# Homework Help: Thermodynamics / micro/macro connection problem

1. Nov 20, 2009

### greenskyy

1. The problem statement, all variables and given/known data
A 10 cm * 10 cm * 10 cm box contains .01 mol of nitrogen at 20 degrees C.
What is the rate of collisions (collisions/s) on one wall of the box?

2. Relevant equations
$$\frac{N_{coll}}{t}=\frac{NAv_{x}}{2V}$$

$$v_{rms}=\sqrt{\frac{3k_{B}T}{m}}$$

3. The attempt at a solution
I'm really banging my head on the desk with this one. So far, I've calculated the following information.

T = 20 degrees C = 293 degrees K

N = .01 mol * avogadros constant = $$6.02*10^{21}$$ particles.

Nitrogen is diatomic, so the molecular mass is 28u.

m = molecular mass / avogadros constant = $$4.65*10^{-26} kg$$

$$v_{rms}=\sqrt{\frac{3k_{B}T}{m}} = 511 m/s$$

Plugging this into the first equation given

$$\frac{N_{coll}}{t}=\frac{NAv_{x}}{2V}$$ = $$\frac{Nv_{x}}{2*.1} = 1.54*10^{25}$$

This answer is incorrect according to masteringphysics, and I have no ideas from here. If anyone could offer some suggestions I would be very grateful.

2. Nov 20, 2009

### greenskyy

Thanks for the help everyone... =\

I figured it out after talking to some other students today. I was lacking one equation:

$$(v_{x}^{2})_{avg}=\frac{v_{rms}^{2}}{3}$$

Good luck on this problem for whoever needs this :D

3. Nov 20, 2009

### Gotejjeken

I happen to also be working on this problem. In your last equation it is possible to get:

$$(v_{x})_{avg}=sqrt(\frac{v_{rms}^{2}}{3})$$

Do you plug this in to this equation:

$$\frac{N_{coll}}{t}=\frac{NAv_{x}}{2V}$$

for $$(v_{x})$$? I tried doing this and came up with the wrong answer. I assume the number of collisions per second would be an average of all of the molecules hitting the wall, however Mastering Physics would have me believe that I am wrong in this regard.

4. Nov 20, 2009

### greenskyy

Yes, that's what you do. The mastering physics system has the incorrect answer stored as well, as if the number of moles were .1 rather than .01. The "true" answer should be to the magnitude of 10^24, while the one accepted by mastering physics is 10^25.

5. Nov 20, 2009

### Gotejjeken

Alright, thanks. That problem was giving me many problems as well, and I am glad that forums such as these exist!