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Partial Derivatives for an Ideal Gas

  1. Oct 3, 2013 #1
    The question is:
    a) Find explicit expressions for an ideal gas for the partial derivatives:
    (∂P/T)T, (∂V/∂T)P and (∂T/∂P)V

    b) use the results from a) to evaluate the product
    (∂P/V)T*(∂V/∂T)P*(∂T/∂P)V

    c) Express the definitions of V(T,P) KT(T,P)an BT(T,V) in terms of the indicated independent variables, where each property depends on only two independent variables.

    My problem is I don't fully understand what these partial derivatives are actually describing to set up expression for them.
     
    Last edited: Oct 3, 2013
  2. jcsd
  3. Oct 3, 2013 #2
    The ideal gas is described by three parameters: p,V,T
    Any of the three can be expressed as a function of the other two.
    For example, pressure is a function of V and T: p(V,T).
    Then you can have two derivatives of this function: either you keep V constant and take the derivative in respect to T or you keep T constant and take the derivative in respect to V.
    The first one will be written as (∂p/∂T)V, the second one (∂p/∂V)T

    Your first derivative (in the problem) has an error. It does not make much sense the way you wrote it. I mean the (∂P/T)T.
     
  4. Oct 3, 2013 #3
    Oops, thanks for catching that - I corrected it above. If you can write it two ways, would their solutions be the same? Say if the function with respect to V is messy but that with respect to T is much easier to deal with, could you choose which one you want to use?
     
  5. Oct 3, 2013 #4

    DrClaude

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    Do you know the equation of state for an ideal gas?
     
  6. Oct 3, 2013 #5
    PV=nRT
     
  7. Oct 3, 2013 #6

    DrClaude

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    So you see that you won't get anything "messy".

    The use of one derivative or another usually depends on the physical situation you are considering: processes at constant temperature, constant pressure, etc.
     
  8. Oct 3, 2013 #7
    No, they are different, in general.
    It's very easy to check this. Assume that you have F(x,y)= x^2*y and calculate the two partial derivatives.

    And in physics the two have different units and different physical significance.
     
  9. Oct 3, 2013 #8
    Thanks, I see that now that I've worked them out that they couldn't be the same. I questioned if that was possible in general but suspected for the example here they shouldn't be equal since they have different meaning.

    So, just to check that my math is correct, for (∂P/T)T, (∂V/∂T)P and (∂T/∂P)V
    (∂P/V)T=(-nRT/V^2)
    (∂P/T)V=nR/V
    (∂V/∂T)P = (nR/P)
    (∂V/∂P)T= (nRT)
    (∂T/∂P)V= (V/nR)
    (∂T/∂V)P= (P/nR)
     
    Last edited: Oct 3, 2013
  10. Oct 3, 2013 #9

    DrClaude

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    That one is incorrect. The others are ok.
     
  11. Oct 3, 2013 #10
    Hi DrClaude. I'm having trouble figuring out what you find incorrect about this. It looks OK to me.

    V = (nR/P)T
    [tex]\left(\frac{\partial V}{\partial T}\right)_P=\frac{nR}{P}[/tex]
     
  12. Oct 4, 2013 #11

    DrClaude

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    Sorry, I took the wrong one. It is

    (∂V/∂P)T= (nRT)

    that is incorrect.
     
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