A balloon contains 0.20 m^3 of helium at a pressure of 1.0 × 10^5 Pa and a temperature of 300 K. The balloon rises to an altitude where the pressure is 0.80 × 10^5 Pa and the temperature is 250 K. Assuming the amount of helium does not change, what is the new volume of the balloon?
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Correct answer
B. 0.21 m^3
Principle or equation
The combined gas law for a fixed amount of ideal gas: P1V1/T1 = P2V2/T2.
Why this answer is correct
Given P1=1.0×10^5 Pa, V1=0.20 m^3, T1=300 K, P2=0.80×10^5 Pa, T2=250 K. Solving for V2: V2 = P1V1T2/(T1P2) = (1.0×10^5 × 0.20 × 250)/(300 × 0.80×10^5) = (0.20 × 250)/(300 × 0.80) = 50/240 = 0.2083 m^3 ≈ 0.21 m^3.
Example
If a gas at 1 atm, 300 K, and 2 L is cooled to 200 K and pressure drops to 0.5 atm, the new volume is (1×2×200)/(300×0.5)=2.67 L.
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