Ideal gas law: A weather balloon contains 12.0 m³ of hydrogen gas when the balloon is released from a location at which the temperature is 22.0°C and the pressure is 101 kPa. The balloon rises to a location where the temperature is -30.0°C and the pressure is 20.0 kPa. If the balloon is free to expand so that the pressure of the gas inside is equal to the ambient pressure, what is the new volume of the balloon? Assume that in both cases the hydrogen gas is in thermal equilibrium with the outside air. A) 14.0 m³ B) 2.38 m³ C) 49.9 m³ D) 82.6 m³ E) 4.16 m³

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Quizplus · Accepted Answer

To solve this problem, we use the ideal gas law in the form of $P_1V_1/T_1 = P_2V_2/T_2$, where $P$ is pressure, $V$ is volume, and $T$ is temperature in Kelvin. We need to convert temperatures from Celsius to Kelvin by adding 273.15. Thus, $T_1 = 22.0 + 273.15 = 295.15$ K and $T_2 = -30.0 + 273.15 = 243.15$ K. Given $P_1 = 101$ kPa, $V_1 = 12.0$ m$^3$, and $P_2 = 20.0$ kPa, we solve for $V_2$ as follows:$$V_2 = \frac{P_1V_1T_2}{P_2T_1} = \frac{101 \times 12.0 \times 243.15}{20.0 \times 295.15} \approx 49.9 \text{ m}^3.$$Thus, the new volume of the balloon is approximately 49.9 m$^3$, making choice C correct.

Ideal Gas Law: a Weather Balloon Contains 12