Entropy and Enthelpy for adiabatic process

[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.

 

[Chestermiller]

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...

As you showed, it's actually not equal to zero. What made you think that dH = 0 for an adiabatic reverisble process?

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?

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

 

[WannabeNewton]

However, I do not understand it conceptually...

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.

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?

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 $\Omega_f > \Omega_i$.