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.
Context question:
A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$.
Context answer:
\boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$}


Context question:
A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water.
Context answer:
\boxed{$a_{w}=0.56$ , $b_{w}=3.1 \cdot 10^{-5} $}


Context question:
A4 Estimate the diameter of water molecules $d_{w}
Context answer:
\boxed{$4 \cdot 10^{-10}$}


Extra Supplementary Reading Materials:

Part B. Properties of gas and liquid

This part of the problem deals with the properties of water in the gaseous and liquid states at temperature $T=100{ }^{\circ} \mathrm{C}$. The saturated vapor pressure at this temperature is known to be $p_{L G}=p_{0}=1.0$. $10^{5} \mathrm{~Pa}$, and the molar mass of water is $\mu=1.8 \cdot 10^{-2} \frac{\mathrm{kg}}{\mathrm{mole}}$.

Gaseous state

It is reasonable to assume that the inequality $V_{G} \gg b$ is valid for the description of water properties in a gaseous state.

Question: B1 Derive the formula for the volume $V_{G}$ and express it in terms of $R, T, p_{0}$, and $a$.

Solution:
Usingtheinequality $V_{G} \gg b$, the van der Waals equation of state can be written as

$$
\left(p_{0}+\frac{a}{V_{G}^{2}}\right) V_{G}=R T
\tag{B1.1}
$$

which has the following solutions

$$
V_{G}=\frac{R T}{2 p_{0}}\left(1 \pm \sqrt{1-\frac{4 a p_{0}}{R^{2} T^{2}}}\right)
\tag{B1.2}
$$



Smaller root in (B1.2) gives the volume in an unstable state on the rising branch of thevan der Waals isotherm. The volume of gas is given by the larger root, since at $a=0$ an expression for the volume of an ideal gasshould be obtained, i.e.

$$
V_{G}=\frac{R T}{2 p_{0}}\left(1+\sqrt{1-\frac{4 a p_{0}}{R^{2} T^{2}}}\right)
\tag{B1.3}
$$

For given values of the parameters the value $\frac{a p_{0}}{(R T)^{2}}=5.8 \cdot 10^{-3}$. It can therefore be assumed that $\frac{a p_{0}}{(R T)^{2}} \ll 1$, then (B1.3)takes the form

$$
V_{G} \approx \frac{R T}{p_{0}}\left(1-\frac{a p_{0}}{R^{2} T^{2}}\right)=\frac{R T}{p_{0}}-\frac{a}{R T}
\tag{B1.4}
$$

Final answer: $V_{G}=\frac{R T}{2 p_{0}}(1+\sqrt{1-\frac{4 a p_{0}}{R^{2} T^{2}}})$