# Is constant pressure heat capacity = entropy during adiabatic reversible expansion of perfect gas?

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 - ST

The process is reversible, so

G = 0

and hence

H = ST

For perfect gas, assume heat capacity independent of temperature,

H = CpT

and so

CpT = ST.

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?

• dH = Cp.dT is valid for isobaric process, but you consider isentopic adiabatic process. Apr 5, 2020 at 15:44
• @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! Apr 5, 2020 at 15:47
• 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 ? Apr 5, 2020 at 15:52
• @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. Apr 5, 2020 at 15:59
• Who says dG is zero for a reversible adiabatic expansion? Apr 6, 2020 at 11:30