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Chemical Potential

  1. Oct 20, 2015 #1
    1. The problem statement, all variables and given/known data
    I just have a question about chemical potential for ideal monatomic gas. I see that by definition [tex] \mu = \frac{\partial U}{\partial N} [/tex]

    2. Relevant equations
    [tex] \mu = \frac{\partial U}{\partial N} [/tex]

    3. The attempt at a solution
    I was wondering why it is wrong to use [tex] U=3/2NkT [/tex] take the partial with respect to N and get [tex] \mu = 3/2kT [/tex].

    I know this isn't right, but what is exactly wrong with it?

    The correct equation for the chemical potential of a monatomic ideal gas by the way is
    [tex] \mu = -kT \ln({\frac{V}{N}(\frac{4 \pi mU}{3h^2}})^\frac{3}{2}) [/tex]
     
    Last edited: Oct 20, 2015
  2. jcsd
  3. Oct 20, 2015 #2
    I'm thinking there is something to do with the variables that need to be held fixed, they are entropy and volume. Does 3/2NkT somehow not keep these values fixed maybe?
     
  4. Oct 21, 2015 #3

    DrClaude

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    Staff: Mentor

    Indeed. You need to start from the thermodynamic identity
    $$
    dU = T dS - PdV + \mu dN
    $$
    from which you will see that
    $$
    \mu = \left( \frac{\partial U}{\partial N} \right)_{S,V}
    $$
     
  5. Oct 21, 2015 #4
    Yes, I most definitely see that.

    My biggest issue right now is why is it not correct to use [tex] U=\frac {f}{2}NkT [/tex] take that partial with respect to N and obtain an answer? My thought is that this equation doesn't hold S and V constant inherently, but how would I see that? Just that fact that T is in the equation maybe?
     
  6. Oct 21, 2015 #5

    DrClaude

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    Staff: Mentor

    Yes, you would have a term in ##\partial T / \partial N## in there.
     
  7. Oct 21, 2015 #6
    Thank you very much. Little simple things like this keep holding me up a little. Hopefully my math skills tighten up in these last few years of my undergraduate studies.
     
  8. Oct 21, 2015 #7

    DrClaude

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    Staff: Mentor

    You can take comfort in the fact that you had the right intuition as to why simply taking the derivative of U didn't work. The more you use math, the better your skills.
     
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