The way I prove that entropy is a state function is that I consider a carnot cycle which consists of an isothermal expansion followed by adiabatic expansion and then isothermal and adiabatic contraction.

This gives the relation $ \frac {Q_c}{Q_h} = \frac { T_c} { T_h}$

So when calculating entropy, $ S = \int_C \frac {dQ}{T} = \frac {Q_h}{T_h} - \frac {Q_c}{T_c} = 0$

So I proved that the closed integral of $ \frac {dQ}{T}$ is equal to zero and so it is path independent and that means that $dS$ is a perfect differential so $S$ is a state function that only depends on the initial and final points.

Is using a carnot cycle enough to prove this or am I wrong?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.