On a hot day, a party balloon is filled with helium. The following day is much colder, but the air pressure is the same. Assuming that no helium escapes from the balloon in that time, will the balloon have a larger volume than, a smaller volume than, or the same volume as that it had on the previous day?
Okay, so in this question, we are dealing with a party balloon. We’ve been told that this party balloon is filled with helium This can be our molecules of helium gas inside our balloon. And we’ve been told that the day on which the balloon was filled with helium was a hot day. Now, as well as this, we’ve been told that the next day the temperature is much lower because we’ve been told that the day after is a cold today. And we have to find out whether the balloon on this cold day has a larger volume, a smaller volume, or the same volume as the day before. Extra information we’ve been given includes the fact that the air pressure is the same on both days.
Now, the air pressure is simply the pressure exerted on the balloon from the air outside the balloon. And in order for a balloon to remain at a constant size or in other words not increase or decrease in volume over time, the pressure exerted on the balloon from the outside, which in other words, is the air pressure has to be the same as the pressure exerted on the inside of the balloon by the gas inside the balloon, which is helium. And so, what we’ve just realized is that the pressure exerted from the outside — the air pressure — is the same as the pressure exerted from the inside — the pressure of the helium inside the balloon.
And since on the hot day and the cold day, the air pressure is the same, this must mean that the pressure inside the balloon on the cold day is also the same. Because on the cold day, regardless of whether the balloon is smaller or the same size as the day before or even larger, once the balloon has reached this new volume because of the change in temperature overnight, we’ve not been told that the balloon continues to expand or shrink or whatever it’s going to be doing. We’re assuming that on the cold day, the balloon stays the same size after it has changed its volume because of the change in temperature, if indeed it does change in volume due to the change in temperature.
And hence, we know that the pressure of the helium gas inside the balloon on the cold day is the same as the pressure inside the balloon due to the helium gas on the hot day. In other words, if we say that the pressure exerted by the helium gas inside the balloon on the hot day is 𝑃, then we can say that the pressure exerted by the helium gas on the cold day is also 𝑃. As well as this, to make life easier for ourselves, let’s say that the temperature of the balloon on the hot day was 𝑇 one. And let’s also say that the temperature of the balloon on the cold day was 𝑇 two. Now, we know that because the first day was the hot day, the temperature 𝑇 one is greater than the temperature 𝑇 two. This is what makes 𝑇 one the hotter temperature and 𝑇 two the colder temperature. And we’ve been asked to find how the volume of the balloon changes.
So if we say that the volume of the balloon on the first day — on the hot day — is 𝑉 subscript one and the volume on the second day or the cold day is 𝑉 subscript two, then to see how this changes, we need to recall something known as Charles’s law. Charles’s law tells us that the volume of an ideal gas at constant pressure is directly proportional to its temperature. In other words, we can say using symbols that for an ideal gas, the volume 𝑉 is directly proportional to the temperature 𝑇, but only if the pressure 𝑃 is constant.
Now, we can apply this law to our scenario here because the pressure is indeed constant. We’ve seen that the pressure before on the hot day is 𝑃 and the pressure afterward of the helium gas inside the balloon on the cold day is 𝑃 as well. And additionally, we can assume that the helium guess inside the balloon is indeed an ideal gas because the behaviour of helium gas at atmospheric pressures and average air temperatures is very close to that of an ideal gas. So, we can confidently say that for the gas inside the balloon, the volume of the gas, and therefore the volume of the balloon, is directly proportional to the temperature of the gas, which is also the air temperature. This means that if the temperature of the gas inside the balloon changes, the volume changes in the same way. And what we mean by this is that if the temperature of the gas inside the balloon increases, the volume of the balloon increases proportionally as well.
And in our case, we’ve seen that the temperature decreases from the first day to the second. The first day is hotter and the second day is colder. And if the temperature decreases, then the volume of the gas decreases as well. And therefore, from Charles’s law, we can gather that the volume 𝑉 one of the balloon on the hot day is greater than the volume 𝑉 two of the balloon on the cold day. In other words, on the cold day, the balloon has a smaller volume than on the hot day.
And note that the additional sentence in our question statement telling us that no helium escapes from the balloon was helpful because it meant that we were dealing with the same amount of gas in both cases, both on the hot day and the cold day. This is important because if the amount of gas had changed, if some had escaped, or if some extra gas had been added to our balloon, then Charles’s law would no longer hold. Because Charles’s law not only applies to a gas of constant pressure, but it also implicitly assumes — that is it makes an assumption without stating it — that the amount of gas does not change. It talks about the same gas throughout the process.
So anyway, our final answer is that the balloon will have a smaller volume on the cold day than it did on the hot day.