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# Ideal gas. watch

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1. okay, from the expression pV = NKT

I'm supposed to show <kinetic energy> = (3/2) KT

p = (1/3)rho <c^2>

(1/3)rho <c^2> V = NKT
(1/3)m <c^2> = NKT
(2/3)(1/2)m<c^2> = NKT

(1/2)m<c^2> = (3/2)NKT

How do I get rid of the N?
2. N is the number of gas molecules i believe, so for a single molecule N=1 and the average kinetic energy will be (3/2)[1]KT.

You might also want to check this:
http://en.wikipedia.org/wiki/Kinetic...kinetic_energy
3. (Original post by andrewlee89)
okay, from the expression pV = NKT

I'm supposed to show <kinetic energy> = (3/2) KT

p = (1/3)rho <c^2>

(1/3)rho <c^2> V = NKT
(1/3)m <c^2> = NKT
(2/3)(1/2)m<c^2> = NKT

(1/2)m<c^2> = (3/2)NKT

How do I get rid of the N?
I'm used to calling K R so in this post I shall refer to K as R, just because I'll get confused if I don't!

First recall that:

Now:

Therefore:

Now, note that:

where is the mass of one of the particles.

Since is constant...

Phew!
4. hey thanks guys
5. very irrelevant but how do people post messages in that kind of text as shown by IChem?
6. Its LaTeX
http://www.thestudentroom.co.uk/wiki/LaTeX

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