Question

when designing a weather balloon, molecular - level structure is important since it will dictate the functionality of the design of the materials used. a weather balloon is filled with 3.76 x102 l of helium at 1.01 atm at 30.0 °c. it is released, and rises to an altitude where the pressure is 0.806 atm with a temperature of - 1.1 °c. calculate the volume of the weather balloon at the new altitude.
○ 422 l
○ 3.76 × 10²l
○ 17 l
○ 420 l

Explanation:

Step1: Convert temperatures to Kelvin

To convert Celsius to Kelvin, we use the formula \( T(K) = T(^\circ C) + 273.15 \).

For the initial temperature: \( T_1 = 30.0 + 273.15 = 303.15 \, K \)

For the final temperature: \( T_2 = -1.1 + 273.15 = 272.05 \, K \)

Step2: Apply the combined gas law

The combined gas law is \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \), where \( P_1 = 1.01 \, atm \), \( V_1 = 3.76 \times 10^2 \, L \), \( T_1 = 303.15 \, K \), \( P_2 = 0.806 \, atm \), \( T_2 = 272.05 \, K \), and we need to find \( V_2 \).

Rearranging the formula to solve for \( V_2 \): \( V_2 = \frac{P_1V_1T_2}{P_2T_1} \)

Substitute the values:

\( V_2 = \frac{1.01 \times 3.76 \times 10^2 \times 272.05}{0.806 \times 303.15} \)

First, calculate the numerator: \( 1.01 \times 3.76 \times 10^2 \times 272.05 \approx 1.01 \times 376 \times 272.05 \approx 1.01 \times 102290.8 \approx 103313.708 \)

Then, calculate the denominator: \( 0.806 \times 303.15 \approx 244.34 \)

Now, divide the numerator by the denominator: \( V_2 \approx \frac{103313.708}{244.34} \approx 422.8 \, L \approx 422 \, L \) (rounded to a reasonable number of significant figures)

Answer:

422 L