# Homework Help: Kinetic-Molecular Model of Ideal Gas: vrms/vav

1. Feb 13, 2006

### verd

Hey,

This should be a pretty simple problem to answer... I'm just a bit confused on this, and want to make sure I'm right. It's an easy problem:

Molecules in a gas can only move in the x direction (i.e., $$v_{y}=v_{z}=0)$$. You set up an experiment in which you measure the velocity of a few molecules and the result that you obtain is the following (expressed in m/s):

2, -4, 6, 1, -3, -2, -5, 2, -1, 4, 3, -5

Calculate: a) the average x-component of the velocity $$(v_{x})_{av}$$, b) the average speed $$(v)_{av}$$, and c) the root mean square of the velocity $$v_{rms}$$

For a), the x-component of velocity is literally just the average, right? No absolute values b/c we're not talking about speed here.

For b) because I'm being asked for the average speed, here is where I take the absolute values of all of these and average them together, right?

For c) This is where I'm most confused... Here, wouldn't I just square what I got for b) and then take the square root of it? That seems to make absolutley no sense. Would I then use the formula below?

I noticed something in the book: $$(v_{x}^2)_{av}, (v_{y}^2)_{av}, (v_{z}^2)_{av}$$ must all be equal. Hence: $$(v_{x}^2)_{av} = \displaystyle{\frac{1}{3}}(v^2)_{av}$$

This wouldn't apply for this situation, correct? As the y and the z components are 0, right?

Last edited: Feb 13, 2006
2. Feb 13, 2006

### finchie_88

Its been a while since I did some gas theory, but isn't the rms of the velocities, simply $$v_{rms} = \frac{A}{\sqrt{2}}$$ (A = Ans to part b.)
Don't quote me on that, I'm not sure, but I think its right since rms voltage in electronics is defined in a similar way.

Last edited by a moderator: Feb 13, 2006
3. Feb 13, 2006

### verd

Thanks for responding, heh

I don't quite understand why you get that, could you explain it further?
(I'm not doubting you, I just don't understand)

Thanks again

4. Feb 13, 2006

### verd

5. Feb 13, 2006

### verd

Would this be the same as the root-mean-square in mathematics?

$$\sqrt{\displaystyle{\frac{\sum_{i=1}^{n}x^2}{n}}}$$

So my problem would look like this:
$$\sqrt{\displaystyle{\frac{2^2+(-4)^2+6^2+1^2+(-3)^2+(-2)^2+(-5)^2+2^2+(-1)^2+4^2+3^2+(-5)^2}{12}}}$$

Is this correct?