Enthalpy of an ideal gas

  1. fluidistic

    fluidistic 3,357
    Gold Member

    I've read on Wikipedia that the enthalpy of an ideal gas does not depend on pressure (http://en.wikipedia.org/wiki/Enthalpy:
    ).
    However mathematically I get that it's false. So I'm wondering weather I'm making some error(s) which seems likely or the wiki article is wrong.
    Here's my work:
    I consider a monoatomic ideal gas.
    From the equations of state ##PV=nRT## and ##U=\frac{3nRT}{2}##, one can obtain the fundamental equation ##S(U,V,n)=nS_0+nR \ln \left [ \left ( \frac{U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac{n}{n_0} \right ) ^{-5/2} \right ]##. I've done it myself and it can be found in Callen's book.
    Now using the definition of the enthalpy, I get that ##H=U+PV## where the independent variables of H are S, P and n. So that ##H(S,P,n)##. In order to get the enthalpy I must get ##U(S,P,n)## and ##V(S,P,n)##.
    From the equations of state, I get that ##U=\frac{2PV}{3}##. Plugging that into the enthalpy, I reach that ##H=\frac{5}{3}PV##. So if I can find ##V(S,P,n)## I'm done.
    I take the exponential in both sides of the fundamental equation to get ##e^S=\exp \{ nS_0+nR \ln \left [ \left ( \frac{U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac{n}{n_0} \right ) ^{-5/2} \right ] \}##
    ##\Rightarrow e^S = e^{nS_0} \left [ \left ( \frac{U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac{n}{n_0} \right ) ^{-5/2} \right ] ^{nR}##
    ##\Rightarrow \left [ \left ( \frac{U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac{n}{n_0} \right ) ^{-5/2} \right ] ^{nR} =e^{S-nS_0}##
    ##\Rightarrow \left ( \frac{U}{U_0} \right ) ^{3nR/2} \left ( \frac{V}{V_0} \right ) ^{nR} \left ( \frac{n}{n_0} \right ) ^{-5nR/2} =e^{S-S_0n}##
    ##\Rightarrow \left ( \frac{2PV}{3U_0} \right )^{3nR/2} \left ( \frac{V}{V_0} \right ) ^{nR} n^{-5nR/2} =e^{S-nS_0} ##
    ##\Rightarrow \left ( \frac{2}{3U_0} \right ) ^{3nR/2} \cdot \frac{1}{V_0 ^{nR}} \cdot n ^{-5nR/2 } \cdot V^{5nR/2} \cdot P ^{3nR/2} =e^{S-nS_0}##
    ##\Rightarrow V(S,P,n) = e^{\frac{2(S-nS_0)}{5nR}}P^{-3/5}nV_0^{2/5} \left ( \frac{2}{3U_0} \right )^{3/5}##
    Which makes [tex]H(S,P,n)= e^{\frac{2(S-nS_0)}{5nR}}P^{2/5}nV_0^{2/5} \left ( \frac{2}{3U_0} \right )^{3/5}= c_1 \cdot e^{c_2(S-nS_0)}P^{2/5}n[/tex] where there's a dependence of the enthalpy on the pressure and ##c_1## and ##c_2## are positive constants.
     
  2. jcsd
  3. DrClaude

    DrClaude 2,327
    Science Advisor
    Homework Helper
    Gold Member

    This should be ##U=\frac{3PV}{2}##.
     
  4. fluidistic

    fluidistic 3,357
    Gold Member

    True. Also in my last expression for H(S,P,n) I forgot to multiply by that constant. Nevertheless this doesn't change the general expression I reached, namely ##H(S,P,n)=c_1 \cdot e^{c_2(S-nS_0)}P^{2/5}n##. The dependence of H on P remains.
     
  5. Let me simplify your claim:

    claim 0. the enthalpy of an ideal gas doesn't depend on the pressure.
    but from the second law dH = TdS + VdP
    we have [itex]\frac{\partial H}{\partial P} = V \neq 0[/itex].

    All of the extra junk is just working with the integrated forms
    of the entropy expression.

    Here's a similar line of reasoning for internal energy.
    claim 1. the internal energy of an ideal gas doesn't depend on the volume.
    but from the second law TdS = dU + pdV
    we have [itex]\frac{\partial U}{\partial V} = -p \neq 0[/itex].

    So what gives?

    The claim isn't about what is the dependence of H on p
    when entropy is constant - but when the temperature is held constant.

    From your first expressions
    [itex] H = U + pV = \frac{3}{2}nR T + nRT = C_p T [/itex],
    from whence it's obvious
    [tex] \left( \frac{\partial H}{\partial p} \right)_T = 0, [/tex]
    and so the enthalpy is independent of the pressure.
     
  6. fluidistic

    fluidistic 3,357
    Gold Member

    Thanks for the reply.
    If I understand well the claim of Wikipedia is not really accurate. They should have added "when the temperature is held constant", which isn't a given beforehand. But indeed, if one fixes T, then H doesn't depend on P. I didn't know this, good to know.
    Have I understood you well?
     
  7. Quote by WikiTheGreat
    Enthalpies of ideal gases and incompressible solids and liquids do not depend on pressure, unlike entropy and Gibbs energy.

    This is incorrect. The enthalpies of incompressible solids and liquids depend on pressure:

    [tex]dH = C_pdT+(V-T\frac{\partial V}{\partial T}) dP[/tex]

    This is a general thermodynamic relationship the applies to any material.
     
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