1.4 The Coexistence of Phases and the Critical Point 31
Because the entire fluid (liquid and gas) surface in Figure 1.7 is connected, a com-
pletely successful equation of state should represent the entire surface. The equations
of state that we have discussed yield surfaces that resemble the true surface in the
liquid region as well as in the gas region, although they do not represent the tie lines.
In Chapter 5 we will discuss a technique for constructing the tie lines for a particu-
lar equation of state. The modified Redlich–Kwong–Soave equation of Gibbons and
Laughton seems to be fairly accurate in representing both the liquid and the gas, and the
van der Waals equation is often used to give qualitative information. For any equation
of state, we can obtain equations that locate the critical point.EXAMPLE1.11
Derive formulas for the critical temperature and critical molar volume for a gas obeying the
van der Waals equation of state.
Solution
We seek the point at which(∂P/∂Vm)T 0 (1.4-1)
(∂^2 P/∂Vm^2 )T 0 (1.4-2)The first derivative of Eq. (1.3-2) with respect toVmis
(
∂P
∂Vm)T−
RT
(Vm−b)^2+
2 a
Vm^3(1.4-3)and the second derivative is
(
∂^2 P
∂Vm^2)T−
2 RT
(Vm−b)^3+
6 a
Vm^4(1.4-4)Setting the right-hand side of each of these two equations equal to zero gives us two simul-
taneous algebraic equations, which are solved to give the values of the critical temperature
Tcand the critical molar volumeVmc:Tc
8 a
27 Rb
, Vmc 3 b (1.4-5)Exercise 1.10
Solve the simultaneous equations to verify Eq. (1.4-5). One way to proceed is as follows: Obtain
Eq. (I) by setting the right-hand side of Eq. (1.4-3) equal to zero, and Eq. (II) by setting the
right-hand side of Eq. (1.4-4) equal to zero. Solve Eq. (I) forTand substitute this expression
into Eq. (II).When the values ofTcandVmcare substituted into the van der Waals equation of
state the value of the critical pressure for a van der Waals gas is obtained:Pca
27 b^2