Subfield: Thermodynamics

Van der Waals equation of state

In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered.

Part A. Non-ideal gas equation of state

Taking into account the finite size of the molecules, the gaseous equation of state takes the form

$$
P(V-b)=R T
\tag{1}
$$

where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume.
Context question:
A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$.
Context answer:
\boxed{$b=N_{A} d^{3}$}


Extra Supplementary Reading Materials:

With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter

$$
\left(P+\frac{a}{V^{2}}\right)(V-b)=R T
\tag{2}
$$

where $a$ is another specific constant.

At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal.

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Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2).

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Figure 2. Several isotherms for van der Waals equation of state.

With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy.

Question: A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$.

Solution:
In the most general case thevan der Waals equation of state can be rewritten as

$$
P_{c} V^{3}-\left(R T_{c}+b P_{c}\right) V^{2}+a V-a b=0
\tag{A2.1}
$$

Since at the critical values of the gas parameters the straight line disappears, then, the solution of (A2.1) must have one real triple root, i.e. it can be rewritten as follows

$$
P_{c}\left(V-V_{c}\right)^{3}=0
\tag{A2.2}
$$

obtained

Comparing the coefficients of expression (A2.1) and (A2.2), the following set of equations is

$$
\left\{\begin{array}{c}
3 P_{c} V_{c}=R T_{c}+b P_{c} \\
3 P_{c} V_{c}^{2}=a \\
P_{c} V_{c}^{3}=a b
\end{array}\right.
\tag{A2.3}
$$

Solution to the set (A2.3) is the following formulas for the van der Waals coefficients

$$
a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}
\tag{A2.4}
$$
$$
b=\frac{R T_{c}}{8 P_{c}}
\tag{A2.5}
$$

Final answer: $a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$