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Homework Help: Change in entropy, Gibbs and Helmholtz in an isothermal compression

  1. Nov 3, 2007 #1


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    1. The problem statement, all variables and given/known data

    Determine the change in the entropy, Helmholtz free energy, and Gibbs free energy, when a mole of ideal gas is compressed from 1atm to 100atm at 20C.

    3. The attempt at a solution

    I am not entirely convinced by my attempt below -- can anyone spot something wrong? (I thought it odd that the entropy change should be temperature-independent...)

    From the first law

    [tex]dU = dQ - dW = dQ - PdV[/tex]

    From the second law,

    [tex]dS = dQ / T \Rightarrow dQ = TdS[/tex]


    [tex]dU = TdS - PdV[/tex]

    Transposing to get dS:

    [tex]dS = \frac{dU}{T} + \frac{P}{T}dV[/tex]

    But the compression takes place at 20C => isothermal => dT = 0 => dU = 0. Thus:

    [tex]dS = \frac{P}{T}dV[/tex]

    From the ideal gas law,

    [tex]PV = RT \Rightarrow V = \frac{RT}{P} \Rightarrow dV = -\frac{RT}{P^{2}}dP[/tex]


    [tex]dS = -R\frac{dP}{P} \Rightarrow \Delta S = Rln(P_{1}/P_{2})[/tex]

    For the Helmholtz free energy, I reason:

    [tex]dF = -SdT - PdV[/tex]

    isothermal => dT = 0. Thus:

    [tex]dF = -PdV[/tex]

    Again, using the ideal gas law, rewrite as:

    [tex]dF = RT\frac{dP}{P} \Rightarrow \Delta S = RTln(P_{2}/P_{1})[/tex]

    And finally, for the Gibbs Free energy, I reason:

    [tex]dG = -SdT + VdP = VdP = \frac{RT}{P}dP[/tex]

    So, in this case, [tex]\Delta G = \Delta F[/tex]

    Is this the right approach?

    Last edited: Nov 3, 2007
  2. jcsd
  3. Nov 4, 2007 #2


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    Homework Helper

    I took a slightly different approach which otherwise parallels yours and got the same result. If we start from

    [tex]PV = RT[/tex] , with T constant, and differentiate, we have

    [tex] P dV + V dP = 0 \Rightarrow P dV = - V dP[/tex].

    For an isothermal process,

    [tex]dU = dQ - PdV = 0 \Rightarrow dQ = P dV = -V dP[/tex] ,

    so we can write

    [tex]dS = dQ / T = -(V dP/T) = -(RT/PT) dP = - R (dP/P)[/tex] ,

    using the ideal gas law. This yields the result you found,

    [tex]\Delta S = R ln(P_{1}/P_{2})[/tex] .

    If we start from the definitions for the free energies,

    [tex]F = U - TS [/tex] and [tex] G = U + PV - TS [/tex] and differentiate, we have

    [tex]dF = dU - T dS - S dT[/tex] and

    [tex]dG = dU + P dV + V dP - T dS - S dT [/tex].

    If we strip away all the terms equal to zero for an isothermal process (dU = 0 , dT = 0), these reduce to

    [tex]dF = - T dS [/tex] and [tex]dG = P dV + V dP - T dS [/tex] ;

    but, as we showed above, [tex] P dV + V dP = 0[/tex] , so

    [tex]dF = - T dS = dG [/tex] .

    Thus, applying our result for [tex]\Delta S[/tex],

    [tex]\Delta F = \Delta G = RT ln(P_{2}/P_{1})[/tex].

    [Almost missed that: the signs flip again going from dS to dF or dG, so the order of the integration limits doesn't reverse this time...]

    I suppose that, given the definitions of what F and G are (see Wiki, for instance), it perhaps isn't surprising that [tex]\Delta F[/tex] and [tex]\Delta G[/tex] are the same for this process.
    Last edited: Nov 4, 2007
  4. Nov 5, 2007 #3


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    Thanks very much for the confirmation :smile:
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