and Delta U is still Q minus W this equation applies and it's also equal to

equation applies and it's also equal to n cv.

t now for an isobaric process since the pressure is constant when you increase

pressure is constant when you increase the

the temperature the volume will

temperature the volume will increase and according to Charles's Law

increase and according to Charles's Law you can use this equation V1 / T1 is

you can use this equation V1 / T1 is equal to V2 /

process which looks like this as you mentioned before the work

this as you mentioned before the work performed by the gas is equal to the

performed by the gas is equal to the area of the Shad of region

and since we have a shape of a rectangle the area is basically the length times

the area is basically the length times the width the length is the change in

the width the length is the change in pressure the width is the change in

pressure the width is the change in volume so in this instance the work is

volume so in this instance the work is equal to the change in pressure times

equal to the change in pressure times the change in

volume so that's a combination of an isobaric and isocoric

isobaric and isocoric process now if you have a PB diagram

process now if you have a PB diagram that looks like this let's say

that looks like this let's say if you start at an initial pressure and

if you start at an initial pressure and it increases to a final pressure and

it increases to a final pressure and also the volume changes from an initial

also the volume changes from an initial value to a final

value to a final value the work perform is still the area

value the work perform is still the area of the Shaded

of the Shaded region but a simple way to calculate

region but a simple way to calculate this area is to use this equation it's

this area is to use this equation it's to use an average pressure to find the

to use an average pressure to find the average

pressure you still use the equation P Delta V but to find the average pressure

Delta V but to find the average pressure you simply add up the initial and the

you simply add up the initial and the final pressure and divide it by two so

final pressure and divide it by two so the work performed is2 the sum of the

the work performed is2 the sum of the two pressures times the change in

two pressures times the change in volume that's if the pressure changes at

volume that's if the pressure changes at a constant rate it can increase or

a constant rate it can increase or decrease as long as that line is a

decrease as long as that line is a straight line you can use this

straight line you can use this equation if you want to you can

equation if you want to you can calculate the area by turning this into

calculate the area by turning this into a rectangle and triangle so if you find

a rectangle and triangle so if you find the area of the rectangle and and the

the area of the rectangle and and the triangle um you can also calculate the

triangle um you can also calculate the area which is equal to the

area which is equal to the work the area of a rectangle is length

work the area of a rectangle is length time width the area of a triangle 1 12

time width the area of a triangle 1 12 base time

height now there's one more thing we need to talk

need to talk about so let's say if we have a cyclic

process so you know how to find the work which is the area of the Shaded region

which is the area of the Shaded region but you need to know if it's positive or

but you need to know if it's positive or negative so if the process is moving in

negative so if the process is moving in the clockwise Direction the work

the clockwise Direction the work performed is

performed is positive but if you have a cyclic

positive but if you have a cyclic process moving in the counterclockwise

process moving in the counterclockwise Direction then the work performed is

Direction then the work performed is negative so keep that in

negative so keep that in mind and

mind and also the net heat flow during a cyclic

also the net heat flow during a cyclic process is equal to W because as you

process is equal to W because as you start from position a going to B to C to

start from position a going to B to C to D and back to

D and back to a going from a to

a going from a to a the change in internal energy is zero

a the change in internal energy is zero and since Delta U is Q minus

and since Delta U is Q minus W if Delta U is equal to zero that means

W if Delta U is equal to zero that means Q has to equal W so for any cyclic

Q has to equal W so for any cyclic process the total heat flow or the net

process the total heat flow or the net heat flow is

heat flow is basically W the net heat flow

basically W the net heat flow is the amount of heat that enters the

is the amount of heat that enters the system minus the heat that leaves the

system minus the heat that leaves the system qh represents the energy that

system qh represents the energy that enters the system that's how much that's

enters the system that's how much that's the total heat absorbed by the system ql

