Question

A helium-filled balloon of the type used in long-distance flying contains 1.5 ✕ 107 L of helium. Let us say you fill the balloon with helium on the ground where the pressure is 837 mm Hg and the temperature is 18.4°C. When the balloon ascends to a height of 6 miles where the pressure is only 707. mm Hg and the temperature is -31°C, what volume is occupied by the helium gas? Assume the pressure inside the balloon matches the external pressure.

Answer 1

We can use the combined gas law to solve this problem:

(P1V1/T1) = (P2V2/T2)

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

We are given that the initial pressure is P1 = 837 mm Hg and the initial volume is V1 = 1.5 × 10^7 L. The initial temperature is T1 = 18.4°C, which we need to convert to Kelvin by adding 273.15:

T1 = 18.4°C + 273.15 = 291.55 K

We are also given that the final pressure is P2 = 707 mm Hg and the final temperature is T2 = -31°C, which we need to convert to Kelvin:

T2 = -31°C + 273.15 = 242.15 K

Now we can solve for the final volume, V2:

(P1V1/T1) = (P2V2/T2)

V2 = (P1V1T2) / (P2T1)

V2 = (837 mm Hg * 1.5 × 10^7 L * 242.15 K) / (707 mm Hg * 291.55 K)

V2 = 5.26 × 10^6 L

Therefore, the volume occupied by the helium gas at the higher altitude is 5.26 × 10^6 L.