So, the force on each wall of the box is 138 N.

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The discussion revolves around calculating the force on the walls of a sealed cubic metal box due to increased internal pressure after heating. Initially, the pressure inside the box was calculated using the ideal gas law, but the external atmospheric pressure of 1 atm was not properly accounted for, leading to incorrect force calculations. The correct approach involves determining the pressure difference between the inside and outside of the box to find the net force. Participants emphasized the importance of using SI units for volume and pressure to ensure accurate results. Ultimately, the calculations need to be redone with proper unit conversions to obtain the correct force on the walls.
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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
 
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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?
 
jennypear said:
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.
 
OlderDan said:
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
 
jennypear said:
so that would be
F=deltaP*A
.35*.32^2=.03584

that came up as incorrect

What have you done about units? What are the units of the answer you are trying to find? How do you express one atmosphere in those units?
 
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.
 
Like Dex said...units!

Since you are working in SI units, your volume must be in meters and your pressure must be in Pa (n/m^2). Only then will your answer be in Newtons.
 

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