Help with the computations

Question

A hot air balloon with an internal pressure of 1.50 atm and a volume of 14.50 L at 20.00C is released. What volume will the balloon occupy at an altitude where the pressure is 0.600 atm and the temperature is –20.00C?

Answer 1

Given:

Initial pressure = P1 = 1.50 atm

Initial volume = V1 = 14.50 L

Initial temperature = T1 = 20.00ºC + 273.15 = 293.15K

Final pressure = P2 = 0.600 atm

Final temperature = T2 = -20.00ºC + 273.15 = 253.15K

Asked for:

Final volume = V2 = ?

Procedure:

Use the Combined Gas Law P1V1/T1 = P2V2/T2

(1.50 atm)(14.50 L) / 293.15K = (0.600 atm)(V2) / 253.15K

V2 = (1.50 atm)(14.50 L)(253.15K) / (293.15K)(0.600 atm)

V2 = final volume = 31.3 L (3 signifiant figures)