Equilibrium of Constant-Temperature Constant-Pressure System

Another case of physical interest is that where a system, $A$ , is maintained under conditions of both constant temperature and constant pressure. This is a situation that occurs frequently in the laboratory, where we might carry out an experiment in a thermostat at atmospheric pressure (say). Therefore, let us suppose that $A$ is in thermal contact with a heat reservoir, $A'$ , that is held at the constant temperature $T_0$ , and at the constant pressure $p_0$ . The system $A$ can exchange heat with the reservoir, $A'$ , but the latter is so large that its temperature, $T_0$ , remains constant. Likewise, the system $A$ can change its volume, $V$ , at the expense of that of the reservoir, $A'$ , doing work on the reservoir in the process. However, $A'$ is so large that its pressure, $p_0$ , remains unaffected by this relatively small volume change.

The analysis of the equilibrium conditions for system $A$ follows similar lines to the that in the previous section. The entropy, $S^{(0)}$ , of the combined system $A^{(0)} = A + A'$ satisfies the condition that in any spontaneous process

$${\mit\Delta}S^{(0)} = {\mit\Delta}S +{\mit\Delta}S'\geq 0.$$ (9.19)

If $A$ absorbs heat $Q$ from $A'$ in this process then ${\mit\Delta}S'=-Q/T_0$ . Furthermore, the first law of thermodynamics yields

$$Q = {\mit\Delta}\overline{E} + p_0 {\mit\Delta}V + W^{ \ast},$$ (9.20)

where $p_0 {\mit\Delta}V$ is the work done by $A$ against the constant pressure, $p_0$ , of the reservoir, $A'$ , and where $W^{ \ast}$ denotes any other work done by $A$ in the process. (For example, $W^{ \ast}$ might refer to electrical work.) Thus, we can write

$${\mit\Delta}S^{(0)} ={\mit\Delta}S-\frac{Q}{T_0}=\frac{T_0 {\mit... ...,\ast})}{T_0} =\frac{{\mit\Delta}(T_0 S-\overline{E}-p_0 V)-W^{ \ast}}{T_0},$$ (9.21)

or

$${\mit\Delta} S^{(0)} = \frac{-{\mit\Delta}G_0-W^{ \ast}}{T_0}.$$ (9.22)

Here, we have made use of the fact that $T_0$ and $p_0$ are constants. We have also introduced the definition

$$G_0\equiv \overline{E} - T_0 S+p_0 V,$$ (9.23)

where $G_0$ becomes the Gibbs free energy, $G=\overline{E}-T S+p V$ , of system $A$ when the temperature and pressure of this system are both equal to those of the reservoir, $A'$ .

The fundamental condition (9.19) can be combined with Equation (9.22) to give

$$W^{ \ast}\leq (-{\mit\Delta} G_0)$$ (9.24)

(assuming that $T_0$ is positive). This relation implies that the maximum work (other than the work done on the pressure reservoir) that can be done by system $A$ is $(-{\mit\Delta}G_0)$ . (Incidentally, this is the reason that $G$ is also called a ``free energy''.) The maximum work corresponds to the equality sign in the preceding equation, and is obtained when the process used is quasi-static (so that $A$ is always in equilibrium with $A'$ , and $T=T_0$ , $p=p_0$ ).

If the external parameters of system $A$ (except its volume) are held constant then $W^{ \ast}=0$ , and Equation (9.24) yields the condition

$${\mit\Delta} G_0\leq 0.$$ (9.25)

Thus, we arrive at the following statement:

If a system is in contact with a reservoir at constant temperature and pressure then the stable equilibrium state is such that

$$G_0={\rm minimum}.$$

The preceding statement can again be phrased in more explicit statistical terms. Suppose that the external parameters of $A$ (except its volume) are fixed, so that $W^{ \ast}=0$ . Furthermore, let $A$ be described by some parameter $y$ . If $y$ changes from some standard value $y_1$ then the corresponding change in the total entropy of $A^{(0)}$ is

$${\mit\Delta}S^{(0)} =-\frac{{\mit\Delta}G_0}{T_0}.$$ (9.26)

But, in an equilibrium state, the probability, $P(y)$ , that the parameter lies between $y$ and $y+\delta y$ is given by

$$P(y)\propto \exp\left[\frac{S^{(0)}(y)}{k}\right].$$ (9.27)

Now, from Equation (9.26),

$$S^{(0)}(y) = S^{(0)}(y_1)-\frac{{\mit\Delta}G_0}{T_0} = S^{(0)}(y_1) - \left[\frac{G_0(y)-G_0(y_1)}{T_0}\right].$$ (9.28)

However, because $y_1$ is just some arbitrary constant, the corresponding constant terms can be absorbed into the constant of proportionality in Equation (9.27), which then becomes

$$P(y)\propto \exp\left[-\frac{G_0(y)}{k T_0}\right].$$ (9.29)

This equation shows explicitly that the most probable state is one in which $G_0$ attains a minimum value, and also allows us to determine the relative probability of fluctuations about this state.