How is entropy affected in a reversible process for an ideal gas?

[Pouyan]

Homework Statement

Determine the maximum work that can be recovered by obtaining a mole of ideal gas with constant heating capacity Cv, temperature T and pressure p to reversibly adopt the ambient temperature T0 and pressure p0.

Relevant Equations

ΔU=ΔW + ΔQ
S(p,T) = (Cv+nR)ln(T)-nRln(p)
pV=nRT

What do I see in my solution is :

ΔW + ΔQ = W_pv + W' + ΔQ (A little difficult to perceive the useful work )

Work on the environment : -p0*(-ΔV) (WHY negative sign?, Is this the work ON the gas?)
ΔV=nR (T0/p0 -T/p)
By TdS = dQ
ΔS + ΔS0 =0

Reversible case:
ΔU= -T0ΔS - (-p0(-ΔV)) + W' (WHY T0ΔS is negative here?)

-W'=-ΔU-T0ΔS-p0ΔV

The rest is not difficult to calculate, but I still do not understand how we can get the useful work in such a process. I am used to the ideal gas equation dU = TdS-pdV. Here I see an extra parameter W', (I know it is the useful work, but it is confusing when I see it with pdV ). I think I don't understand well when a GAS do work on it's environment or vice versa !

 

[Chestermiller]

What they are doing here is to assume that there is some unspecified reversible process that is applied to an ideal gas to get the maximum amount of useful work out of it, using only a reservoir at To and an environment at Po. The process could even partially consist of using the ideal gas as the high temperature reservoir that supplies heat to a second ideal gas (working fluid) that executes reversible Carnot cycles.

The focus of the calculation is the ideal gas under consideration. Because the unspecified process that it is involved with is reversible, there is no entropy generated within the gas, and its change in entropy during the process is only the result from exchange of heat with the reservoir at To. So the heat transferred from the reservoir to the gas is $$Q=T_0\Delta S$$ where $$\Delta S = (C_v+R)\ln((T_0/T))+R\ln{(P_0/P)}$$The change in internal energy of the gas between its initial state and its final state is: $$\Delta U=C_v(T_0-T)$$The first law tells us that $$\Delta U=Q-W$$where W is the total work done by the gas. So, $$W=Q-\Delta U$$The work done by the gas in pushing back the surroundings $P_0\Delta V$ is not considered useful work. So the net useful work is $$W_{useful}=W-P_0\Delta V$$

 

[Pouyan]

What they are doing here is to assume that there is some unspecified reversible process that is applied to an ideal gas to get the maximum amount of useful work out of it, using only a reservoir at To and an environment at Po. The process could even partially consist of using the ideal gas as the high temperature reservoir that supplies heat to a second ideal gas (working fluid) that executes reversible Carnot cycles.

The focus of the calculation is the ideal gas under consideration. Because the unspecified process that it is involved with is reversible, there is no entropy generated within the gas, and its change in entropy during the process is only the result from exchange of heat with the reservoir at To. So the heat transferred from the reservoir to the gas is $$Q=T_0\Delta S$$ where $$\Delta S = (C_v+R)\ln((T_0/T))+R\ln{(P_0/P)}$$The change in internal energy of the gas between its initial state and its final state is: $$\Delta U=C_v(T_0-T)$$The first law tells us that $$\Delta U=Q-W$$where W is the total work done by the gas. So, $$W=Q-\Delta U$$The work done by the gas in pushing back the surroundings $P_0\Delta V$ is not considered useful work. So the net useful work is $$W_{useful}=W-P_0\Delta V$$

Thanks so much for the help but one thing, when we talk about reversible process, what happens to entropy?
I know that for adiabatic process the entropy change is 0 but too reversible then?

 

[Chestermiller]

Thanks so much for the help but one thing, when we talk about reversible process, what happens to entropy?
I know that for adiabatic process the entropy change is 0 but too reversible then?

The entropy change is zero for a system experiencing an adiabatic reversible process. For a closed system experiencing an arbitrary process, the entropy change is equal to the entropy transferred into and out of the system via heat transfer at the system boundaries (Q/T at the boundary) plus the entropy increase resulting from irreversible entropy generation within the system. For a closed system experiencing an arbitrary reversible process, the entropy generation within the system is negligible, and only the entropy exchanges at the boundary are significant. For more on this, see Fundamentals of Engineering Thermodynamics by Moran et al.