A weather balloon is inflated to a volume of 28.5 L at a pressure of 748 mmHg and a temperature of 28.0 °C. The balloon rises in the atmosphere to an altitude of approximately 25,000 ft, where the pressure is 385 mmHg and the temperature is -15.0 °C. Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

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Accepted Answer

All right. Hi, everyone. So this question says that a hot air balloon is inflated to a volume of 34.5 L at a pressure of 678 mg of mercury and a temperature of 32.0 °C. The balloon rises in the atmosphere to an altitude of approximately 30,000 ft where the pressure is 425 millimeters of mercury and the temperature is negative 10.0 °C. Assuming the b the balloon can't freely expand, calculate the volume of the balloon at this altitude. And here we have four different answer choices, proposing different values for the volume. So we know for a fact that the balloon is not going to leak. So this means that the A mt of helium inside of the balloon is going to be fixed, that part is not going to change. So when we recall the gas constant or excuse me, the ideal gas law, that's PV equals N RT R is the gas constant and N which is the number of moles is also going to stay the same. So we can rearrange this expression to physically separate those constants. This means that NR is equal to PV divided by T and so for two different states of an ideal gas, we can say that P one V one divided by T one is equal to P two V two divided by T two. Now, because we're provided the information about the balloon before it rises in the atmosphere, we're given P one V one and T one. So in this case, P one is equal to 678 millimeters of mercury V one is equal to 34.5 L and T one is equal to 32 °C. But in this case, let's go ahead and convert our temperature into Kelvin. And so we'll do that by adding 273 to the temperature in degrees Celsius. And so T one equals 305 Kelvin. So because we're being told to solve for the volume of the balloon after it rises, what we're solving for is V two. And so we're given P two and T two as well because the pressure at 30,000 ft in altitude, that's P two is 425 millimeters of mercury and T two is equal to negative 10.0 °C, which we can once again convert into Kelvin by adding 273 to that number. Equaling 263 Kelvin. So when rearranging our expression for the two states of an ideal gas, we can solve four V two, right? So V two is equal to P one V one divided by T one mortified buy T two divided by P two. And so from here, we can go ahead and plug in the information provided previously, right? So for this first term, P one is 678 millimeters of mercury multiplied by T one excuse me V one, which is 34.5 L and divided by T one which is 305 K. So for a second term T two is 263. Kelvin divided by P two which is 425 millimeters of mercury. And notice how when considering our multiplication, right, our product, all of our units cancel out with the exception of leaders. So our answer for the volume after evaluating this expression is equal to 47.5 L and this matches option C in the multiple choice. So there you have it. And with that being said, thank you so very much for watching. And I hope you found this helpful.

Verified Solution

Key Concepts

Ideal Gas Law

Charles's Law

Boyle's Law