Question

a gas - filled weather balloon with a volume of 52.0 l is held at ground level, where the atmospheric pressure is 757 mm hg and the temperature is 21.6 °c. the balloon is released and rises to an altitude where the pressure is 0.0741 atm and the temperature is - 28.1 °c. what is the volume of the weather balloon at the higher altitude?
options:
641 l
762 l
6.56×10⁴ l
355 l

Explanation:

Step1: Convert pressure to atm and temperature to Kelvin

First, convert the initial pressure from mmHg to atm. We know that 1 atm = 760 mmHg. So, \( P_1 = \frac{757\ \text{mmHg}}{760\ \text{mmHg/atm}} \approx 0.996\ \text{atm} \).
Convert initial temperature \( T_1 = 21.6^\circ\text{C} + 273.15 = 294.75\ \text{K} \).
Convert final temperature \( T_2 = -2.81^\circ\text{C} + 273.15 = 270.34\ \text{K} \). Final pressure \( P_2 = 0.0741\ \text{atm} \), initial volume \( V_1 = 52.0\ \text{L} \).

Step2: Apply the combined gas law \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)

Rearrange to solve for \( V_2 \): \( V_2 = \frac{P_1V_1T_2}{P_2T_1} \).
Substitute the values:
\( V_2 = \frac{0.996\ \text{atm} \times 52.0\ \text{L} \times 270.34\ \text{K}}{0.0741\ \text{atm} \times 294.75\ \text{K}} \).

Step3: Calculate the numerator and denominator

Numerator: \( 0.996 \times 52.0 \times 270.34 \approx 0.996 \times 14057.68 \approx 14001.45 \).
Denominator: \( 0.0741 \times 294.75 \approx 21.84 \).

Step4: Divide to find \( V_2 \)

\( V_2 \approx \frac{14001.45}{21.84} \approx 641\ \text{L} \).

Answer:

641 L (corresponding to the first option: 641 L)