The Student Room Group
Reply 1
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_theory#Temperature_and_kinetic_energy
Reply 2
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:

n=mMn=\frac{m}{M}

ρ=mV\rho = \frac{m}{V}

N=n×NAN= n \times N_A where NAN_A is the Avogadro constant.

Now:

p=13ρ<c2>p=\frac{1}{3} \rho <c^2>

p=13mV<c2>p=\frac{1}{3} \frac{m}{V} <c^2>

pV=13m<c2>=nRTpV=\frac{1}{3} m <c^2> = nRT

Therefore:

<c2>=3nMRT=3RTM<c^2> = 3\frac{n}{M} RT = \frac{3RT}{M}

Now, note that:

NA=MmpN_A = \frac{M}{m_p} where mp m_p is the mass of one of the particles.

<Ek>=12mp<c2><E_k> = \frac{1}{2} m_p <c^2>

<Ek>=12mp3RTM=32RNAT<E_k> = \frac{1}{2} m_p \frac{3RT}{M} = \frac{3}{2} \frac{R}{N_A}T

Since RNA\frac{R}{N_A} is constant...

<Ek>=32kT<E_k> = \frac{3}{2} kT

Phew!:smile:
Reply 3
hey thanks guys :smile:
Reply 4
very irrelevant but how do people post messages in that kind of text as shown by IChem?
Reply 5
Its LaTeX
Look here for more info:
http://www.thestudentroom.co.uk/wiki/LaTeX

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