## force on each wall of a box

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

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 Recognitions: Homework Help Science Advisor There is no need to calculate n, since it is constant, but it is not wrong to do so. Assuming your calculation of the pressure is correct, it looks like you have just neglected to account for the 1atm pressure still on the outside.
 how do you account for that?

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## force on each wall of a box

 Quote by jennypear how do you account for that?
The pressure on the outside is still 1 atm. It is the pressure difference between inside and outside that produces the net force on the walls.

 Quote by OlderDan The pressure on the outside is still 1 atm. It is the pressure difference between inside and outside that produces the net force on the walls.
so that would be
F=deltaP*A
.35*.32^2=.03584

that came up as incorrect

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 Blog Entries: 9 Recognitions: Homework Help Science Advisor Your whole numerical computations need to be redone,as they're wrong...The volume of the box,for example $$V_{\mbox{box}}=(32\cdot 10^{-2}\mbox{m})^{3}=2^{15}\cdot 10^{-6}\mbox{m}^{3}=32.784\cdot 10^{-6}\mbox{m}^{3}\simeq 32.8\cdot 10^{-3}\mbox{m}^{3}$$ Use SI-mKgs units... Daniel.