# Van der Waals Equation

• ### Conventional Derivation of the Van der Waals Equation

The state of a given amount of any substance can be described by three parameters: pressure $$p,$$ volume $$V,$$ and temperature $$T.$$ These parameters are related to each other. Their relationship is described by the equation of state, which in the general case has the form:

$F\left( {p,V,T} \right) = 0.$

The specific form of the equation depends on the substance. For example, a rarefied gas at a sufficiently high temperature is well described by the ideal gas model. Its equation of state is the well-known ideal gas law stated by Emile Clapeyron $$\left({1799-1864}\right)$$ in $$1834:$$

$pV = \frac{m}{M}RT.$

Here $$m$$ is the mass of the gas, $$M$$ is the molar mass (i.e. the mass of one mole of the gas), $$R$$ is the universal gas constant. For one mole of gas, this equation takes the following form:

$pV = RT.$

Subsequent experiments revealed deviations in the behavior of real gases from the ideal gas law. These results were summarized by the Dutch physicist Johannes Diderik van der Waals $$\left({1837-1923}\right)$$ who in $$1873$$ proposed a more accurate equation of state for a real gas.

It is called the Van der Waals equation. For one mole of a gas, it can be written as

$\left( {p + \frac{a}{{{V^2}}}} \right)\left( {V – b} \right) = RT.$

This equation takes into account the attractive and repulsive forces between molecules. The attractive forces are taken into account through the near-wall effect.

Indeed, for the particles located in the inner region, the attractive forces from other molecules are compensated on the average. However, for the particles near the walls of the container, there is an uncompensated attractive force $$f$$ directed inside the container. This force, on the one hand, is proportional to the concentration of particles $$n$$ in the container. On the other hand, it is proportional to the concentration of particles in the boundary layer. As a result, we obtain:

$f \sim {n^2} \sim \frac{1}{{{V^2}}},$

where $$n$$ is the concentration of molecules in the container, $$V$$ is the volume of one mole of gas.

The effect of attraction of the molecules of the near-wall layer reduces the pressure on the walls of the container. In the formal transition from the ideal gas law to the Van der Waals equation, this corresponds to the replacement

$p \to p + \frac{a}{{{V^2}}},$

where $$a$$ is the coefficient depending on the particular gas and the size of the container.

The repulsive forces between the molecules in the Van der Waals model are taken into account as follows: The molecules are assumed to have a spherical shape of radius $$r$$ and their centers can not come closer to each other than $$2r.$$ We can suppose that there exists a “forbidden” (excluded) volume around one of the two molecules (Figure $$1$$), equal to

$\frac{4}{3}\pi {\left( {2r} \right)^3} = 8 \cdot \frac{4}{3}\pi {r^3}.$

Hence, the excluded volume per one molecule is given by

${b_0} = 4 \cdot \frac{4}{3}\pi {r^3} = 4{V_0},$

where $${V_0}$$ is the volume of one molecule.

As a result, if for an ideal gas, the available void space where the molecules can move was equal to $$V,$$ now it becomes equal to

$V – {N_A}{b_0} = V – b,$

where $${N_A}$$ is the Avogadro’s number (the number of molecules in one mole of any substance), $$b$$ is the excluded volume caused by repulsion of the particles.

Assuming that the attractive and repulsive forces between the molecules act together and replacing $$p$$ and $$V$$ with the new expressions, we can convert the ideal gas law into Van der Waals equation:

$\left( {p + \frac{a}{{{V^2}}}} \right)\left( {V – b} \right) = RT.$

### Van der Waals Isotherms

At a fixed temperature, the Van der Waals equation describes the dependence $$p\left( V \right).$$ In the $$pV$$-plane, this dependence is represented as a family of isotherms, each of which corresponds to a certain temperature. To investigate this dependency in more detail, we transform the Van der Waals equation to the following form:

${\left. {\left( {p + \frac{a}{{{V^2}}}} \right)\left( {V – b} \right) = RT\;} \right| \cdot {V^2},\;\;}\Rightarrow {\left( {p{V^2} + a} \right)\left( {V – b} \right) = RT{V^2},\;\;}\Rightarrow {p{V^3} + aV – pb{V^2}} – {ab – RT{V^2} = 0,\;\;}\Rightarrow {{p{V^3} – \left( {pb + RT} \right){V^2}}} + {{aV – ab = 0\;|:p,}\;\;}\Rightarrow {{{V^3} – \left( {b + \frac{{RT}}{p}} \right){V^2}} + {\frac{a}{p}V – \frac{{ab}}{p} = 0.}}$

For a fixed value of $$p$$, the resulting equation is a third degree equation with respect to the variable $$V.$$ It is known that a cubic equation can have $$1$$ or $$3$$ real roots. The first case occurs at high temperatures $$T$$ (the green isotherm $$AB$$ in Figure $$2$$).

