Question

1. choose the best answer.

the helium inside a balloon has a pressure of 1.05 atm, a volume of 5.00 l and the temperature is 20.0 °c. if the balloon is released outdoors, what will the new volume be when the balloon rises to an altitude where the pressure is 0.65 atm and the temperature is 5.00 °c?
8.5 l
7.7 l
32.3 l

Explanation:

Step1: Convert temps to Kelvin

$T_1 = 20.0 + 273.15 = 293.15\ \text{K}$
$T_2 = 5.00 + 273.15 = 278.15\ \text{K}$

Step2: Use combined gas law

The combined gas law is $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$. Rearrange to solve for $V_2$:
$V_2 = \frac{P_1V_1T_2}{P_2T_1}$

Step3: Substitute given values

$V_2 = \frac{1.05\ \text{atm} \times 5.00\ \text{L} \times 278.15\ \text{K}}{0.65\ \text{atm} \times 293.15\ \text{K}}$

Step4: Calculate the result

$V_2 = \frac{1.05 \times 5.00 \times 278.15}{0.65 \times 293.15} \approx 7.7\ \text{L}$

Answer:

7.7 L