Van der Waals Gas

Consider a non-ideal gas whose equation of state takes the form

$$\left(p +\frac{a}{v^{ 2}}\right)(v-b) = R T,$$ (6.156)

where $a$ and $b$ are positive constants. (See Section 8.11.) Such a gas is known as a van der Waals gas. The previous approximate equation of state attempts to take into account the existence of long-range attractive forces between molecules in real gases, as well as the finite volume occupied by the molecules themselves. The attractive forces gives rise to a slight compression of the gas (relative to an ideal gas)--the term $a/v^{ 2}$ represents this additional positive pressure. The parameter $b$ is the volume occupied by a mole of gas molecules. Thus, $v\rightarrow b$ as $T\rightarrow 0$ . Of course, in the limit $a\rightarrow 0$ and $b\rightarrow 0$ , the van der Waals equation of state reduces to the ideal gas equation of state.

Table 6.2: Critical pressures and temperatures, and derived van der Waals parameters, for common gases.

Gas	Symbol	$p_c$ (bar)	$T_c$ (K)	$a$ (SI)	$b$ (SI)
Helium	He	2.27	5.19	$3.46\times 10^{-3}$	$2.38\times 10^{-5}$
Hydrogen	H	13.0	33.2	$2.47\times 10^{-2}$	$2.65\times 10^{-5}$
Nitrogen	${\rm N}_2$	33.9	126.2	$1.37\times 10^{-1}$	$3.87\times 10^{-5}$
Oxygen	${\rm O}_2$	50.5	154.6	$1.38\times 10^{-1}$	$3.18\times 10^{-5}$

The van der Waals equation of state can be written

$$p = \frac{R T}{v-b} -\frac{a}{v^{ 2}}.$$ (6.157)

At fixed temperature, the previous equation yields $p(v)$ curves that exhibit a maximum and a minimum at two points where $(\partial p/\partial v)_T=0$ . At a particular temperature, the maximum and minimum coalesce into a single inflection point where $(\partial^{ 2} p/\partial v^{ 2})_T=0$ , in addition to $(\partial p/\partial v)_T=0$ . This point is called the critical point, and its temperature, pressure, and molar volume are denoted $T_c$ , $p_c$ , and $v_c$ , respectively. It is readily demonstrated that

$$T_c = \frac{8 a}{27 R b},$$ (6.158)

$$p_c = \frac{a}{27 b^{ 2}},$$ (6.159)

$$v_c = 3 b.$$ (6.160)

(See Exercise 13.) The critical temperature, $T_c$ , and the critical pressure, $p_c$ , of a substance are easily determined experimentally, because they turn out to be the maximum temperature and pressure, respectively, at which distinct liquid and gas phases exist. (See Section 9.10.) This allows a determination of the constants $a$ and $b$ in the van der Waals equation of state. In fact,

$$a = \frac{27}{64} \frac{(R T_c)^{ 2}}{p_c},$$ (6.161)

$$b = \frac{R T_c}{8 p_c}.$$ (6.162)

Table 6.2 shows the experimentally measured critical pressures and temperatures, as well as the derived van der Waal parameters, for some common gases.

Let us calculate the Joule coefficient for a van der Waals gas. It follows from Equation (6.156) that

$$\left(\frac{\partial p}{\partial T}\right)_V(v-b) = R,$$ (6.163)

which implies that

$$T\left(\frac{\partial p}{\partial T}\right)_V=\frac{R T}{v-b} = p+ \frac{a}{v^{ 2}}.$$ (6.164)

Substitution into Equation (6.155) gives

$$\eta = -\frac{a}{v^{ 2} c_V}.$$ (6.165)

Thus, the Joule coefficient for a van der Waals gas is negative. This implies that the temperature of the gas always decreases as it undergoes free expansion. Of course, this temperature decrease is a consequence of the work done in overcoming the inter-molecular attractive forces. Over a relatively small temperature range, $T_2<T<T_1$ , any possible temperature dependence of $c_V$ is negligibly small. Thus, $c_V$ can be regarded as essentially constant, and Equations (6.149) and (6.165) yield

$$T_2 -T_1 =-\frac{a}{c_V}\left(\frac{1}{v_1}-\frac{1}{v_2}\right).$$ (6.166)

For an expansion, where $v_2>v_1$ , this equation confirms that $T_2<T_1$ . In other words, the temperature of non-ideal gas that undergoes free expansion is reduced.

In principle, it appears that the free expansion of a gas could provide a method of cooling the gas to low temperatures. In practice, a substantial difficulty arises because of the appreciable heat capacity, $C_c$ , of the container. Because the container's internal energy changes by an amount $C_c (T_2-T_1)$ , the molar heat capacity, $c_V$ , in Equation (6.166), must be replaced by the total molar heat capacity, $c_V + C_c/\nu$ . Given that the heat capacity of the container is generally much greater than that of the gas (i.e., $C_c\gg \nu c_V$ ), it follows that the actual temperature reduction is much smaller than that predicted by Equation (6.166).