A certain flexible weather balloon contains helium gas at a volume of 855 L. Initially, the balloon is at sea level where the temperature is 25°C and the barometric pressure is 728 torr. The balloon then rises to an altitude of 6000 ft, where the pressure is 605 torr and the temperature is 15°C. What is the change in volume of the balloon (in L) as it ascends from sea level to 6000 ft?

The volume at 6000 ft is 994.3 L.

The change in volume is 139.3 L

Explanation:

Step 1: Data given

Volume of the gas = 855L

Temperature = 25.0 °C = 273 + 25 = 298 Kelvin

barometric pressure = 728 torr

When the balloon rises 6000 ft, the pressure is 605 torr

temperature = 15.0°C = 273 + 15 = 288 Kelvin

Step 2: Calculate the new volume

(p1*V1) / T1 = (P2*V2) / T2

⇒ with P1 = the initial pressure of gas in the balloon = 728 torr

⇒ with V1 = the initial volume = 855 L

⇒ with T1 = the initial temperature = 298 Kelvin

⇒ with P2 = the pressure after rising to 6000 ft = 605 torr

⇒ with V2 = the volume of the gas at 6000 ft

⇒ with T2 = the temperature at 6000 ft = 288 Kelvin

V2 = (P1 * T2 * V1) / (P2*T1)

V2 = (728*288*855) / (605*298)

V2 = 994.3 L

The change in volume = 994.3 - 855 = 139.3 L

The volume at 6000 ft is 994.3 L.

The change in volume is 139.3 L