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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?

Respuesta :

Answer:

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