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For any process, ∆G = ∆H - ∆(TS).

Apply this on reversible adiabatic expansion of perfect gas, since the entropy of system does not change during this process, S is constant so

∆G = ∆H - S∆T

The process is reversible, so

∆G = 0

and hence

∆H = S∆T

For perfect gas, assume heat capacity independent of temperature,

∆H = Cp∆T

and so

Cp∆T = S∆T.

∆T is not zero during adiabatic reversible expansion- temperature drops. So divide both sides by ∆T gives

Cp = S

Is this legit? If not where is my mistake? I have a hunch that I did something terribly stupid. The final result seems to be so counter-intuitive because

The Cp and S are constant throughout the process, so at the end of the process the two will still be the same. By measuring Cp, even though your system is not doing adiabatic reversible expansion, you will get S. So we are able to measure S?

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  • $\begingroup$ dH = Cp.dT is valid for isobaric process, but you consider isentopic adiabatic process. $\endgroup$
    – Poutnik
    Commented Apr 5, 2020 at 15:44
  • $\begingroup$ @Poutnik I thought: dH always = CpdT for perfect gas. Just that dH not the same as dq if process is not isobaric. Enthalpy is a state function! $\endgroup$
    – TheLearner
    Commented Apr 5, 2020 at 15:47
  • $\begingroup$ Yes, H is a state function. but dH=Cp.dT describes the state path via the constant pressure path. If the pressure changes with T, the final state will be different and so does H. Additionally, what does the index p in Co mean ? $\endgroup$
    – Poutnik
    Commented Apr 5, 2020 at 15:52
  • $\begingroup$ @Poutnik The index means p is constant. But for a perfect gas, enthalpy is independent of pressure. For perfect gas, enthalpy depends on Tp only. If Tx2 and px2, let enthalpy change be a. If Tx2 and Vx2, the enthalpy change will still be a. Correct me if I am wrong. $\endgroup$
    – TheLearner
    Commented Apr 5, 2020 at 15:59
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    $\begingroup$ Who says dG is zero for a reversible adiabatic expansion? $\endgroup$
    – Chet Miller
    Commented Apr 6, 2020 at 11:30

1 Answer 1

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Inspired by Chet Miller.

The fallacy in the argument is that, ΔG = 0 because the process is reversible. This is not true and so leading to wrong final result.

ΔG is a signpost for spontaneity of a process only if the process occurs through constant temperature and pressure. Likewise, that ΔG = 0 means reversibility, is true only if the process happens through constant temperature and constant pressure.

Adiabatic expansion clearly involves change in temperature, so ΔG is not a signpost for spontaneity of adiabatic change, and hence ΔG ≠ 0 for reversible adiabatic expansion.

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