A weather balloon has a volume of 0.50 m³ at ground level where atmospheric pressure is 100 kPa. The balloon rises to a height where the atmospheric pressur...
Question 1 Report
A weather balloon has a volume of 0.50 m³ at ground level where atmospheric pressure is 100 kPa. The balloon rises to a height where the atmospheric pressure is 40 kPa. The temperature of the gas inside remains constant and the balloon can expand freely.
What is the new volume of the balloon?
Answer Details
Boyle's law applies to a fixed mass of gas at constant temperature: \( P_1 V_1 = P_2 V_2 \). As the balloon rises, the atmospheric pressure decreases, allowing the gas inside to expand.
\[ 100 \times 0.50 = 40 \times V_2 \]
\[ V_2 = \frac{100 \times 0.50}{40} = \frac{50}{40} = 1.25 \text{ m}^3 \]
The pressure drops to \( \frac{40}{100} = \frac{2}{5} \) of its original value, so the volume increases to \( \frac{5}{2} = 2.5 \) times the original. \( 0.50 \times 2.5 = 1.25 \text{ m}^3 \). Selecting 0.20 m\(^3\) would result from multiplying pressure ratio by volume instead of dividing. Selecting 2.50 m\(^3\) would come from applying the factor of 2.5 to 1.0 rather than 0.50, or confusing the multiplier.
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