Video Transcript
A balloon is inflated by adding more air to it, increasing its volume, as shown in the diagram. The pressure of the air in the balloon is 101 kilopascals before it is inflated and is 121 kilopascals after it is inflated. The temperature of the air in the balloon does not change when it is inflated. The air can be assumed to behave as an ideal gas. Find the volume of the balloon when it is inflated as a percent of the volume of the balloon before it is inflated. Give your answer to the nearest percent.
In our diagram, we see the balloon before and after being inflated with more air. We’re also told the volume the balloon occupies at each of these two stages. To answer this first part of our question, let’s clear some space on screen. And let’s say that the volume of the balloon before it is inflated is represented by 𝑉 sub b, while the volume of the balloon after it’s inflated is represented by 𝑉 sub a. We want to find the volume of the balloon when it is inflated, that’s 𝑉 sub a, as a percent of the volume of the balloon before it is inflated. That’s 𝑉 sub b. In other words, we want to express 𝑉 sub a as a percent of 𝑉 sub b.
To think of how to do this, let’s imagine two numbers. Choosing to make these numbers easier to work with, we’ll choose one and 0.5. Say that we want to know what this number, 0.5, is as a percent of this one. We can do that by first taking that first number, 0.5, and dividing it by the second one, one. That’s equal to 0.5. But then, to make this answer into a percent, we’ll need to move the decimal place one, two spots to the right. We can see that so because 0.5 is 50 percent of one. Our formula then is to divide 0.5 by the first number and then multiply that fraction by 100 percent. And that gives us the percent of one that 0.5 is.
This equation is specific to the two numbers we picked, but the form of this equation is actually a general one that we can apply to 𝑉 sub a and 𝑉 sub b. What we want to do is find 𝑉 sub a as a percentage of 𝑉 sub b. We’ll then take the volume of the balloon after it’s inflated, divide it by the balloon’s volume before it’s inflated, and multiply that fraction by 100 percent. The volume of the balloon after it’s inflated is given as 0.033 cubic meters, and that before it’s inflated is 0.012 cubic meters. To the nearest percent, this expression equals 275 percent. This is the volume of the balloon when it’s inflated as a percent of the volume of the balloon before it’s inflated. Knowing this, let’s now move on to part two of our question.
This part says, find the mass of the air in the balloon when it is inflated as a percent of the mass of the balloon before it is inflated. Give your answer to the nearest percent.
Unlike with volume, we’re not told the mass of the air in the balloon before and after it’s inflated. But we are told that we can treat that air as an ideal gas. This means we can describe it using the ideal gas law. This law says that a gas’s pressure times its volume is equal to the number of moles of the gas times a constant multiplied by the gas temperature. Even though this equation doesn’t have mass anywhere in it, it does have the number of moles of the gas in. We can think of this as a representative, a proxy for the mass of the air. The number of moles 𝑛 represents the number of particles of air, each particle having the same amount of mass.
We’re told that air is added to our balloon when it’s inflated. This means that the number of moles of air in the balloon before inflation, we’ll call that 𝑛 sub b, is not equal to the number of moles of air in the balloon after inflation. We’ll call that 𝑛 sub a. And specifically, because the balloon is being inflated, we know that 𝑛 sub a is greater than 𝑛 sub b. Because these numbers of moles of air are stand-ins, as we’ve said, for the mass of the air in the balloon, if we find the number of moles of air in the balloon after inflation as a percent of the number of moles of air in the balloon before, then that will be equal to the mass of air in the balloon after inflation as a percent of the mass of air in the balloon beforehand.
So, to solve for this, which is what our question asks about, we can calculate this. Let’s recall that the air in our balloon is to be modeled as an ideal gas. Before the balloon is inflated, it has some pressure, some volume, and some temperature, as well as some number of moles of air inside. We can write the ideal gas law for the air in the balloon before inflation like this. All of these b subscripts stand for before. And notice that this value of 𝑅 doesn’t have one, and that’s because 𝑅 is a constant value called the gas constant. If we were to do something similar for the balloon after inflation, the ideal gas law equation would look like this, the same as above, except now with a subscripts.
In order to calculate our value of interest, we’ll need to know the ratio 𝑛 sub a over 𝑛 sub b. We can begin solving for that by dividing this entire equation by this equation. We can do this because since both sides of this equation in the denominator are equal, by definition, we’re dividing both sides of the equation in the numerator by the same value.
Another way we could write this fraction then is as this single equation. What we’ve done is we’ve taken the equation in the numerator. And we’ve divided the left side of that equation by a value that’s equal to the value by which we’ve divided the right side. This is why writing the ratio of these two equations as a single equation as we’ve done here is valid. On the right-hand side of this expression, notice that the gas constant 𝑅 appears in both numerator and denominator and therefore cancels out. Along with this, in our original problem statement, we’re told that as the balloon is inflated from a smaller to a larger volume, its temperature stays constant. Using our symbols, that means that 𝑇 sub a is equal to 𝑇 sub b. That equality means that 𝑇 sub a divided by 𝑇 sub b equals one.
We now have an expression for 𝑛 sub a divided by 𝑛 sub b in terms of the pressure and the volume of the balloon before and after it’s inflated. From our diagram, we’re given the volumes 𝑉 sub a and 𝑉 sub b. And in our problem statement, we’re given values of 𝑃 sub a and 𝑃 sub b. The pressure of the air in the balloon before inflation, 𝑃 sub b, is 101 kilopascals. And the pressure of this air after inflation, 𝑃 sub a, is 121 kilopascals.
Having values for all four of the variables on the left-hand side of this equation, we’re ready to plug in and solve for 𝑛 sub a over 𝑛 sub b. To do that, and indeed to solve for 𝑛 sub a as a percentage of 𝑛 sub b, let’s clear some space at the top of our screen. Knowing that this value will equal the mass of the air in the balloon after inflation as a percentage of the mass of the air in the balloon before inflation, we substitute in the given values for 𝑃 sub a, 𝑉 sub a, 𝑃 sub b, and 𝑉 sub b. Notice that all the units in the numerator of this fraction also appear in the denominator. All the units therefore cancel out. When we calculate this expression to the nearest percent, we get 329.
So, the mass of the air in the balloon after it’s inflated as a percentage of the mass of the air in the balloon before it’s inflated is 329 percent. This means that there is 3.29 times as much air in the balloon after inflation as before.
