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Free energy Helmholtz using only the equation of state

  1. Jul 15, 2017 #1
    1. The problem statement, all variables and given/known data
    This is a state ecuation of a gas:
    PV=AT+B/V, where A and B there are constants.
    First: Demonstrate that ##c_V## depends only of T
    Second: Find U(T,V) and S(T,V)

    2. Relevant equations
    ##\left(\frac{\partial U}{\partial S}\right)_V=T\text{ (1)}##
    ##\left(\frac{\partial U}{\partial V}\right)_S=-p\text{ (2)}##
    ##F=U-TS\text{ (3)}##
    ##\left(\frac{\partial F}{\partial V}\right)_T=-p\text{ (4)}##
    ##\left(\frac{\partial F}{\partial T}\right)_V=-S\text{ (5)}##

    3. The attempt at a solution
    For second item, I think use first (2) and then integrate respect to V using the state ecuation, I got:
    ##U(S,V)=-AT\ln V+B/V+u(S)##
    when u(S) is a constant integration, then I use the (1) for find u(S) and the complete expression for U(S,V):
    ##U(S,V)=-AT\ln V+B/V+TS+U_0##
    now using (4) I would like to obtain the free energy of Helmholtz:
    \begin{equation}
    F(T,V)=-ATlnV+B/V+f(T)
    \end{equation}
    when f(T) it is a new constant integration, then I use the (5) I got:
    -S(T,V)=-AlnV+f'(T) then S(T,V)=AlnV-f'(T)
    if F=U(S,V)-TS(T,V) and derivate respect to T, this it is the same
    -AlnV=-AlnV+f'(T)
    therefore f(T)=c with c constant, then
    S(T,V)=AlnV-c
    and
    U(T,V)=-ATlnV+B/V-cT+U0

    Well, I doubt that the entropy don't depends of T, I haven't see the error, it should be something like to
    ##S(T,V)=S_0+c_V\ln\frac{T}{T_0}+R\ln\frac{V}{V_0}## for a van der Waals gases
    (Sears, Thermodynamics, chapter 9)
     
  2. jcsd
  3. Jul 15, 2017 #2
    What is the equation for dU in terms of dT, Cv, P, and dV?
     
  4. Jul 15, 2017 #3
    Hi, do you mean at this?
    $$dU=TdS-pdV$$ or another form
    $$dU=c_vdT-pdV$$
    but I can't the real form for ##c_v##, except of course $$c_V=T\left(\frac{dS}{dT}\right)_V$$ but I haven't if the expression for S(T,V) found is correct.
     
  5. Jul 15, 2017 #4
    This equation is incorrect. It should read $$dU=c_vdT+\left(T\frac{\partial S}{\partial V}-P\right)dV$$From a Maxwell relation involving dF=-SdT-PdV, what is ##\partial S/\partial V## at constant T in terms of P, V, and T?
     
    Last edited: Jul 16, 2017
  6. Jul 16, 2017 #5
    Yeah! thanks a lot, reviewing a lot of writings, I see the same but I do not complete de entropy. My doubt now is the ##c_V## because except that the problem say it is constant, good! but if ##c_V=c_V(T)## there will be a integral in the expression for energy and entropy.

    For the first issue, I think this:
    for the first principle ##dU=TdS-pdV## and ##dU## and ##dS## there are exact differential, rewriting:
    $$\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV=T\left(\frac{\partial S}{\partial T}\right)_VdT+T\left(\frac{\partial S}{\partial V}\right)_TdV-pdV$$
    gathering differentials for ##dT##
    $$\left(\frac{\partial U}{\partial T}\right)_V=T\left(\frac{\partial S}{\partial T}\right)_V$$
    when the term ##T\left(\frac{\partial S}{\partial T}\right)_V## is the specific heat at V constant $c_v$, therefore only depend of T

    This is correct?
     
  7. Jul 16, 2017 #6
    Well, your math is correct. But you still haven't proven that Cv depends only on T for this particular equation of state. Do you know how to determine ##\left(\frac{\partial S}{\partial V}\right)_T## in terms of P, V, and T? Do you know how to show that
    $$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$
     
  8. Jul 16, 2017 #7
    I think I understand what you say:
    I must prove that:
    $$
    \frac{\partial c_v}{\partial V}=0$$
    how
    $$\frac{\partial c_v}{\partial V}=T\left[\frac{\partial}{\partial V}\left(\frac{\partial S}{\partial T}\right)_V\right]_T$$
    and how the Maxwell relation say:
    $$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$
    then
    $$\frac{\partial c_v}{\partial V}=T\left[\frac{\partial}{\partial V}\left(\frac{\partial P}{\partial T}\right)_V\right]_T$$
    using the equation of state:
    $$\left(\frac{\partial P}{\partial T}\right)_V=A/V$$
    therefore deriving again respect to T it is null, then ##c_V## it does not depend on ##V##.
     
  9. Jul 16, 2017 #8
    Correct.
     
  10. Jul 16, 2017 #9
    Actually, you have $$dU=C_vdT-\frac{B}{V^2}dV$$. This also shows that Cv is dependent only on temperature.
     
    Last edited: Jul 16, 2017
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