- #1
jennypear
- 16
- 0
A cubic metal box with sides 32.0 cm contains air at a pressure of 1.00 atm and a temperature of 294 K. The box is sealed so that the volume is constant and it is heated to a temperature of 396K. Find the force on each wall of the box due to the increased pressure within the box. [The outside air is at 1 atm of pressure.]
I started out PV=nRT
volume of the box = .32^3=0.328m^3*(1L/1x10^-3)=32.8L
n=PV/RT
1atm(32.8L)/(.08207 atm L/mol*K)(294K)=1.36mol
P=nRT/V
[1.36mol(.08207 atm L/mol*K)396K]/32.8L
=1.35 atm
F=P*A
=1.35 atm*.32^2=.138
**that isn't the correct answer
I started out PV=nRT
volume of the box = .32^3=0.328m^3*(1L/1x10^-3)=32.8L
n=PV/RT
1atm(32.8L)/(.08207 atm L/mol*K)(294K)=1.36mol
P=nRT/V
[1.36mol(.08207 atm L/mol*K)396K]/32.8L
=1.35 atm
F=P*A
=1.35 atm*.32^2=.138
**that isn't the correct answer