Stability of Single-Phase Substance

Consider a system in a single phase (e.g., a liquid or a solid). Let us focus attention on a small, but macroscopic part, $A$ , of this system. Here, $A$ consists of a fixed number of particles, $N\gg 1$ . The remainder of the system is, relatively, very large, and, therefore, acts like a reservoir held at some constant temperature, $T_0$ , and constant pressure, $p_0$ . According to Equation (9.25), the condition for stable equilibrium, applied to $A$ , is

$$G_0 \equiv \overline{E} - T_0 S+p_0 V = {\rm minimum}.$$ (9.30)

Let the temperature, $T$ , and the volume, $V$ , be the two independent parameters that specify the macrostate of $A$ . Consider, first, a situation where $V$ is fixed, but $T$ is allowed to vary. Suppose that the minimum of $G_0$ occurs for $T=\tilde{T}$ , when $G_0=G_{\rm min}$ . Expanding $G_0$ about its minimum, and writing ${\mit\Delta}T=T-\tilde{T}$ , we obtain

$${\mit\Delta}_m G_0\equiv G_0-G_{\rm min} = \left(\frac{\partial G... ...frac{\partial^{ 2}G_0}{\partial T^{ 2}}\right)_V({\mit\Delta}T)^{ 2}+\cdots.$$ (9.31)

Here, all derivatives are evaluated at $T=\tilde{T}$ . Because $G_0=G_{\rm min}$ is a stationary point (i.e., a maximum or a minimum), ${\mit\Delta}_m G_0=0$ to first order in ${\mit\Delta} T$ . In other words,

$$\left(\frac{\partial G_0}{\partial T}\right)_V = 0 \mbox{for $T=\tilde{T}$} .$$ (9.32)

Furthermore, because $G_0=G_m$ is a minimum, rather than a maximum, ${\mit\Delta}_m G_0\geq 0$ to second order in ${\mit\Delta} T$ . In other words,

$$\left(\frac{\partial^{ 2} G_0}{\partial T^{ 2}}\right)_V \geq 0 \mbox{for $T=\tilde{T}$} .$$ (9.33)

According to Equation (9.30), the condition, (9.32), that $G_0$ be stationary yields

$$\left(\frac{\partial G_0}{\partial T}\right)_V = \left(\frac{\par... ...ne{E}}{\partial T}\right)_V-T_0 \left(\frac{\partial S}{\partial T}\right)_V=0.$$ (9.34)

However, the fundamental thermodynamic relation

$$T dS = d\overline{E} +\bar{p} dV$$ (9.35)

implies that if $V$ is fixed then

$$T \left(\frac{\partial S}{\partial T}\right)_V= \left(\frac{\partial \overline{E}}{\partial T}\right)_V.$$ (9.36)

Thus, Equation (9.34) becomes

$$\left(\frac{\partial G_0}{\partial T}\right)_V = \left(1-\frac{T_0}{T}\right)\left(\frac{\partial \overline{E}}{\partial T}\right)_V=0.$$ (9.37)

Recalling that the derivatives are evaluated at $T=\tilde{T}$ , we get

$$\tilde{T} = T_0.$$ (9.38)

Thus, we arrive at the rather obvious conclusion that a necessary condition for equilibrium is that the temperature of the subsystem $A$ be the same as that of the surrounding medium.

Equations (9.33) and (9.37) imply that

$$\left(\frac{\partial^{ 2} G_0}{\partial T^{ 2}}\right)_V= \frac... ...ight)\left(\frac{\partial^{ 2} \overline{E}}{\partial T^{ 2}}\right)_V\geq 0.$$ (9.39)

Recalling that the derivatives are evaluated at $T=\tilde{T}$ , and that $\tilde{T}=T_0$ , the second term on the right-hand side of the previous equation vanishes, and we obtain

$$\left(\frac{\partial \overline{E}}{\partial T}\right)_V\geq 0.$$ (9.40)

However, this derivative is simply the heat capacity, $C_V$ , at constant volume. Thus, we deduce that a fundamental criterion for the stability of any phase is

$$C_V\geq 0.$$ (9.41)

The previous condition is a particular example of a very general rule, known as Le Chátelier's principle, which states that

If a system is in a stable equilibrium then any spontaneous change of its parameters must give rise to processes that tend to restore the system to equilibrium.

