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