Question Video: Determining the Mass of an Inflated Balloon Containing an Ideal Gas | Nagwa Question Video: Determining the Mass of an Inflated Balloon Containing an Ideal Gas | Nagwa

Question Video: Determining the Mass of an Inflated Balloon Containing an Ideal Gas Physics • Second Year of Secondary School

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 kPa before it is inflated and is 121 kPa 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 mass of 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.

04:36

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 mass of 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.

In this question, we have a balloon that has been inflated so that its volume increases from 0.012 cubic meters to 0.033 cubic meters. We need to find the balloon’s final mass as a percentage of its initial mass. This is quite an unusual question. Usually when we think about ideal gasses, we have a fixed amount of gas, and we’re thinking about changes to its pressure, volume, or temperature. However, in this question, the amount of gas we’re thinking about changes. As the balloon is inflated, the amount of gas it contains increases, and so does its mass.

Clearing some space on screen, let’s start by looking at the ideal gas law for a fixed amount of gas. This is often written in the form 𝑃 times 𝑉 equals 𝑘 times 𝑇, where 𝑃 is the pressure of the gas, 𝑉 is the volume, and 𝑇 is the temperature. Here, 𝑘 is a constant of proportionality. If we divide both sides of the equation by 𝑇, we can make 𝑘 the subject, 𝑘 equals 𝑃 times 𝑉 divided by 𝑇. When we’re looking at a fixed amount of gas, 𝑘 has a constant value. But it turns out that 𝑘 is actually proportional to the total number of particles in the gas, 𝑁. So, if we’re not looking at a fixed amount of gas, 𝑘 isn’t a constant number. Using the fact that 𝑘 is proportional to the number of gas particles, 𝑁, we can rewrite the ideal gas equation in a more helpful way. 𝑃 times 𝑉 divided by 𝑇 is proportional to 𝑁.

So, how do we link this to the mass of the gas? Well, the total mass of a volume of gas, which we’ll call 𝑚 sub gas, is equal to the number of particles in the gas, 𝑁, multiplied by the mass of each individual gas particle. If we divide both sides of this equation by the mass of an individual gas particle, we find that the number of particles, 𝑁, is equal to the total mass of the gas divided by the mass of each particle. Now let’s substitute this back into our expression for the ideal gas law. This gives us 𝑃 times 𝑉 divided by 𝑇 is proportional to the mass of the gas divided by the mass of the single gas particle. Because the mass of the particle has a constant value, we can absorb it into the proportionality relation. This gives us 𝑃 times 𝑉 divided by 𝑇 is proportional to the mass of the gas.

With this mass relation known, we can now begin to answer this question. We need to find the mass of the balloon after it’s been inflated as a percentage of its initial mass. Clearing space to work, we’ll call the balloon’s initial mass 𝑚 one and its final mass 𝑚 two. As a percentage of the initial mass, the final mass is equal to 𝑚 two divided by 𝑚 one multiplied by 100. Using our expression for the mass of a gas, we can rewrite this in terms of the pressure, volume, and temperature of the air in the balloon. As a percentage of 𝑚 one, 𝑚 two is equal to 𝑃 two times 𝑉 two divided by 𝑇 two all divided by 𝑃 one times 𝑉 one divided by 𝑇 one all multiplied by 100.

We’re told that before the balloon is inflated, the pressure of the air in the balloon is 101 kilopascals and its volume is 0.012 cubic meters. We aren’t told the initial temperature of the air in the balloon, so we’ll just call this 𝑇. After the balloon is inflated, the pressure of the air in the balloon is 121 kilopascals and its volume is 0.033 cubic meters. We’re told that the temperature of the balloon doesn’t change, so 𝑇 two is also just 𝑇.

Now we can substitute these values into our equation. We see that these two 𝑇 terms cancel. So, as a percentage of 𝑚 one, 𝑚 two is equal to 121 kilopascals times 0.033 cubic meters divided by 101 kilopascals times 0.012 cubic meters. Working through the calculation, we get a value of 329.455 percent. When we round this to the nearest percent, we get 329 percent.

So, we have found that the mass of the inflated balloon is equal to 329 percent of its original mass. The answer to this question is then 329 percent.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy