Quasi-static processes

Consider the special case of an interaction of the system $A$ with its surroundings which is carried out so slowly that $A$ remains arbitrarily close to equilibrium at all times. Such a process is said to be quasi-static for the system $A$. In practice, a quasi-static process must be carried out on a time-scale which is much longer than the relaxation time of the system. Recall that the relaxation time is the typical time-scale for the system to return to equilibrium after being suddenly disturbed (see Sect. 3.5).

A finite quasi-static change can be built up out of many infinitesimal changes. The infinitesimal heat $\,{\mathchar'26\mskip-12mud}Q$ absorbed by the system when infinitesimal work $\,{\mathchar'26\mskip-12mud}W$ is done on its environment and its average energy changes by $d \bar{E}$ is given by

$${\mathchar'26\mskip-12mud}Q \equiv d\bar{E} + {\mathchar'26\mskip-12mud}W.$$ (124)

The special symbols $\,{\mathchar'26\mskip-12mud}W$ and $\,{\mathchar'26\mskip-12mud}Q$ are introduced to emphasize that the work done and heat absorbed are infinitesimal quantities which do not correspond to the difference between two works or two heats. Instead, the work done and heat absorbed depend on the interaction process itself. Thus, it makes no sense to talk about the work in the system before and after the process, or the difference between these.

If the external parameters of the system have the values $x_1$, $\cdots, x_n$ then the energy of the system in a definite microstate $r$ can be written

$$E_r = E_r(x_1,\cdots, x_n).$$ (125)

Hence, if the external parameters are changed by infinitesimal amounts, so that $x_\alpha \rightarrow x_\alpha + dx_\alpha$ for $\alpha$ in the range 1 to $n$, then the corresponding change in the energy of the microstate is

$$d E_r = \sum_{\alpha =1}^n \frac{\partial E_r}{\partial x_\alpha}\,dx_\alpha.$$ (126)

The work $\,{\mathchar'26\mskip-12mud}W$ done by the system when it remains in this particular state $r$ is

$${\mathchar'26\mskip-12mud}W_r = - d E_r = \sum_{\alpha=1}^n X_{\alpha\,r}\,dx_\alpha,$$ (127)

where

$$X_{\alpha\,r} \equiv -\frac{\partial E_r}{\partial x_\alpha}$$ (128)

is termed the generalized force (conjugate to the external parameter $x_\alpha$) in the state $r$. Note that if $x_\alpha$ is a displacement then $X_{\alpha\,r}$ is an ordinary force.

Consider now an ensemble of systems. Provided that the external parameters of the system are changed quasi-statically, the generalized forces $X_{\alpha\,r}$ have well defined mean values which are calculable from the distribution of systems in the ensemble characteristic of the instantaneous macrostate. The macroscopic work $\,{\mathchar'26\mskip-12mud}W$ resulting from an infinitesimal quasi-static change of the external parameters is obtained by calculating the decrease in the mean energy resulting from the parameter change. Thus,

$${\mathchar'26\mskip-12mud}W = \sum_{\alpha =1}^n \bar{X}_\alpha\, dx_\alpha,$$ (129)

where

$$\bar{X}_\alpha \equiv -\overline{ \frac{\partial E_r}{\partial x_\alpha}}$$ (130)

is the mean generalized force conjugate to $x_\alpha$. The mean value is calculated from the equilibrium distribution of systems in the ensemble corresponding to the external parameter values $x_\alpha$. The macroscopic work $W$ resulting from a finite quasi-static change of external parameters can be obtained by integrating Eq. (129).

The most well-known example of quasi-static work in thermodynamics is that done by pressure when the volume changes. For simplicity, suppose that the volume $V$ is the only external parameter of any consequence. The work done in changing the volume from $V$ to $V + dV$ is simply the product of the force and the displacement (along the line of action of the force). By definition, the mean equilibrium pressure $\bar{p}$ of a given macrostate is equal to the normal force per unit area acting on any surface element. Thus, the normal force acting on a surface element $d{\bf S}_i$ is $\bar{p}\,\,d{\bf S}_i$. Suppose that the surface element is subject to a displacement $d{\bf x}_i$. The work done by the element is $\bar{p}\,\,d{\bf S}_i \!\cdot\! d{\bf x}_i$. The total work done by the system is obtained by summing over all of the surface elements. Thus,

$${\mathchar'26\mskip-12mud}W = \bar{p}\, \,dV,$$ (131)

where

$$dV = \sum_i d{\bf S}_i\!\cdot\! d{\bf x}_i$$ (132)

is the infinitesimal volume change due to the displacement of the surface. It follows from (130) that

$$\bar{p} = - \frac{ \partial\bar{E} }{\partial V},$$ (133)

so the mean pressure is the generalized force conjugate to the volume $V$.

Suppose that a quasi-static process is carried out in which the volume is changed from $V_i$ to $V_f$. In general, the mean pressure is a function of the volume, so $\bar{p} = \bar{p}(V)$. It follows that the macroscopic work done by the system is given by

$$W_{if} = \int_{V_i}^{V_f} {\mathchar'26\mskip-12mud}W = \int_{V_i}^{V_f} \bar{p}(V)\, dV.$$ (134)

This quantity is just the ``area under the curve'' in a plot of $\bar{p}(V)$ versus $V$.