S1.5.3 The gas laws (part 2) | Mike Sugiyama Jones (MSJ Chem) | YouTubeToText
YouTube Transcript: S1.5.3 The gas laws (part 2)
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this is msj chem in this video i'll be
looking at
examples involving the gas laws in our
first example we have a sample of gas
with a volume of 15 cubic centimeters
at a pressure of 575 kilopascals
assuming that temperature remains
constant what volume will the gas occupy
at a pressure of 968 kilopascals
to solve this problem we'll use the
equation p1
times v1 is equal to p2 times v2
which is boyle's law in the question we
are given
p1 v1 and p2 so we need to solve for v2
so here we have the equation rearranged
to solve for v2
next we substitute the values into the equation
equation
p1 is 575 kilopascals
and v1 is 15.0 cubic centimeters
p2 is 968 kilopascals
and this gives us a value for v2 of 8.91
cubic centimeters in our next example we
have a sample of gas with a volume of
32.0 cubic decimeters
at a temperature of 256 kelvin
assuming the pressure remains constant
what volume
will the gas occupy at a temperature of
391 kelvin
to solve this we'll use this equation
which is charles's law
in the question we're given v1 t1
and t2 so we need to rearrange this
equation to solve for v2
and v2 is equal to v1 times t2
divided by t1 so next we substitute the values
values
into the equation v1 is 32.0
cubic decimeters and t2 is 391 kelvin
t1 is 256 kelvin and this gives us a
value for v2
of 48.9 cubic decimeters
in our next example we have a sample of
gas with a pressure of 73.9 kilopascals
at a temperature of 347 kelvin
assuming that volume remains constant
what will be the pressure of the gas
at a temperature of 602 kelvin so here
we have the equation for gaylor sac's law
law
in the question we're given p1 t1
and t2 so we need to solve for p2
and here we have the equation rearranged
to solve for p2
next we substitute the values into the equation
equation
p1 is 73.9 kilopascals
t2 is 602 kelvin and t1
is 347 kelvin and this gives us a value
for p2 of 128 kilopascals
in the next example a sample contains
5.13 moles of gas
with a volume of 1.28 cubic meters
assuming that temperature and pressure
remain constant what volume will the gas occupy
occupy
if 3.49 moles of gas are added
so here we have the equation for
avogadro's law
in the question we're given v1 and n1
to find n2 we need to add the two amount
in moles of gas
which is 5.13 plus 3.49
to find v2 we need to rearrange the equation
equation
so v2 equals v1 times n2
divided by n1 v1 is 1.28 cubic meters
n2 is 8.62 moles and
n1 is 5.13 moles and this gives us
a value for v2 of 2.15 cubic meters
in our last example a sample of gas has
a volume of 1.54 cubic meters
at a temperature of 447 kelvin
and a pressure of 12.0 kilopascals
if the temperature and pressure are
changed to 561 kelvin
and 15.7 kilopascals respectively
what volume will the gas occupy to solve
this we'll use the combined gas law
in the question we're given p1 v1
t1 p2 and t2
so we need to rearrange the equation to
solve for v2
and v2 equals v1 times p1
times t2 divided by t1
times p2 v1 is 1.54 cubic meters
p1 is 12.0 kilopascals
t2 is 561 kelvin t1
is 447 kelvin and p2
is 15.7 kilopascals so this gives us a
value for v2
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