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Ch 8.1 - Gas pressure | General Chemistry | YouTubeToText
YouTube Transcript: Ch 8.1 - Gas pressure
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This content explains the concept of gas pressure, its definition, various units of measurement, and how it is measured using barometers and manometers.
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this video is on the topic of gas
pressure the learning objectives are
on your screen so go ahead and pause the
video and write those down in your notes
to begin the topic of gas pressure uh
it's useful to
first discuss pressure in terms of
something we might have heard of
atmospheric pressure
and atmospheric pressure is uh
you know we define it as the force
exerted by a column of air
uh in the atmosphere um that force on one
one
square inch of space that's one way to
think about atmospheric pressure
and the amount of uh of force exerted on
a given space
actually turns out to be about 14.7 pounds
pounds
as a as a weight um of force
on one square inch of uh
space and so that's equal to
one atmosphere of pressure
so in general sense and as you'll see in
a second we actually i already mentioned
units of
of atmospheres and i've also mentioned
pounds per square inch
so really there are a number of units
of pressure and the the main thing to
keep in mind is
the general definition of pressure
uh defined as capital p is going to be force
force
over area and so pressure is
is directly proportional to force and
inversely proportional to area
the pressure units i mentioned are right
here i'm going to quickly breeze through these
these
um definitely use this table um
from uh uh use this table to help
in in in your problem solving um uh practice
practice
so uh the pascal is the si unit um or
the iu
the recommended iupac unit um it is the
only one here you'll notice that's
defined in terms of
si units of force and and sort of length so
so
newton per meter squared um uh so
the kilo pascal's just you know the the
same prefix multiplier kilo so
i'm not going to go into that psi pounds
per square inch i just i already
mentioned that
that's at sea level which is about one
atmosphere it's about 14.7 pounds per
square inch
um this is a really useful uh
relationship here between atmospheres we
have one atmosphere of pressure
is the equivalent of 101 325 pascals
and that's equal to 760 torr so um
one of the common ones that i remember
definitely is that one atmosphere is
equal to 760 torr that one definitely is
a very
common um conversion we also have bar
this is
uh specific to not specific but commonly
used in meteorology
one bar is equal to exactly 100 000 pascals
pascals
a millibar is just you know the prefix
multi uh
modifier milli um
inches of mercury is one of my favorite
ones just because it's literally
the displacement of liquid mercury
measured in inches
um and that's typically um
uh resulting from devices
that have been designed to measure
pressure barometers and manometers and
we'll talk about those in a little bit
one inch of mercury is equal to 3 386 pascals
pascals
um the tour itself i mentioned tor up above
above
that one atmosphere is equal to 760
taurus this is just the tour
defined as in terms of just one tour
it's equal to
um one 760th atmospheres
um i just find it easier to remember
that one tour is equal to 700
um sorry that one atmosphere is equal to
760 torr millimeters of mercury um
is is quite similar to inches of mercury
and that it's the displacement of liquid mercury
mercury
one millimeter of mercury's is is about
one torr
so the other way to say that is that 760
millimeters of mercury is equal to
760 torr and both of these are equal to
one atmosphere um
so interconverting these uh is pretty
much dimensional analysis
so i'll leave it to you to to to
practice that we'll get
practice in class what i want to jump
into is how to measure
um pressure so this is an example of a
barometer this is one uh one type of
instrument used to measure pressure and
a barometer is typically used to
measure atmospheric pressure and change
its atmospheric pressure
you can use a barometer that has
liquid water in the bottom in this in
this container that's open to atmosphere
but you can also use liquid mercury
that's in the container that's open to atmosphere
atmosphere
the benefit of using mercury liquid
mercury is that it's
much uh more dense as a liquid with a
much higher density than water
so that means that if there is um if
there are you know
kind of big changes in pressure you uh
the the the the vertical displacement up
in the the tube
uh of the barometer um won't be that high
high
because it's such a dense liquid and
i'll and i'll show i'll give you an
expression for that accounts for liquid
density in a second
but you'll see that um water you know
has much lower density it has
it in order to measure pressure changes
we actually have to have much
higher displacements of water in the
tube so for practicality reasons
mercury makes a lot of sense for
measuring changes in pressure
we don't want to have have to make this gigantic
gigantic
