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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

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

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

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


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

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

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

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

$\displaystyle {\mit\Delta} G_0\leq 0.$ (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

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

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

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

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

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

Now, from Equation (9.26),

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

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

next up previous
Next: Stability of Single-Phase Substance Up: Multi-Phase Systems Previous: Equilibrium of Constant-Temperature System
Richard Fitzpatrick 2016-01-25