# Force on each wall of a box

1. May 4, 2005

### jennypear

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

2. May 4, 2005

### OlderDan

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.

3. May 4, 2005

### jennypear

how do you account for that?

4. May 4, 2005

### 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.

5. May 4, 2005

### jennypear

so that would be
F=deltaP*A
.35*.32^2=.03584

that came up as incorrect

6. May 4, 2005

### OlderDan

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?

7. May 4, 2005

### dextercioby

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.

8. May 4, 2005

### FredGarvin

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.