Question

question 49
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 if 0.0741 atm and the temperature is - 2.81 °c. what is the volume of the weather balloon at the
○ 3.55 l
○ 762 l
○ 6.56×10⁶ l
○ 641 l

Explanation:

Step1: Convert units to consistent ones

First, convert pressure at ground level 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 temperatures to Kelvin: \( T_1 = 21.6 + 273.15 = 294.75\ \text{K} \), \( T_2 = -2.81 + 273.15 = 270.34\ \text{K} \).
Initial volume \( V_1 = 52.0\ \text{L} \), \( P_2 = 0.0741\ \text{atm} \).

Step2: Apply Combined Gas Law

The Combined Gas Law is \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \). Solve for \( V_2 \):
\( V_2 = \frac{P_1V_1T_2}{P_2T_1} \)
Substitute 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 result

First, calculate numerator: \( 0.996 \times 52.0 \times 270.34 \approx 0.996 \times 14057.68 \approx 14001.45 \)
Denominator: \( 0.0741 \times 294.75 \approx 21.84 \)
Then, \( V_2 \approx \frac{14001.45}{21.84} \approx 641\ \text{L} \)

Answer:

641 L (corresponding to the option "641 L")