- 23

- 2

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.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.

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}##.

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...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....

$$R_{min} = \Delta E - T_0 \Delta S +P_0 \Delta V$$

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: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

$$dR_{min}= (T-T_0) dS - (P-P_0) dV$$

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?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.