### Video Transcript

A gas cylinder holds 3.25 meters
cubed of gas at a pressure of 520 kilopascals and a temperature of 300 kelvin. At what temperature would the gas
pressure in the cylinder become 865 kilopascals? Answer to three significant
figures.

So this question describes a
cylinder of gas with a volume of 3.25 meters cubed, a pressure of 520 kilopascals,
which is the same as 520,000 pascals, and a temperature of 300 kelvin. Weβre being asked to calculate the
temperature at which the gas pressure in the cylinder would become 865 kilopascals
or 865,000 pascals. So we could imagine the same
cylinder of gas at sometime later where the pressure has changed to 865 kilopascals
or 865,000 pascals and the temperature has taken some unknown value that we need to
work out. And because itβs the same cylinder,
we can assume that the volume would go unchanged.

At this point, because we have two
different sets of values for the quantities volume, pressure, and temperature, letβs
call the initial set of values on the left π one, π one, and π one and the
changed values on the right can be π two, π two, and π two. So our challenge in this question
is to find π two. So to start with, letβs think of
the variables that we have to deal with in this question and see if we can think of
any equations that might help us.

Well, this question asks us to
consider the volume, pressure, and temperature of the gas. One equation that gives us the
relationship between these quantities is the ideal gas equation. This tells us that the pressure of
an ideal gas multiplied by its volume is equal to some constant multiplied by its
absolute temperature. We can recall that the ideal gas
equation makes the assumptions that gas molecules have negligible size and donβt
interact with each other. Even though this question talks
about a real gas, itβs still reasonable to use the ideal gas equation as it can
still give us accurate answers without being too complicated.

Now, in this question, weβre trying
to find the value of π. So letβs rearrange the ideal gas
equation to make π the subject and do this by just dividing both sides of the ideal
gas equation by π, giving us ππ over π equals π. Since we need to find the
temperature of the gas after the pressure has increased, it seems reasonable that we
can just plug these values, π two and π two, into this equation. And it will tell us π two. However, while this is technically
true, if we tried doing this, weβll quickly realize that we donβt know what the
value of π is. This is because π actually takes
different values depending on the number of gas molecules that weβre dealing
with. So itβs not something we can just
look up.

So as it stands, we canβt
substitute π two and π two straight into the equation and just get a value of π
two. However, because we have a set of
initial conditions for the volume, pressure, and temperature of the gas, there is
something else we can do with the ideal gas equation to find the answer. If we start with ππ equals ππ
and divide both sides of the equation by π, we get ππ over π equals π. Because the amount of gas molecules
in this question is constant, this means that π is constant. So regardless of how we try to
change the pressure, volume, or temperature of a fixed amount of gas, we find that
the value of ππ over π is always constant.

In other words, π one times π one
divided by π one is equal to π two times π two divided by π two. Writing the ideal gas equation in
this way is actually equivalent to its more familiar form ππ equals ππ. However, it makes it easy to
calculate how variables change without needing to know π. And we can use this formulation for
this question. Since weβre trying to find π two,
letβs first rearrange this to make π two the subject. First, we can multiply both sides
of the equation by π two, then multiply both sides of the equation by π one, and
finally divide both sides of the equation by π one π one.

With the equation in this form
because we know the initial temperature, initial pressure, and initial volume before
the pressure increased and we know the volume and the pressure after the pressure
increased. We can just substitute all of these
values in and calculate π two. Letβs just give ourselves some more
space. And we know that π one is 300
kelvin, π two is 865,000 pascals, π two is 3.25 meters cubed, π one is 520,000
pascals, and π one is 3.25 meters cubed as well. Plugging all this into our
calculator, we get an answer of 499.04 which has units of kelvin since itβs a
temperature. And rounding our answer to three
significant figures gives us a final answer of 499 kelvin.