barometer tube in which um just so that
we can allow
the water or some not very dense liquid to
to
shoot up very uh and have very big displacements
displacements
okay so it's strictly um practicality purposes
purposes
certainly not for safety reasons to use
mercury right
um okay so the premise here is that this
that there's a secondary container open
to atmosphere and we have this closed end
end
tube that is vacuum sealed on uh that's
inverted so what happens is
there's some equilibrium that's
maintained um that's reached when
atmospheric pressure is pushing down
but if atmospheric pressure increases
and it pushes down
on that liquid surface um then the uh
displacement uh in within this tube will
go up
and it will push if it's higher
atmospheric pressure it will push the mercury
mercury
up into the tube and we can measure that
displacement in inches of mercury or
millimeters of mercury whatever we want
and also if the if the pressure drops
back down below
atmospheric pressure or there's just a
decrease in atmospheric pressure
we would see that vertical displacement
go back down in the barometer
the mathematical expression that i mentioned
mentioned
for a barometer is that we have um
the pressure that we can measure is
going to be equal to this um this is
supposed to be a capital p
is equal to h times uh this is not a p
this is the greek letter rho
i did not really it's kind of difficult
to draw greek letters sometimes maybe
i'll draw it more like this
to emphasize that it's not a standard p
times g
and we can also think about the height
displacement of the fluid rho here
so what i'm saying row it's rho is a typical
typical
um greek letter used for
density and this is where the density of
the liquid comes into play and then g
here is actually the
acceleration due to
gravity but if we use a a single liquid
then that means that the density is a
constant um
the acceleration due to gravity is a
constant so really the pressure is
directly proportional then to the height
of the liquid that we can measure in the barometer
barometer
but a barometer is useful for
atmospheric pressure
what's more useful in a laboratory
setting is to be able to actually
measure the pressure of
a sample for example not just the open atmosphere
atmosphere
and a manometer actually does not it's
not opened atmosphere
you have a sample uh reservoir that we
can sort of
in a controlled manner introduce to this um
um
either a closed tube system with a with
a youtube
shape down here an inverted tube filled
with liquid mercury there's also another
design where we have an open atmosphere
a open-ended tube where now we have to
also account for the pressure
of of the atmosphere in our calculations
so the main premise here is that if we introduce
introduce
um to our manometer uh a sample of gas that
that
has um any really measurable uh
uh well gases will have measurable
pressure but if it's a volatile liquid
that has a substantial vapor pressure
what will happen is that when we open up
this valve here
the pressure will force the the liquid mercury
mercury
down right there's a greater pressure in
the side of the manometer
and that will cause this sort of
vertical displacement
on the other side of this inverted tube
since this is closed to the atmosphere
it's a closed system we don't have to
account for the atmospheric pressure in
our calculations
the pressure of the gas is going to be
equal to the height times rho
times or height times density times the
acceleration due to gravity
so essentially that vertical
displacement is directly proportional
to the amount of pressure from our gas sample
sample
and i'll look at one practice problem in
a second uh that deals with that
in the open-ended cases so both of these
on the right are open-ended cases
but uh uh the what's the the situation
over here is when
the pressure of our sample is less than
that of the atmospheric pressure okay
and so uh what happens is since the
other end is not vacuum sealed
um it's open to the atmosphere then we
also have to always account for the fact
that there's atmospheric pressure pushing
pushing
down on the open end of the tube so if
we open up this valve to our sample gas
and that sample gas is not pushing with
one atmosphere of pressure
or 760 millimeters of mercury or um
760 torr whatever unit you want to
describe one atmosphere as then that
means that um
if it's less than one atmosphere the
atmospheric pressure will push down the mer
mer
the liquid mercury and will actually
measure a net negative
displacement in the in the liquid height
um in our uh manometer um
and you can imagine a case it doesn't
show it here but imagine a case where
the gas sample pressure is exactly equal
to one atmosphere
then you would measure a net zero displacement
displacement
in the liquid height okay that means
that the sample pressure is equal to or
the gas pressure is equal to that of
atmospheric pressure
now in the case where we have the gas sample
sample
when opened to the atmosphere actually
is uh
or open to the manometer is greater than
one atmosphere of pressure
um then that means that this will not
