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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

    (h2-h1)T=4C/T3(p1-p2)T

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

    (S2-S1)T=Rln(p1/p2)+3C/T4(p1-p2)T


    3. The attempt at a solution

    taking h=U+pv

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

    dh=d(pv)=-Cdp/T3

    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

    physicsworks

    User Avatar
    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
    [tex]p(V,T)=\frac{RT}{V}[/tex]
    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]
    Then
    [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:
    AND:
    [tex]\Delta H_{12} = - \frac{4C}{T^3} \Delta p_{12}[/tex]
     
    Last edited: May 7, 2010
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