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Prove dQ is an inexact differential

  1. Oct 14, 2016 #1
    1. The problem statement, all variables and given/known data
    ##dz=Mdx+Ndy## is an exact differential if ##(\frac{\partial M}{\partial y})_x=(\frac{\partial N}{\partial x})_y##.
    By invoking the condition for an exact differential, demonstrate that the
    reversible heat ##Q_R## is not a thermodynamic property.

    2. Relevant equations
    ##dQ=dU+PdV##
    ##dU=C_VdT##

    3. The attempt at a solution
    ##dQ=C_VdT+PdV##
    So, we have to show that, ##(\frac{\partial C_V}{\partial V})_T \neq (\frac{\partial P}{\partial T})_V##
    Now,
    ##LHS=(\frac{\partial C_V}{\partial V})_T##
    ##=(\frac{\partial}{\partial V}((\frac{\partial U}{\partial T})_V))_T~~~~##[##C_V=(\frac{\partial U}{\partial T})_V##]
    ##=(\frac{\partial}{\partial T}((\frac{\partial U}{\partial V})_T))_V~~~~## [Symmetry of second derivatives]
    So, we need to show that
    ##(\frac{\partial U}{\partial V})_T \neq P+F(V)~~~## [where, ##F(V)## is an arbitary function of ##V##]
    ##(\frac{\partial U}{\partial V})_T - P \neq F(V)##

    If the equation of state is given, this can easily be proved. How can I prove this in general? I also do not understand what would be the difference for irreversible heat.
     
    Last edited: Oct 14, 2016
  2. jcsd
  3. Oct 14, 2016 #2

    Charles Link

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    Homework Helper

    Just an input for you=I don't have the solution, but ## TdS=dQ_{reversible} ## so that according to the above, you should be able to show that the functional property holds for ## S ## but not for ## Q_{reversible} ##.
     
  4. Oct 14, 2016 #3
    I think you should work with $$C_V=T\left(\frac{\partial S}{\partial T}\right)_V$$ and determine the partial with respect to V at constant T. This can be expressed in terms of the equation of state parameters by applying a Maxwell relationship.
     
  5. Oct 14, 2016 #4
    I tried as you mentioned,
    ##\left(\frac{\partial C_V}{\partial V}\right)_T##
    ##=\left(\frac{\partial}{\partial V}(T\left(\frac{\partial S}{\partial T}\right)_V)\right)_T##
    ##=T\left(\frac{\partial}{\partial V}(\left(\frac{\partial S}{\partial T}\right)_V)\right)_T## ... (i)

    Now,
    ##\left(\frac{\partial S}{\partial T}\right)_V=\left(\frac{\partial S}{\partial P}\right)_V \left(\frac{\partial P}{\partial T}\right)_V=-\left(\frac{\partial V}{\partial T}\right)_S \left(\frac{\partial P}{\partial T}\right)_V##...(ii) [Using Maxwell relationship]
    Using (i) and (ii),
    ##\left(\frac{\partial C_V}{\partial V}\right)_T##
    ##=-T\left(\frac{\partial}{\partial V}(\left(\frac{\partial V}{\partial T}\right)_S \left(\frac{\partial P}{\partial T}\right)_V)_V)\right)_T##
    What should be the next steps?
     
  6. Oct 14, 2016 #5
    I didn't notice before that, in your original post, your equation for dU is incorrect. Do you know what the correct relationship is?
     
  7. Oct 14, 2016 #6
    You are going to kick yourself. This problem is much simpler than what you've been trying to do.

    For a reversible process, dQ=TdS

    Now, if you express dS in terms of dT, dV, and partials of S with respect to T and V, what do you get? What do you get if you substitute that relationship into dQ=TdS?
     
  8. Oct 14, 2016 #7

    Charles Link

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    Homework Helper

    @Chestermiller I don't see an additional response from the OP yet, but your input was useful. I see when testing the criteria given by the OP in post #1, there is an extra term so that the two sides are not equal. In the case of ## dS=(1/T)(dU+PdV) ## though, the criterion is met, at least in the case of an ideal gas.
     
  9. Oct 14, 2016 #8
    Thanks for pointing out the mistake. I am used to solve problems for ideal gases, that's why I have made the mistake, I guess.

    Thanks for the hints!
    ##dQ_R=TdS=T\left(\frac{\partial S}{\partial T}\right) _V dT + T\left(\frac{\partial S}{\partial V}\right) _T dV ##
    So, we have to show that,
    ##\left(\frac{\partial}{\partial V} \left( T\left(\frac{\partial S}{\partial T}\right) _V \right)\right)_T \neq \left(\frac{\partial}{\partial T} \left(T\left(\frac{\partial S}{\partial V}\right) _T \right)\right )_V##
    ##RHS=T\left(\frac{\partial}{\partial T} \left(\frac{\partial S}{\partial V}\right) _T \right )_V + \left(\frac{\partial S}{\partial V}\right)_T##
    And,
    ##LHS=T \left(\frac{\partial}{\partial V} \left(\frac{\partial S}{\partial T}\right) _V \right)_T##

    But ##S(V,T)## is a state function of ##V## and ##T##. So ##\left(\frac{\partial S}{\partial V}\right)_T \neq 0## in general.
     
    Last edited: Oct 14, 2016
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