To show how this principle applies to the present case, suppose that the temperature, $T$ , of the subsystem $A$ has increased above that of its surroundings, $A'$ , as a result of a spontaneous fluctuation. The process brought into play is the spontaneous flow of heat from system $A$ , at the higher temperature, to its surroundings $A'$ , which brings about a decrease in the energy, $\overline{E}$ , of $A$ (i.e., ${\mit\Delta}\overline{E}<0$ ). By Le Chátelier's principle, this energy decrease must act to decrease the temperature of $A$ (i.e., ${\mit\Delta}T<0$ ), so as to restore $A$ to its equilibrium state (in which its temperature is the same as its surroundings). Hence, it follows that ${\mit\Delta} \overline{E}$ and ${\mit\Delta} T$ have the same sign, so that ${\mit\Delta}\overline{E}/{\mit\Delta}T>0$ , in agreement with Equation (9.40).

Suppose, now, that the temperature of subsystem $A$ is fixed at $T=T_0$ , but that its volume, $V$ , is allowed to vary. We can write

$${\mit\Delta}_m G_0 = \left(\frac{\partial G_0}{\partial V}\right)... ...frac{\partial^{ 2}G_0}{\partial V^{ 2}}\right)_T({\mit\Delta}V)^{ 2}+\cdots,$$ (9.42)

where ${\mit\Delta} V = V-\tilde{V}$ , and the expansion is about the volume $V=\tilde{V}$ at which $G_0$ attains its minimum value. The condition that $G_0$ is stationary yields

$$\left(\frac{\partial G_0}{\partial V}\right)_T=0.$$ (9.43)

The condition that the stationary point is a minimum gives

$$\left(\frac{\partial^{ 2} G_0}{\partial V^{ 2}}\right)_T\geq 0.$$ (9.44)

Making use of Equation (9.30), we get

$$\left(\frac{\partial G_0}{\partial V}\right)_T = \left(\frac{\par... ...{E}}{\partial V}\right)_T-T_0 \left(\frac{\partial S}{\partial V}\right)_T+p_0.$$ (9.45)

However, Equation (9.35) implies that

$$T \left(\frac{\partial S}{\partial V}\right)_T= \left(\frac{\partial \overline{E}}{\partial V}\right)_T + \bar{p}.$$ (9.46)

Hence,

$$\left(\frac{\partial G_0}{\partial V}\right)_T = T \left(\frac{\p... ...rtial V}\right)_T-\bar{p}-T_0 \left(\frac{\partial S}{\partial V}\right)_T+p_0,$$ (9.47)

or

$$\left(\frac{\partial G_0}{\partial V}\right)_T =-\bar{p}+p_0,$$ (9.48)

because $T=T_0$ . Thus, the condition (9.43) yields

$$\bar{p}= p_0.$$ (9.49)

This rather obvious result states that in equilibrium the pressure of subsystem $A$ must be equal to that of the surrounding medium.

Equations (9.44) and (9.48) imply that

$$\left(\frac{\partial^{ 2} G_0}{\partial V^{ 2}}\right)_T= -\left(\frac{\partial \bar{p}}{\partial V}\right)_T\geq 0.$$ (9.50)

When expressed in terms of the isothermal compressibility,

$$\kappa_T =-\frac{1}{V}\left(\frac{\partial V}{\partial \bar{p}}\right)_T,$$ (9.51)

the stability criterion (9.50) is equivalent to

$$\kappa_T\geq 0.$$ (9.52)

To show how the preceding stability condition is consistent with Le Chátelier's principle, suppose that the volume of subsystem $A$ has increased by an amount ${\mit\Delta V}>0$ , as a result of a spontaneous fluctuation. The pressure, $\bar{p}$ , of $A$ must then decrease below that of its surroundings (i.e., ${\mit\Delta}\bar{p}<0$ ) in order to ensure that the net force exerted on $A$ by its surroundings is such as to reduce the volume of $A$ back towards its equilibrium value. In other words, ${\mit\Delta V}$ and ${\mit\Delta}\bar{p}$ must have opposite signs, implying that $\kappa_T>0$ .

