Video Transcript
A balloon holds 0.012 cubic meters
of air at a pressure of 101 kilopascals and a temperature of 300 kelvin. The air from this balloon is
transferred to another balloon that is half the volume of the first one. 125 kilopascals of external
pressure is required to transfer the air. What is the air temperature in the
new balloon? Give your answer to the nearest
kelvin.
Let’s say that this is our first
balloon and that it’s attached by a tube to a second balloon that has half the
volume of the first one. Let’s say further that this tube
has a valve in it that can be opened or closed. With the valve closed, we’re told
how much air is in this first balloon, its pressure, as well as its temperature. Let’s refer to that volume pressure
and temperature as 𝑉 one, 𝑃 one, and 𝑇 one, respectively.
If we open this valve and our whole
system was at the same pressure 𝑃 one, then some of the air in the first balloon
would naturally flow into the second one. But we want to transfer all of the
air from the first balloon to the second. What’s needed, our problem
statement tells us, is 125 kilopascals of external pressure to fully transfer all
the air from the first to the second balloon. When the air has moved into the
second balloon, it has a volume we’ll call 𝑉 two, a pressure we’ll call 𝑃 two, and
a temperature 𝑇 two. It’s that temperature 𝑇 two that
we want to solve for.
As we get started doing this, let’s
assume that the air we’re working with is an ideal gas. That means it follows the ideal gas
law that a gas’s pressure times its volume equals the number of moles of the gas
times the gas constant multiplied by the gas’s temperature. In our scenario, as air is
transferred from the first to the second balloon, the amount of that air,
represented by the number of moles of the air 𝑛, stays the same. We could say then that in our case
𝑛 times 𝑅 in the ideal gas law is a constant value. If 𝑛 doesn’t change and 𝑅 is a
constant as it is, then 𝑛 times 𝑅 must be a constant.
By dividing both sides of the
equation by the temperature 𝑇, we can arrive at a form of the ideal gas law where
in our case all the quantities on the left change and all the quantities on the
right are constant. This implies that if we calculated
𝑃 times 𝑉 divided by 𝑇 for the gas in our first balloon, that would equal 𝑃
times 𝑉 divided by 𝑇 for the gas in our second balloon. In this equation, we can recall
that it’s the quantity 𝑇 two that we want to solve for.
To help us do that, let’s write
down some of the information given to us in our problem. We’re told that the initial volume
of the air in the balloon is 0.12 cubic meters. The initial pressure of that air is
101 kilopascals, and its initial temperature is 300 kelvin. In terms of the volume 𝑉 two,
we’re not told directly what this is, but we are told that our second balloon has a
volume equal to half that of the first one. We could write then that 𝑉 two is
equal to 𝑉 one divided by two.
Similarly, we’re not told the
pressure 𝑃 two, but we are told that 125 kilopascals of external pressure is
required to transfer the air from the first to the second balloon. Therefore, the pressure of the air
in the second balloon will equal the air pressure in the first balloon plus this
external pressure of 125 kilopascals. Knowing all this, we now know all
the values in this expression except the one we want to solve for, 𝑇 two. To find that out, let’s first clear
some space off the top of our screen.
Currently, we’re working with this
equation that we want to solve for 𝑇 two. If we multiply both sides of the
equation by 𝑇 two and then by 𝑇 one divided by 𝑃 one times 𝑉 one, we find that
on the right-hand side 𝑇 two cancels out. And on the left, the first pressure
𝑃 one, the first volume 𝑉 one, and the first temperature 𝑇 one cancel. That gives us this equation where
𝑇 two is the subject.
We can now substitute in for all
the values on the right-hand side. 𝑇 one is 300 kelvin; 𝑃 one is 101
kilopascals. That means that 𝑃 two is 101
kilopascals plus 125 kilopascals, or 226 kilopascals. And then for our volumes, we
actually don’t need to write in those specific values. And that’s because we have 𝑉 two,
which is 𝑉 one divided by two, all divided by 𝑉 one.
Thinking of this expression in
parentheses overall, we can multiply both the numerator and the denominator by one
over 𝑉 one. The result of that is to cancel 𝑉
one in the denominator and in the numerator. We’re left with the value just of
one-half. That’s the ratio of 𝑉 two to 𝑉
one. These values all multiplied
together will give us the temperature 𝑇 two. Before we calculate that value,
notice that our units of kilopascals in numerator and denominator will cancel,
leaving us with final units of kelvin. Rounding our answer to the nearest
kelvin, 𝑇 two is 336 kelvin. This is the temperature of the air
in the new or second balloon.
