A weather balloon is filled with helium to a volume of 12.0 L at sea level, where the pressure is 1.00 atm and the temperature is 20.0°C. The balloon rises to an altitude where the pressure is 0.60 atm and the temperature is -10.0°C. Assuming the balloon is flexible and the amount of helium remains constant, what is the new volume of the balloon?
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Correct answer
A. 18.0 L
Principle or equation
Combined gas law: P1V1/T1 = P2V2/T2, with temperatures in Kelvin.
Why this answer is correct
Convert temperatures to Kelvin: T1 = 20.0 + 273.15 = 293.15 K, T2 = -10.0 + 273.15 = 263.15 K. Use P1V1/T1 = P2V2/T2. Solve for V2 = P1V1T2 / (P2T1) = (1.00 atm × 12.0 L × 263.15 K) / (0.60 atm × 293.15 K) ≈ 17.95 L, which rounds to 18.0 L.
Example
If a gas at 1.0 atm and 300 K occupies 2.0 L, and it is changed to 2.0 atm and 450 K, then V2 = (1.0 × 2.0 × 450) / (2.0 × 300) = 1.5 L.
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