Question

A weather balloon is inflated with helium to a volume of 825 L at sea level, where the temperature is 30°C and the barometric pressure is 760 torr. The balloon rises to an altitude of 4500 feet where the pressure is 645 torr and the temperature is 6°C. What is the change in volume of the balloon in going from sea level to 4500 feet?

Answer 1

Answer:

The volume increased with 70.1L, from 825 to 895.1 L

Explanation:

Step 1: Data given

Volume at sea level = 825 L

Temperature = 30°C = 303 K

Pressure = 760 torr = 1 atm

Temperature at 4500 feet = 6 °C

Pressure = 645 torr = 645/760 = 0.84868421 atm

Step 2:

(P1 *V1)/T1  = (P2*V2)/T2

⇒ with P1 = the pressure at sea level = 1 atm

⇒ with V1 = the volume at sea level = 825 L

⇒ with T1 = the temperature at sea level = 303 Kelvin

⇒ with P2 = the pressure at 4500 feet = 0.84868421 atm

⇒ with V2 = The volume at 4500 feet = TO BE DETERMINED

⇒ with T2 = the temeprature at 4500 feet = 6 °C = 279 Kelvin

(1*825)/303 = (0.84868421 * V2)/279

V2 = 895.1 L

The volume at 4500 feet is 895.1 L

This means the volume increased with 70.1L