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.
Context question:
B1 Derive the formula for the volume $V_{G}$ and express it in terms of $R, T, p_{0}$, and $a$.
Context answer:
\boxed{$V_{G}=\frac{R T}{2 p_{0}}(1+\sqrt{1-\frac{4 a p_{0}}{R^{2} T^{2}}})$}


Extra Supplementary Reading Materials:

Almost the same volume $V_{G 0}$ can be approximately evaluated using the ideal gas law.
Context question:
B2 Evaluate in percentage the relative decrease in the gas volume due to intermolecular forces, $\frac{\Delta V_{G}}{V_{G 0}}=\frac{V_{G 0}-V_{G}}{V_{G 0}}$.
Context answer:
\boxed{0.58 %}


Extra Supplementary Reading Materials:

If the system volume is reduced below $V_{G}$, the gas starts to condense. However, thoroughly purified gas can remain in a mechanically metastable state (called supercooled vapor) until its volume reaches a certain value $V_{G \mathrm{~min}}$.

The condition of mechanical stability of supercooled gas at constant temperature is written as: $\frac{d P}{d V}<$ 0.

Question: B3 Find and evaluate how many times the volume of water vapor can be reduced and still remains in a metastable state. In other words, what is $V_{G} / V_{G \text { min }}$ ?

Solution:
Mechanical stability of a thermodynamic system is inpower provided that

$$
\left(\frac{d P}{d V}\right)_{T}<0
\tag{B3.1}
$$

The minimum volume, in which the mattercan still exist in the gaseous state, corresponds to a point in which

$$
V_{G \min } \rightarrow\left(\frac{d P}{d V}\right)_{T}=0
\tag{B3.2}
$$

Using the van der Waals equation of state (B3.2) is written as

$$
\left(\frac{d P}{d V}\right)_{T}=-\frac{R T}{(V-b)^{2}}+\frac{2 a}{V^{3}}=0
\tag{B3.3}
$$

From (B3.2) and (B3.3), and with the help of $V_{G \min } \gg b$, it is found that

$$
V_{Gmin}=\frac{2a}{RT}
\tag{B3.4}
$$

Thus,

$$
V_{G \min }=\frac{2 a}{R T}
\tag{B3.5}
$$

$$
\frac{V_{G}}{V_{G \min }}=\frac{R^{2} T^{2}}{2 a p_{0}}=86
$$

Final answer: 86