Question

when you heat the air inside a hot air balloon, it rises. why does this occur?
a. the mass of cold air displaced by the balloon is less than the mass of hot air inside the balloon.
b. the hot air inside the balloon becomes denser than the air outside the balloon.
c. the volume of the balloon decreases.
d. the hot air inside the balloon becomes less dense than the air outside the balloon.
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Explanation:

Step1: Analyze option A

The mass of cold air displaced by the balloon is not less than the mass of hot air inside the balloon. According to Archimedes' principle, the buoyant force is related to the density and volume of the displaced fluid (air in this case), not directly in this mass - comparison way as described in option A.

Step2: Analyze option B

When air is heated, its density decreases. Hot air inside the balloon does not become denser than the outside air. Density is inversely related to temperature (for an ideal gas, \(PV = nRT\), and density \(
ho=\frac{m}{V}\), \(m = nM\), so \(
ho=\frac{PM}{RT}\), as \(T\) increases, \(
ho\) decreases).

Step3: Analyze option C

When air is heated, for a balloon (assuming pressure inside and outside is approximately the same, \(P_{in}\approx P_{out}\)), using the ideal gas law \(PV = nRT\), as \(T\) increases, \(V\) increases (if \(n\) is constant). So the volume of the balloon does not decrease.

Step4: Analyze option D

Using the density formula \(
ho=\frac{m}{V}\), and the ideal gas law \(PV = nRT\) (\(n=\frac{m}{M}\)), so \(
ho=\frac{PM}{RT}\). When the air inside the balloon is heated (\(T\) increases), \(
ho\) (density of hot - air) decreases. Since the buoyant force \(F_b=
ho_{outside}gV\) and the weight of the balloon \(W=
ho_{inside}gV + W_{balloon\ material}\), when \(
ho_{inside}<
ho_{outside}\), the balloon rises.

Answer:

D. The hot air inside the balloon becomes less dense than the air outside the balloon.