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Ideal gas. watch

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    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?
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    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
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    (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:

    n=\frac{m}{M}

    \rho = \frac{m}{V}

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

    Now:

    p=\frac{1}{3} \rho &lt;c^2&gt;

    p=\frac{1}{3} \frac{m}{V} &lt;c^2&gt;

    pV=\frac{1}{3} m &lt;c^2&gt; = nRT

    Therefore:

    &lt;c^2&gt; = 3\frac{n}{M} RT = \frac{3RT}{M}

    Now, note that:

    N_A = \frac{M}{m_p} where  m_p is the mass of one of the particles.

    &lt;E_k&gt; = \frac{1}{2} m_p &lt;c^2&gt;

    &lt;E_k&gt; = \frac{1}{2} m_p \frac{3RT}{M} = \frac{3}{2} \frac{R}{N_A}T

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

    &lt;E_k&gt; = \frac{3}{2} kT

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