# Question Video: Balloon Air Transfer Physics • 9th Grade

A balloon holds 0.012 m³ of air at a pressure of 101 kPa and a temperature of 300 k. The air from this balloon is transferred to another balloon that is half the volume of the first one. 125 kPa 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.

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### 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.