Video Transcript
A weather balloon has an initial volume of 1.2 cubic meters, an initial pressure of 102 kilopascals, and an initial temperature of 290 kelvin. When high in the air, the balloon has a volume of 1.0 cubic meters, a pressure of 96 kilopascals, and a temperature of 240 kelvin. What percentage of the gas in the balloon has leaked out? Give your answer to one decimal place.
Let’s say that at the outset, our weather balloon is here at ground level. At this spot, the balloon occupies a volume, we’ll call 𝑉 one. It has a pressure, we’ll call 𝑃 one, and a temperature, we’ll label 𝑇 one. Later on, when the balloon has moved high up in the air, it has a new volume, pressure, and temperature, we’ll call 𝑉 two, 𝑃 two, and 𝑇 two, respectively. In the process of going through this change in elevation, some of the gas in the balloon has leaked out. To begin finding out just how much, let’s assume that the air in the balloon can be treated as an ideal gas. That means this air is described by the ideal gas law. That says that the pressure of a gas multiplied by its volume equals the number of moles of the gas times a constant multiplied by the gas temperature.
Let’s apply the ideal gas law to our balloon at two different instants in time. At the first instant, when the balloon was at ground level, we have 𝑃 one times 𝑉 one equals 𝑛 one times 𝑅 times 𝑇 one. Note that all of these values have a subscript one except 𝑅 which is the gas constant and therefore always the same. We can write a similar equation but now for when the balloon is high up in the air. This equation we see makes use of 𝑉 two, 𝑃 two, and 𝑇 two. In this question, we’re looking to solve for the percentage of the gas that has leaked out. The variable in the ideal gas law that tells us about gas quantity is 𝑛, the number of moles of the gas.
To answer our question then, we’ll want to compare 𝑛 one, the number of moles of gas in the balloon when the balloon was at ground level, to 𝑛 two, the number of moles of gas in the balloon when it was at elevation. With that goal in mind, let’s rearrange these two equations so that 𝑛 one and 𝑛 two are the respective subjects. In the equation on the left, we’ll divide both sides by 𝑅 times 𝑇 one. This causes that gas constant 𝑅 and the temperature 𝑇 one to cancel on the right. We find then that 𝑛 one equals 𝑃 one 𝑉 one divided by 𝑅 times 𝑇 one.
Similarly, we divide both sides of the equation on the right now by 𝑅 times 𝑇 two. This cancels both of those factors on the right-hand side and gives us an expression for 𝑛 two. Now that we have expressions for 𝑛 one and 𝑛 two, let’s consider how to compare these values so we can express the difference between them as a percent. Since we’re told that gas has leaked out of the balloon as it rose from its lower to its higher elevation, we know that 𝑛 one, the number of moles of gas in the balloon at ground level, will be greater than 𝑛 two, the number of moles of gas in the balloon when it’s in the air.
We can think then of 𝑛 one as representing 100 percent of the gas. That is, that’s our baseline against which we’ll compare 𝑛 two. What we’re going to do is find the percentage difference between 𝑛 one and 𝑛 two, where 𝑛 one, as we’ve said, is our standard. We can write an equation for percent difference this way: the larger value minus the smaller value all divided by our baseline value 𝑛 one, then multiplied by 100 percent. By using this relationship, we’re calculating the percent change between 𝑛 one and 𝑛 two.
To start doing that, let’s record some of the information given to us in our problem. We’re told, for example, that this balloon has an initial volume of 1.2 cubic meters, that’s 𝑉 one, an initial pressure of 102 kilopascals, that’s 𝑃 one, and an initial temperature 𝑇 one of 290 kelvin. Once it’s high in the air, the balloon has a new volume 1.0 cubic meters, that’s 𝑉 two, a new pressure of 96 kilopascals, that’s 𝑃 two, and a new temperature 𝑇 two of 240 kelvin. Clearing some space at the top of our screen, we can rewrite our equation for percent difference and then substitute in 𝑃 one 𝑉 one over 𝑅 times 𝑇 one for 𝑛 one and 𝑃 two 𝑉 two divided by 𝑅 times 𝑇 two for 𝑛 two.
Notice that both terms in our numerator have one over the gas constant 𝑅. That means we can factor out one over 𝑅. And then we see that one over 𝑅 appears in both numerator and denominator. We can therefore cancel out the gas constant 𝑅. Our equation reduces to this expression. For our next step, we’re going to substitute in all of these known values. When we do that, we get this big equation. In this equation, 𝑉 one is 1.2 cubic meters, 𝑃 one is 102 kilopascals, 𝑇 one is 290 kelvin, 𝑉 two is 1.0 cubic meters, 𝑃 two is 96 kilopascals, and 𝑇 two is 240 kelvin.
When we think about the units in this expression, notice that both terms in our numerator and the term in the denominator all have the same units, kilopascals times meters cubed divided by kelvin. That actually means that all of these units will cancel out. Our final answer will be a percentage with no unit attached. When we calculate this whole expression and round our result to one decimal place, we get 5.2 percent. This is the percentage of gas that leaked out of the balloon as it rose from its lower to its higher elevation.
