Question

A sealed balloon filled with air that has a volume of 6 cubic inches at 99 feet will have a volume of _____ cubic inches at 33 feet.

Answer 1

Let's assume that the gas is behaving ideally. Hence, we can use the ideal gas equation: PV = nRT. We only know the value of the volume. Since it was not mentioned, let's assume that the temperature at those altitude do not have a significant difference. So, we can assume temperature, as well as the number of moles is constant, because it is a closed system. The other parameter, pressure, can be calculated in terms of height:

P = ρgh, where ρ is the air density equal to 0.0765 lb/ft³, g is the acceleration due to gravity equal to 32.2 ft/s², and h is the height in feet. Let's solve for the pressure, P₁.

P₁ = ( 0.0765 lb/ft³)(32.2 ft/s²)(99 feet) = 243.8667 lb/ft-s²
For consistency, let's convert V₁ to ft³ using the conversion 12 inches=1 foot.
V₁ = (6 in³)(1 foot/12 inches)³ = 0.003472 ft³

Then, let's use this to find nRT.

P₁V₁ = nRT
nRT = (243.8667 lb/ft-s²)(0.003472 ft³)
nRT = 0.8467 lb-ft²/s²

Now, let's do the same procedure for P₂:
P₂ = ( 0.0765 lb/ft³)(32.2 ft/s²)(33 feet) = 81.2889 lb/ft-s²
Then, we use the ideal gas equation knowing that nRT=0.8467

P₂V₂ = nRT = 0.8467 lb-ft²/s²
81.2889 lb/ft-s²(V₂) = 0.8467 lb-ft²/s²
V₂ = 0.010416 ft³

Finally, let's convert this to in³:

V₂ = 0.010416 ft³ (12 in/1ft)³
V₂ = 18 ft³