only uh it has to overcome the force of
the the atmospheric pressure
but also it'll push just a little bit
beyond that so if it was exactly one
atmosphere we would measure no
displacement if it's
above one atmosphere it will push the
liquid mercury down over here which
causes a vertical displacement on the
um inverted side of the of the manometer
and then here you can see in these
expressions for the for the
for the first case that i described in
the open-ended denominator
of the gas pressure is actually equal to
the atmospheric pressure
it has to um account for that and then
in this case um it's if we measure that
negative displacement in height
so it's minus height times density times
acceleration due to gravity
so that means that this would indicate a
system or a gas pressure
less than that of atmospheric pressure
um and then the other manometer case
where we have a gas
sample with with greater than
atmospheric pressure it um we have to
say that the gas pressure is equal to
one atmosphere because it overcame that
one atmosphere
the atmospheric pressure plus whatever
that displacement height
was so let's take a quick look um
before uh the video uh before i end the video
video
at two examples the first one is going
to be the closed end
situation so we do not have to account
for atmospheric pressure in our calculations
calculations
the question here that we want to answer
pressure in tor based off of
this diagram that we can see here
so what we can do is and again this is
this is mercury
liquid okay so uh to start we can
uh use our starting value that we're
reading directly off that manometer 26.4
um centimeters of mercury uh height displacement
displacement
and essentially what we can do then is
we can say
that that is a unit of measure of pressure
pressure
centimeters mercury although it's not a
common one so we want to convert it to
tor which is
more common so we what we can do is
actually say okay well if it's a
centimeter of mercury that's just the
length unit we can actually convert
the length unit from one centimeter to 10
10
millimeters of mercury it's that is that straightforward
straightforward
and now that we're in millimeters of
mercury that is a common pressure unit
so now we can say
use a conversion factor here one
millimeter of mercury
is um equal to one torr
and so really this is two 26.4 times 10
so it's 264
torr of um for our gas pressure there
okay and now in the um open end case we
also now we have to
consider the um uh atmospheric pressure
um and i guess i'll point out too that
the benefit of the closed end you notice
that the gas pressure is 264 torr
atmospheric pressure is 760 torque so
that's not greater than atmospheric pressure
pressure
so if this was the open-ended case we
wouldn't get a vertical displacement
in the liquid marker it would actually
go down because atmospheric pressure
is is is greater so
speaking of open-ended cases let's look
at this case here we can make a prediction
prediction
that with this gas sample must have
greater than one atmospheric pressure
because look at this
it was a vertical displacement of liquid
mercury on the other side so
that gas sample must have overcome the
atmospheric pressure and then some
so let's let's do a calculation to see
if that's the case um
what i'm going to do here is essentially
uh jot down uh oh first let's see what
do we want to solve for
let's solve for this pressure in
atmospheres pressure
in atmospheres is it greater than
one so what i'm going to do is actually
take this step by step and say that we
have 13.7
centimeters of mercury similar to before
i'm going to convert that
into millimeters of mercury first
so this is one centimeter of mercury is
equal to
10 millimeters of mercury and i can jump
i can i can keep going from here but i
want to actually just pause
to now talk about what we just
calculated this
137 milligrams of mercury is not the
pressure of the gas sample that's the additional
additional
vertical height gained just right here
but remember this is open to the atmosphere
atmosphere
so this pressure 137 is what
was on top of atmospheric pressure so
really the pressure of the gas
pressure of the gas is going to be equal
to the
uh pressure of the atmosphere plus
whatever we just calculated that
vertical displacement
plus 137 millimeters of mercury so what
this equals is 760
millimeters of mercury for atmospheric
pressure which is approximating about
one atmosphere
um which is 760 ml
millimeters of mercury plus 137 millimeters
millimeters
mercury and that will give us the
pressure of the gas is actually 897
millimeters of mercury okay so that's
that's from the equation that i gave you
above so what is this in atmospheres
we take 897 millimeters of mercury
and we know that the conversion factor
is 760 millimeters of mercury
is equal to one atmosphere
one atmosphere so uh what this gives us then
then
is 1.18 atmospheres
we did predict just based off of looking
at the how the displacement occurred
in the manometer that it's greater than
one atmosphere and we have confirmed that
that
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