the total heat absorbed by the system ql represents the energy that is released

represents the energy that is released by the system and the difference between

by the system and the difference between those two is equal to W the efficiency

those two is equal to W the efficiency is basically the work performed divided

is basically the work performed divided by the total heat that entered the

by the total heat that entered the system these two values are different

system these two values are different this is usually

larger so the efficiency is W minus qh but also times

100% the third process that we need to review is the

review is the ismal

process ISO means the same so isothermal the temperature is going to be the same

the temperature is going to be the same so the change of temperature is

so the change of temperature is zero now the isothermal graph looks like

zero now the isothermal graph looks like this in a PV

this in a PV diagram to calculate the work performed

diagram to calculate the work performed in an isothermal situation you can use

in an isothermal situation you can use this equation it's equal to

this equation it's equal to nrt * Ln V final / V

nrt * Ln V final / V initial you can also use this equation

initial you can also use this equation nrt Ln p P final / P

nrt Ln p P final / P initial now because the change in

initial now because the change in temperature is zero that means Delta U

temperature is zero that means Delta U is equal to zero keep in mind Delta U is

is equal to zero keep in mind Delta U is ncv delta T So if this part is equal to

ncv delta T So if this part is equal to zero the whole thing is equal to Zer now

zero the whole thing is equal to Zer now if Delta U is equal to Z and Delta U is

if Delta U is equal to Z and Delta U is Q minus W that means Q - W is equal to Z

Q minus W that means Q - W is equal to Z which means

which means Q is equal to W for an isothermal

Q is equal to W for an isothermal process so you can also find Q using any

process so you can also find Q using any one of these equations since it is equal

one of these equations since it is equal to

W now since the temperature is constant for an isothermal

for an isothermal process that means that as you increase

process that means that as you increase the volume at constant temperature the

the volume at constant temperature the pressure decreases

pressure decreases proportionally so you can also use this

proportionally so you can also use this equation P1 V1 is equal to P2

equation P1 V1 is equal to P2 V2 that is spoils

V2 that is spoils law so that's it for an isothermal

law so that's it for an isothermal process now let's move on to an

process now let's move on to an adiabatic

process in this process the insulation is either very very good or the process

is either very very good or the process happens so quickly that heat doesn't

happens so quickly that heat doesn't have enough time to enter into or out of

have enough time to enter into or out of the system so for an antibiotic process

the system so for an antibiotic process Q is approximately equal to

Q is approximately equal to zero and if that's the case that means

zero and if that's the case that means that Delta U is equal to W since Q is

that Delta U is equal to W since Q is zero so we know we could find Delta U

zero so we know we could find Delta U using this equation ncv delta T for any

using this equation ncv delta T for any process that means that you can also

process that means that you can also find the work using equation

find the work using equation negative n CV delta

T now keep in mind you can find CV by using this equation it's equal to R over

using this equation it's equal to R over gam minus1 where R is

gam minus1 where R is 8.3145 and CP is equal to R * gamma over

8.3145 and CP is equal to R * gamma over gamma minus one and Gamma is the ratio

gamma minus one and Gamma is the ratio between CP and

CV and then you also have these equations P1

equations P1 V1 raised to the gamma is equal to P2 V2

V1 raised to the gamma is equal to P2 V2 raised to the gamma and also uh T1

raised to the gamma and also uh T1 V1 gamma minus 1 is equal to T2

V1 gamma minus 1 is equal to T2 V2 gamma minus

V2 gamma minus one now you can also find work using

one now you can also find work using some other equations so starting with

some other equations so starting with this

this equation n CV delta

equation n CV delta T it might be useful to write this

T it might be useful to write this equation in terms of pressure and volume

nrt the change in PV is equal to NR time the change in delta

the change in delta T so the change in PV / NR is equal to

T so the change in PV / NR is equal to delta

delta T so we have

T so we have ncv times

ncv times Delta PV over

NR so n cancels so here's another equation that

cancels so here's another equation that you might find useful the work is equal

you might find useful the work is equal to

to negative CV /

negative CV / R

R times

times pfal V

pfal V final minus P initial * V

final minus P initial * V initial so that's another equation that

initial so that's another equation that be useful if you need to calculate the

be useful if you need to calculate the work performed in a adiabatic process

work performed in a adiabatic process and if you don't have the temperature

and if you don't have the temperature but if you have the pressure and the

but if you have the pressure and the volume you can use that as

well you can also use this equation W is equal

to -1/ gamma -1 p P final V final minus P initial V

initial you can get that equation using the fact that CV is R over gamma minus

the fact that CV is R over gamma minus one if you cross multiply CV * gamma

one if you cross multiply CV * gamma minus1 is equal to

R and if you divide both sides by let's say CV

gamma minus one is r/ CV therefore 1 over gamma minus one is CV over R so you

over gamma minus one is CV over R so you replace this with uh one over G minus

replace this with uh one over G minus one and that's how you get the other

one and that's how you get the other equation so basically that's all the

equation so basically that's all the equations that you need for this

equations that you need for this particular topic so that is it for this

particular topic so that is it for this video thanks for watching and have a

video thanks for watching and have a great day

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YouTube Transcript:Thermodynamics, PV Diagrams, Internal Energy, Heat, Work, Isothermal, Adiabatic, Isobaric, Physics

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Thermodynamics, PV Diagrams, Internal Energy, Heat, Work, Isothermal, Adiabatic, Isobaric, Physics