Question

A balloon contains 46.3 L of Helium at 42 oC, and 2.01 atm is released into the air. At a certain altitude, the temperature falls to 21 oC, and the pressure falls to 0.25 atm. What is the volume of the balloon under these conditions

Answer 1

Answer:

347.44 L

Explanation:

Initial Volume, V1 = 46.3 L

Initial Temperature, T1 = 42 oC + 273 = 315 K

Initial Pressure, P1 = 2.01 atm

Final Temperature, T2 = 21 oC + 273 = 294 K

Final Pressure, P2 = 0.25 atm

FInal Volume, V2 = ?

These quantities are related by the general gas equation;

P1V1 / T1  = P2V2 / T2

Inserting the values and solving for V2;

V2 = T2P1V1 / P2T1

V2 = 294 * 2.01 * 46.3 / (0.25 * 315)

V2 = 27360.522 / 78.75 = 347.44 L