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* - *S*∆*T*


The process is reversible, so


∆*G* = 0 


and hence


∆*H* = *S*∆*T*


For perfect gas, assume heat capacity independent of temperature,


∆*H* = *C*p∆*T*


and so


*C*p∆*T* = *S*∆*T*.


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


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




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.


 2