Question

a gas - filled balloon with a volume of 2.20 l at 1.20 atm and 20.0 °c if allowed to rise to the stratosphere (about 30 km above the surface of the earth), where the temperature and pressure are - 23.0 °c and 3.00×10⁻³ atm, respectively. what is the final volume of the balloon, in l?

Explanation:

Step1: Convert temperatures to Kelvin

Initial temperature \( T_1 = 20.0^\circ C + 273.15 = 293.15 \, K \)
Final temperature \( T_2 = -23.0^\circ C + 273.15 = 250.15 \, K \)

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

We need to solve for \( V_2 \), so rearrange the formula: \( V_2 = \frac{P_1V_1T_2}{P_2T_1} \)
Given \( P_1 = 1.20 \, atm \), \( V_1 = 2.20 \, L \), \( P_2 = 3.00 \times 10^{-3} \, atm \)
Substitute the values: \( V_2 = \frac{1.20 \, atm \times 2.20 \, L \times 250.15 \, K}{3.00 \times 10^{-3} \, atm \times 293.15 \, K} \)

Step3: Calculate the numerator and denominator

Numerator: \( 1.20 \times 2.20 \times 250.15 = 1.20 \times 550.33 = 660.396 \)
Denominator: \( 3.00 \times 10^{-3} \times 293.15 = 0.87945 \)

Step4: Divide numerator by denominator

\( V_2 = \frac{660.396}{0.87945} \approx 751 \, L \)

Answer:

\( 751 \, L \) (rounded to three significant figures)