# I Maximum work done by a body in an external medium

#### hgandh

I am reading Landau's Vol 5 on Statistical Physics and have trouble grasping some concepts in Section 20.
Let us now consider a different formulation of the maximum-work problem. Let a body be in an external medium whose temperature $T_0$ and pressure $P_0$ differ from the temperature $T$ and pressure $P$ of the body. The body can do work on some object, assumed thermally isolated both from the medium and from the body. The medium, together with the body in it and the object on which work is done, forms a closed system. The volume and energy of the medium are so large that the change in these quantities due to processes involving the body does not lead to any appreciable change in the temperature and pressure of the medium, which may therefore be regarded as constant.
If I understand this correctly, the body and the medium are in direct contact and can exchange work and heat while the object can only exchange work with the body.
A transition in which the body does the maximum work $|R|_{max}$ corresponds to a reverse transition which requires the external source to do the minimum work $R_{min}$.
Thus the total change $\Delta E$ in the energy of the body in some (not necessarily small) change in its state consists of three parts: the work $R$ done by the body the external source, the work done by the medium, and the heat gained from the medium...
$$\Delta E =R + P_0 \Delta V_0 - T_0 \Delta S_0$$
Since the total volume of the medium and the body remains constant, $\Delta V_0 = -\Delta V$, and the law of increase of entropy shows that $\Delta S + \Delta S_0 \geq 0$...
$$R \geq \Delta E - T_0 \Delta S +P_0 \Delta V$$
The equality occurs for a reversible process....
$$R_{min} = \Delta E - T_0 \Delta S +P_0 \Delta V$$
So the minimum work is realized during a reversible process which means that the system is in thermodynamic equilibrium at every instant, right? That means the body and the medium must have the same temperature and pressure, right? However, Landau then goes on to say...
If the body is in an equilibrium state at every instant during the process (but not, of course, in equilibrium with the medium), then for infinitesimal changes in its state formula (20.2) may be written differently. Substituting $dE = T dS - P dV$ in $dR_{min}-T_0 dS + P_0 dV$, we find
$$dR_{min}= (T-T_0) dS - (P-P_0) dV$$
I don't understand what Landau is doing here. Shouldn't $dR_{min}$ be equal to zero here since it is only for reversible processes so $T=T_0, P=P_0$? Yet, this equation seems to be valid when the body is not equilibrium with the medium. What am I not seeing? Another confusion comes when Landau writes:
Two important cases may be noted. If the volume and temperature of the body remain constant, the latter being equal to the temperature of the medium, (20.2) gives $R_{min}=\Delta (E-TS)$, or
$$R_{min}=\Delta F$$
i.e. the minimum work is equal to the change in the free energy of the body. Secondly, if the temperature and pressure of the body are constant and equal to $T_0$ and $P_0$, we have
$$R_{min} = \Delta \Phi$$
i.e. the work done by the external source is equal to the change in the thermodynamic potential of the body.
It should be emphasized that in both these particular cases the body concerned must be one not in equilibrium, so that its state is not defined by $T$ and $V$ (or $P$) alone; otherwise the constancy of these quantities would mean no process could occur at all. We must consider, for example, a chemical reaction in a mixture or reacting substances, a process of dissolution, or the like.
Again I am confused here. As I am understanding correctly, all of the equations involving $R_{min}$ only apply to reversible processes but Landau says the bodies must not be in equilibrium. Can somebody help me understand these concepts?

• Delta2
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#### tnich

Homework Helper
If I understand this correctly, the body and the medium are in direct contact and can exchange work and heat while the object can only exchange work with the body.
I haven't read your whole post, yet, but right away I noticed that, in contrast to your assumption, the text says: "The body can do work on some object, assumed thermally isolated both from the medium and from the body." If the object is thermally isolated from the medium, then it cannot exchange heat with it.

#### hgandh

I haven't read your whole post, yet, but right away I noticed that, in contrast to your assumption, the text says: "The body can do work on some object, assumed thermally isolated both from the medium and from the body." If the object is thermally isolated from the medium, then it cannot exchange heat with it.
The body and the object are two different things. The body is in an external medium and can exchange both work and heat. The body can only exchange work with the object. So you have 3 things here: the medium, body and the object.

#### tnich

Homework Helper
The body and the object are two different things. The body is in an external medium and can exchange both work and heat. The body can only exchange work with the object. So you have 3 things here: the medium, body and the object.
You are right. I misunderstood your comment.

#### sponteous

I agree, I found this section in Landau and Lifshitz very confusing for the same reasons as the OP when I read it a month or two ago. I just reread it to make sure, and I still find it confusing. It would certainly seem that for a reversible process the temperature of the body and the medium would have to be the same, or only infinitesimally different. It also isn't obvious to me that the work the medium does on the body is $P_0 \Delta V_0$ if the pressures are different. Although, now that I think of it, perhaps the thermally isolated object can also exchange compression/expansion work with the body, and that would account for the discrepancy.

I think what the authors are trying to do is make the analysis as general as possible. For example, to show that change in free energy can give a lower bound to the work required to drive a process even if the system itself is not at constant temperature during the process. As long as the surrounding medium is at constant temperature $T_0$, and the initial and final temperatures of the system (body) are both $T=T_0$, then the total work done on the system is at least $$W \geq \Delta A = \Delta (U-TS).$$ The body is allowed to assume other temperatures in between. This follows from the Clausius Theorem: $$\Delta S \geq \int_A^B dq/T_0 .$$ Since the temperature of the surroundings, $T_0$ is constant, it comes outside the integral and you get $$T_0 \Delta S \geq Q = \Delta U - W$$, where $W$ is the total work done on the system. This can be rearranged to get $$W \geq \Delta U - T_0 \Delta S = \Delta(U-TS),$$ The last equality comes from assuming that the initial and final temperatures of the body are the same as the surroundings.

This is more general than the usual statement that the change in free energy gives a lower bound for _isothermal_ processes. However, it looks like L&L are trying to show something even more general than this, and I don't follow their analysis.

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#### DrDu

It says explicitly that the system doesnt have to be in equilibrium with the surrounding medium. Reversible means here only that the system and source of work can be returned to their initial state, not that all the heat will return from the surrounding. Maybe a trivial example is that of the body driving a paddle in the system, with so much heat transferred to the medium that T and V remain constant.

#### sponteous

It says explicitly that the system doesnt have to be in equilibrium with the surrounding medium. Reversible means here only that the system and source of work can be returned to their initial state, not that all the heat will return from the surrounding. Maybe a trivial example is that of the body driving a paddle in the system, with so much heat transferred to the medium that T and V remain constant.
I am trying to picture this setup but I don't quite follow. Are you saying that some of the work performed by the body is used to stir the surrounding medium with a paddle?

At any rate, I haven't encountered that use of "reversible" before. I thought reversible meant everything, including the surroundings and any lifted weights, can be returned to their initial state. But I'm still new to thermodynamics so that doesn't mean much.

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