Video Transcript
A balloon has a volume of 0.015 cubic meters at sea level, where the air pressure is 101 kilopascals and the temperature is 300 kelvin. Find the volume of this balloon at a point 1.4 kilometers above sea level, where the temperature is 245 kelvin. Model the atmosphere as having a uniform density of 1.225 kilograms per cubic meter. Answer to four decimal places.
Let’s say that this line represents sea level and that our balloon is initially here with a given volume and air pressure and temperature. Over time, though, the balloon moves up to an elevation of 1.4 kilometers. We want to solve for the volume of the balloon at this point. To differentiate this volume from the volume of the balloon at sea level, let’s call this volume 𝑉 two. As we look to solve for this volume, we’re going to treat the air in the balloon as an ideal gas. That means it follows the ideal gas law. This law says that the pressure of a gas times its volume is equal to the number of moles of that gas times a constant 𝑅, called the gas constant, multiplied by the gas’s temperature.
In the case of our balloon, the number of moles of gas in the balloon is constant, and 𝑅, the gas constant, is always the same. And therefore, 𝑛 times 𝑅 in our case is a constant value. If we divide both sides of the ideal gas law by the temperature 𝑇, then that factor cancels out on the right. What we find then is that the gas’s pressure times its volume divided by its temperature equals 𝑛 times 𝑅, which as we’ve seen is constant in our case.
This means if we were to take the pressure of the gas at one moment, call that pressure 𝑃 one, and multiply it by the gas’s volume in that same moment, 𝑉 one, and divide all this by the gas’s temperature at that time, 𝑇 one, then this would equal the pressure of the gas at some other moment, 𝑃 two, times its volume at that other moment, 𝑉 two, divided by its temperature at that second moment, 𝑇 two.
In the case of our balloon, we’ll say that this initial moment in time is when the balloon is at sea level. So, we’ll call the pressure, volume, and temperature of the air in the balloon at that moment 𝑃 one, 𝑉 one, and 𝑇 one. Similarly, we’ll say that this moment when the balloon is 1.4 kilometers up in elevation is the second moment where it has pressure 𝑃 two, volume 𝑉 two, and temperature 𝑇 two. This means that in this equation here, it’s the variable 𝑉 two that we want to solve for. To help us do that, let’s multiply both sides of the equation by 𝑇 two divided by 𝑃 two. This means that on the right-hand side of our expression, that temperature as well as that pressure cancel out.
Rearranging the expression that results, we find that 𝑉 two is equal to 𝑉 one times the pressure ratio 𝑃 one to 𝑃 two multiplied by the temperature ratio 𝑇 two to 𝑇 one. In our problem statement, we’re given values for a number of these variables. We’re told, for example, that the balloon’s volume at sea level — what we’ve called 𝑉 one — is 0.015 cubic meters. The pressure of the air in the balloon there is 101 kilopascals, and this air’s temperature is 300 kelvin. We’re also told that the temperature of the air in the balloon and its elevation of 1.4 kilometers is 245 kelvin; that’s 𝑇 two, which means that we now have values for 𝑉 one, 𝑃 one, 𝑇 one, and 𝑇 two. The one we’re missing though in order to be able to solve for 𝑉 two is that second pressure 𝑃 two.
We can note though that we’re treating the air in the atmosphere as having a uniform density of 1.225 kilograms per cubic meter. We can recall that the pressure in a fluid, for example, air in the atmosphere, is equal to the density of that fluid times the acceleration due to gravity multiplied by the fluid’s height. We need to be careful here though because we might think that the pressure at our elevation of 1.4 kilometers, what we’ve called 𝑃 two, can be calculated by using this equation where the height ℎ is our elevation of 1.4 kilometers. But actually, this span of atmosphere is a span that doesn’t contribute to the atmospheric pressure at this point. Rather, it’s all the air in the atmosphere above this point all the way up to the top of the atmosphere that creates that pressure.
Let’s say that this column of air has a height ℎ two. If we knew this height, we could use this relationship here to solve for the pressure 𝑃 two in our equation for 𝑉 two. To begin doing that, let’s clear some space at the top of our screen. The pressure of the air in the balloon 1.4 kilometers above sea level, we’ve called this 𝑃 two, equals the density of air in the atmosphere times the acceleration due to gravity times ℎ two. We’ve been given the density 𝜌 of air in the atmosphere, and we know that the acceleration due to gravity 𝑔 is 9.8 meters per second squared. We don’t know ℎ two, but we can work towards solving it by considering not the pressure 𝑃 two, but rather the pressure 𝑃 one at sea level. That pressure we can see is due to air of a height ℎ two plus 1.4 kilometers.
If we write out an equation then for the pressure 𝑃 one, that would be equal to 𝜌 times 𝑔 times the quantity ℎ two plus 1.4 kilometers. In this equation, we once again know 𝜌 and 𝑔 but don’t know ℎ two. But the difference now is that we do know 𝑃 one; recall that it’s given to us in our problem statement. What we’ll do then is rearrange this equation so that ℎ two is the subject. To do that, we can first divide both sides by 𝜌 times 𝑔, canceling both of those factors on the right-hand side. Then as a next step, we subtract 1.4 kilometers from both sides so that on the right 1.4 kilometers minus 1.4 kilometers is zero. We now have an expression for ℎ two where all the values in the other side of the equation are known.
As we go to calculate ℎ two, we can substitute in the values for 𝑃 one, 𝜌, and 𝑔. But before we enter this expression on our calculator, we’ll want to make two changes in the units. First, we’ll want to convert kilopascals to units of pascals, and second, we’ll convert units of kilometers to meters. To make these conversions, we can recall that 1000 pascals equals a kilopascal and 1000 meters is equal to a kilometer. In both cases of conversion then, we’ll multiply the values by 1000. 101 times 1000 is 101000, and 1.4 times 1000 is 1400. Because a pascal is equal to a newton per meter squared and therefore expressible all in SI base units, we’re now ready to calculate our height ℎ two.
ℎ two comes out to approximately 7013.2 meters. This then is the height that we can use in our equation to solve for the pressure 𝑃 two. Clearing space to do this, we substitute in the known values for 𝜌, 𝑔, and ℎ two. For our pressure 𝑃 two, we calculate a value of about 84190 pascals, which is equal to 84.190 kilopascals.
Now that we have a value for 𝑃 two, we can move ahead towards solving for the volume 𝑉 two. It’s equal to 𝑉 one, 0.015 cubic meters, multiplied by 𝑃 one divided by 𝑃 two multiplied by 𝑇 two divided by 𝑇 one. Notice that on the right-hand side, we have units of kilopascals in the numerator and denominator. Therefore, these units cancel out, so do the units of kelvin for temperature. Our final answer will have units of cubic meters. 𝑉 two ends up being a long decimal number. But if we round it to four decimal places, we get 0.0147 cubic meters. This is the volume of the balloon, 1.4 kilometers above sea level.