The preceding discussion allows us to characterize the spontaneous volume fluctuations of the small subsystem $A$ . The most probable situation is that where $V$ takes the value $\tilde{V}$ that minimizes $G_0$ , so that $G_0(\tilde{V})=G_{\rm min}$ . Let $P(V) dV$ denote the probability that the volume of $A$ lies between $V$ and $V + dV$ . According to Equation (9.29),

$$P(V) dV \propto \exp\left[-\frac{G_0(V)}{k T_0}\right] dV.$$ (9.53)

However, when ${\mit\Delta} V = V-\tilde{V}$ is small, the expansion (9.42) is applicable. By virtue of Equations (9.43), (9.50), and (9.51), the expansion yields

$$G_0(V)=G_0(\tilde{V}) -\frac{1}{2}\left(\frac{\partial \bar{p}}{\... ...lta}V)^{ 2}=G_{\rm min} -\frac{({\mit\Delta}V)^{ 2}}{2 \tilde{V} \kappa_T}.$$ (9.54)

Thus, Equation (9.53) becomes

$$P(V) dV =B\exp\left[-\frac{(V-\tilde{V})^{ 2}}{2 k T_0 \kappa_T \tilde{V}}\right]dV,$$ (9.55)

where we have absorbed $G_{\rm min}$ into the constant of proportionality, $B$ . Of course, $B$ can be determined via the requirement that the integral of $P(V) dV$ over all possible values of the volume, $V$ , be equal to unity.

The probability distribution (9.55) is a Gaussian whose maximum lies at $V=\tilde{V}$ . Thus, $\tilde{V}$ is equal to the mean volume, $\overline{V}$ . Moreover, the standard deviation of the spontaneous fluctuations of $V$ about its mean value is

$${\mit\Delta}^{\ast} V=\left(k T_0 \kappa_T \overline{V}\right)^{1/2}.$$ (9.56)

(See Section 2.9.) The fluctuations in $V$ imply corresponding fluctuations in the number of particles per unit volume, $n=N/V$ , (and, hence, in the mean density) of the subsystem $A$ . The fluctuations in $n$ are centered on the mean value $\bar{n}=N/\overline{V}$ . For relatively small fluctuations, ${\mit\Delta} n\equiv n-\bar{n}\simeq -(N/\overline{V}^{ 2}) {\mit\Delta V}=-(\bar{n}/\overline{V}) {\mit\Delta V}$ . Hence,

$$\frac{{\mit\Delta}^\ast n}{\bar{n}} = \frac{{\mit\Delta}^\ast V}{\overline{V}}= \left(\frac{k T_0 \kappa_T}{\overline{V}}\right)^{1/2}.$$ (9.57)

Note that the relative magnitudes of the density and volume fluctuations are inversely proportional to the square-root of the volume, $\overline{V}$ , under consideration.

An interesting situation arises when

$$\left(\frac{\partial\bar{p}}{\partial V}\right)_T\rightarrow 0.$$ (9.58)

In this limit, $\kappa_T\rightarrow \infty$ , and the density fluctuations become very large. [They do not, however, become infinite, because the neglect of third-order, and higher, terms in the expansion (9.42) is no longer justified.] The condition $(\partial\bar{p}/\partial V)_T=0$ defines the so-called critical point of the substance, at which the distinction between its liquid and gas phases disappears. (See Section 9.10.) The very large density fluctuations at the critical point lead to strong scattering of light. As a result, a substance that is ordinarily transparent will assume a milky-white appearance at its critical point (e.g., ${\rm CO}_2$ when it approaches its critical point at 304K and 73 atmospheres). This interesting phenomenon is called critical-point opalecance.