How many moles of air must escape from a 10-m  8.0-m  5.0-m room when the temperature is raised from 0c to 20c? assume the pressure remains unchanged at one atmosphere while the room is heated?

Data:

p = 1 atm
V = 10 m * 8 m * 5 m = 400 m^3 = 400,000 liter

To = 0 + 273.15K = 273.15K
Tf = 20 + 273.15K = 293.15K

No - Nf =?

2) Formula

pV = NRT => N = pV / (RT)

3) solution

No = pV / (RTo)

Nf = pV / (RTf)

=> No - Nf = [pv / R] [ 1 / To - 1 / Tf ]

=> No - Nf = [1atm*400,000liter / 0.0821 atm*liter/K*mol ] [ 1 / 273.15 - 1 / 293.15]

No - Nf = 1216.9 moles ≈ 1217 moles

1217 moles

The correct answer would be 1. precipitation