next up previous
Next: Joule-Thompson Throttling Up: Classical Thermodynamics Previous: Free Expansion of Gas

Van der Waals Gas

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

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

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

$\displaystyle T_c$ $\displaystyle = \frac{8 a}{27 R b},$ (6.158)
$\displaystyle p_c$ $\displaystyle = \frac{a}{27 b^{ 2}},$ (6.159)
$\displaystyle v_c$ $\displaystyle = 3 b.$ (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,

$\displaystyle a$ $\displaystyle = \frac{27}{64} \frac{(R T_c)^{ 2}}{p_c},$ (6.161)
$\displaystyle b$ $\displaystyle = \frac{R T_c}{8 p_c}.$ (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

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

which implies that

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

Substitution into Equation (6.155) gives

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

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

next up previous
Next: Joule-Thompson Throttling Up: Classical Thermodynamics Previous: Free Expansion of Gas
Richard Fitzpatrick 2016-01-25