1. Dec 30, 2007

### Rubidium

1. The problem statement, all variables and given/known data
A cubic metal box that has 20-cm-long edges contains air at a pressure of 1.0 atm and a temperature of 300 K. The box is sealed so that the enclosed volume remains constant, and it is heated to a temperature of 400 K. Find the force due to the internal air pressure on each wall of the box.

2. Relevant equations
PV=nRT

3. The attempt at a solution
F=$$\left|$$$$\Delta$$$$\vec{p}$$/$$\Deltat$$$$\right|$$
v=$$\sqrt{3RT/M}$$=$$\sqrt{3(8.314 J/molxK)(300K)/0.029kg/mol)}$$=507.957 m/s
Delta[/tex]$$\vec{p}$$/[tex]=mv^2=(9423.9987g)(507.957m/s)=4.787x10^6
I just need delta t

2. Dec 30, 2007

### Dick

Forget about v. You don't need to count individual particles, PV=nRT will give you the pressure. Pressure is force per unit area.

3. Dec 30, 2007

### Rubidium

How do I use all the information given in the problem to find that?

4. Dec 30, 2007

### Dick

PV=nRT. How many variables in this equation change when the temperature is varied in your problem?

5. Dec 30, 2007

### Rubidium

I think the pressure would change but the volume remains constant in the box. Is the pressure given in the problem negligible or should I be able to use it?

6. Dec 30, 2007

### Dick

Only P and T change, right. But you are given the initial P is 1atm. What's the final P? You certainly shouldn't neglect the pressure change!

7. Dec 30, 2007

### Rubidium

got it...thanks!!!

8. Jul 2, 2008

### MeKnos

I'm having severe trouble on this one:

I know that P1(T1)/T2 = P2

so P2 x (length)^2 x (6 sides) = Internal Pressure Force.

But I'm not getting the right values. Even if I just do P2 x (Length^2) i don't get the right answer.

9. Jul 2, 2008

### Dick

Show your whole solution. We can't really be expected to guess what you did wrong.

10. Jul 2, 2008

### MeKnos

101300 (400) / 300 = P2 = 135093

135093 Pa x (.20^2) x 6 = 32422 N

and the other way 135093 Pa (.2^2) = 5403 N

neither worked

11. Jul 2, 2008