A weather balloon is inflated to a volume of 27.3 L at a pressure of 747 mmHg and a temperature of 28.5 °C. The balloon rises in the atmosphere to an altitude where the pressure is 385 mmHg and the temperature is -13.4 °C. Assuming the balloon behaves as an ideal gas, what is the new volume of the balloon at the higher altitude?

• A. 30.5 L
• B. 22.1 L
• C. 45.8 L
• D. 18.7 L

Answer: (C)

Verified step by step guidance

1. Identify the initial and final conditions of the gas. The initial volume (V1) is 27.3 L, the initial pressure (P1) is 747 mmHg, and the initial temperature (T1) is 28.5 °C. The final pressure (P2) is 385 mmHg, and the final temperature (T2) is -13.4 °C.
2. Convert the temperatures from Celsius to Kelvin, as gas law calculations require absolute temperatures. Use the formula: T(K) = T(°C) + 273.15. Calculate T1 and T2 in Kelvin.
3. Apply the combined gas law, which relates the initial and final states of a gas: \( \frac{P1 \cdot V1}{T1} = \frac{P2 \cdot V2}{T2} \). This equation allows us to solve for the unknown final volume (V2).
4. Rearrange the combined gas law equation to solve for V2: \( V2 = \frac{P1 \cdot V1 \cdot T2}{P2 \cdot T1} \).
5. Substitute the known values into the equation: P1 = 747 mmHg, V1 = 27.3 L, T1 (in Kelvin), P2 = 385 mmHg, and T2 (in Kelvin). Calculate V2 to find the new volume of the balloon at the higher altitude.