# Entropy and Enthalpy for adiabatic process

1. May 28, 2014

### ricard.py

1) What is the reason why dH!=0 for an adiabatic(q=0) reversible process?
The mathematical argument is irrefutable and it is clear that it has to do with the process not being isobaric:
ΔH=ΔU+PΔV+VΔP , ΔU=work=−PΔV
Therefore, ΔH=VΔP and this is not 0.
However, I do not understand it conceptually...

2) Consider an adiabatic irreversible process from state A to B. At first I would expect to have dS=0, as there is no heat transfer, However, if we construct a reverse isothermal path that connects state B and A, then S(B→A)=-S(A→B). The entropy calculated for the isothermal path is positive, so the entropy for the irreversible process is negative, which agrees with the clausius inequality.
My question is: conceptually, why should a reversible adiabatic process has dS=0 and an irreversible adiabatic process have a dS!=0 if there is no heat transfer?

Thanks.

2. May 28, 2014

### Staff: Mentor

As you showed, it's actually not equal to zero. What made you think that dH = 0 for an adiabatic reverisble process?
dS=dq/T only for a reversible path between the initial and final equilibrium states of a system. Otherwise, dS is greater than dq/T. See my blog in my PF area for a more in-depth discussion.

Chet

3. May 28, 2014

### WannabeNewton

What's there to understand conceptually? It's just a consequence of the thermodynamic relation $dH = SdT + Vdp$ which is itself just a change of variables $H = E + pV$ of the conservation of energy equation $dE = \delta Q - \delta W$ when restricted to a quasi-static process.

I'm just adding on to Chet here by looking at things statistically. If the process is not reversible then by definition, for the total number of accessible microstates $\Omega_i(E)$ of the initial equilibrium state and the total number of accessible microstates $\Omega_f(E)$ of the final equilibrium state, we have $\Omega_f > \Omega_i$; here $E$ is the average energy of the system. There clearly need not be any heat transfer in order for $\Omega_f > \Omega_i$ to hold. As a very simple example, just consider a cylinder of $N$ molecules of a certain ideal gas confined to half the volume $V/2$ of the cylinder by means of a piston initially locked in place; then $\Omega_i(E) \propto (V/2)^N$. Imagine now that the piston is removed so quickly (instantaneously) that no work gets done by the gas and no heat gets transferred to the gas; we then wait for the gas to reach a new equilibrium. This final equilibrium state will of course be the one for which the gas occupies the entire volume of the cylinder. But then $\Omega_f(E) \propto (V)^N$ with the exact same proportionality factor since the energy of the gas didn't change at all ($\Delta E = W + Q = 0$) so $\Omega_f > \Omega_i$.