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Homework Help: Enthalpy change and Entropy change; given the equation for the substance

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data

    Equation for the substance given for a certain range of pressure and temperature:

    v=(RT/p) - C/T3

    Where C=Constant

    Show that
    (i) The change of enthalpy for this substance in an isothermal process is


    (ii) The change of entropy for this substance in an isothermal process is


    3. The attempt at a solution

    taking h=U+pv

    dh=du+d(pv) {du=0 for isothermal process}


    but result which i need to show has factor 4 also; which i have no idea how to get.
  2. jcsd
  3. May 7, 2010 #2


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    Gold Member

    Welcome to PF, sac.lalit! :smile:
    Your mistake is to treat [tex]dU=0[/tex] in isothermal process. If it were ideal gas with the usual gas law
    you would be right, but this is not the case.
    In fact, you have
    [tex]V(p,T) = \frac{RT}{p} - \frac{C}{T^3}~~ \text{or}[/tex]
    [tex]p(V,T) = \frac{RT^4}{(VT^3+C)} \left(*\right)[/tex]
    and that's why the usual expression
    [tex]dU = C_V dT[/tex]
    need to be changed (Note also that if [tex]C=0[/tex] in the above equation (*), you get the usual form of equation of state).
    To change U in appropriate way, you must rederive it from the basical first law of thermodynamics in its form:
    [tex]dS = \frac{dU}{T} + \frac{p}{T}dV \left( ** \right)[/tex]
    Now, with [tex]U=U(T,V)[/tex] (caloric equation) you have:
    [tex]dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV[/tex]
    and after substituting this into (**):
    [tex]dS = \frac{1}{T} \left(\frac{\partial U}{\partial T}\right)_V dT + \frac{1}{T} \left[ p + \left(\frac{\partial U}{\partial V}\right)_T \right] dV[/tex]
    From this equation:
    [tex]\left(\frac{\partial S}{\partial T}\right)_V = \frac{1}{T} \left(\frac{\partial U}{\partial T}\right)_V[/tex]
    [tex]\left(\frac{\partial S}{\partial V}\right)_T = \frac{1}{T} \left[ p + \left(\frac{\partial U}{\partial V}\right)_T \right] dV[/tex]
    and remembering that
    [tex]\left(\frac{\partial^2 S}{\partial V \partial T}\right) = \left(\frac{\partial^2 S}{\partial T \partial V}\right)[/tex]
    we finally get:
    [tex]T \left(\frac{\partial p}{\partial T}\right)_V = \left(\frac{\partial U}{\partial V}\right)_T + p \left(***\right)[/tex]

    From this equation we can derive U (caloric equation) if we know thermal equation of state [tex]p=p(V,T)[/tex]
    In fact, from (*):
    [tex]\left(\frac{\partial p}{\partial T}\right)_V = \frac{RT^3 (VT^4 + C)}{(VT^3+C)^2}[/tex]
    and substituting this into (***)
    [tex]\left(\frac{\partial U}{\partial V}\right)_T = \frac{3RCT^4}{(VT^3+C)^2}[/tex]
    [tex]dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV = C_V dT + \frac{3RCT^4}{(VT^3+C)^2}dV[/tex]
    or, using (*) for [tex]V(p,T)[/tex] from which
    [tex]\frac{dV}{dp} = -\frac{RT}{p^2}[/tex]
    and substituting the expression for [tex]p=p(V,T)[/tex] we can rewrite
    [tex]dU = C_V dT - \frac{3C}{T^3}dp[/tex]

    Now, as you wrote, but using the new expression for [tex]dU[/tex]:
    [tex]dH = dU + pdV + Vdp = C_V dT - \frac{3C}{T^3}dp -\frac{RT}{p}dp + \frac{RT}{p}dp - \frac{C}{T^3}dp = C_V dT - \frac{4C}{T^3} dp [/tex]
    For isothermal process
    [tex]dH = - \frac{4C}{T^3}dp[/tex]
    Now the factor 4 pops up :smile:
    [tex]\Delta H_{12} = - \frac{4C}{T^3} \Delta p_{12}[/tex]
    Last edited: May 7, 2010
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