With lowering the temperature, an undulating region appears on the isotherm. In this case, there are three roots (the blue isotherm $$EFGHIJL$$). The transition between the two types of isotherms occurs at a certain temperature $${T_K},$$ which is called the critical temperature.

### Critical Point

At the critical point, the gas is characterized by the critical values of $${T_K}$$, $${p_K}$$ and $${V_K},$$ which are determined only by the gas properties. From the algebraic point of view, the isotherm at the critical point has three real roots, which are equal to each other. This fact makes it easy to calculate the values of $${T_K}$$, $${p_K}$$ and $${V_K}.$$ Indeed, in this case the equation of state of the gas

${V^3} – \left( {b + \frac{{RT}}{p}} \right){V^2} + \frac{a}{p}V – \frac{{ab}}{p} = 0$

should be written as

${\left( {V – {V_K}} \right)^3} = 0.$

Expanding the cube of the difference and equating the coefficients of the terms with equal powers of $$V,$$ we find the expressions for the critical parameters:

${{\left( {V – {V_K}} \right)^3} = 0,\;\;}\Rightarrow {{V^3} – 3{V^2}{V_K} + 3V{V_K} – V_K^3 = 0.}$

Take into account that at the critical point $$p = {p_K}$$, $$T = {T_K}.$$ Consequently, we obtain:

${\left\{ \begin{array}{l} – \left( {b + \frac{{RT}}{p}} \right) = – 3{V_K}\\ \frac{a}{{{p_K}}} = 3V_K^2\\ – \frac{{ab}}{{{p_K}}} = – V_K^3 \end{array} \right.,\;\;}\Rightarrow {\left\{ \begin{array}{l} b + \frac{{RT}}{p} = 3{V_K}\\ \frac{a}{{{p_K}}} = 3V_K^2\\ \frac{{ab}}{{{p_K}}} = V_K^3 \end{array} \right..}$

Divide the third equation by the second:

$\require{cancel} {\frac{{\cancel{a}b \cdot \cancel{p_K}}}{{\cancel{p_K} \cdot \cancel{a}}} = \frac{{V_K^{\cancel{3}}}}{{3\cancel{V_K^2}}},\;\;}\Rightarrow {{V_K} = 3b.}$

Determine ({p_K}) from the second equation:

${{p_K} = \frac{a}{{3V_K^2}} } = {\frac{a}{{3 \cdot {{\left( {3b} \right)}^2}}} } = {\frac{a}{{27{b^2}}}.}$

Finally, we find the critical temperature $${T_K}$$ from the first equation:

${{T_K} = \frac{{\left( {3{V_K} – b} \right){p_K}}}{R} } = {\frac{{\left( {3 \cdot 3b – b} \right) \cdot \frac{a}{{27{b^2}}}}}{R} } = {\frac{{8a}}{{27bR}}.}$

Thus, the critical parameters for a Van der Waals gas depend only on the values of $$a, b$$ and are given by the formulas

${{V_K} = 3b,}\;\;\;\kern-0.3pt {{p_K} = \frac{a}{{27{b^2}}},}\;\;\;\kern-0.3pt {{T_K} = \frac{{8a}}{{27bR}}.}$

Note that we have found the critical values without the use of the derivative. It can be shown that the first and second derivatives of the function $$p\left( V \right)$$ at a critical point are equal to zero (see Example $$1$$).

### Reduced Equation of State

Using the critical parameters $${V_K}$$, $${p_K}$$ and $${T_K}$$, it is convenient to go to the dimensionless variables

${\varphi = \frac{V}{{{V_K}}},}\;\;\;\kern-0.3pt {\pi = \frac{p}{{{p_K}}},}\;\;\;\kern-0.3pt {\tau = \frac{T}{{{T_K}}}}$

and rewrite the Van der Waals equation in reduced form:

${V = \varphi {V_K},\;p = \pi {p_K},\;T = \tau {T_K},\;\;}\Rightarrow { \left( {\pi {p_K} + \frac{a}{{{{\left( {\varphi {V_K}} \right)}^2}}}} \right)\left( {\varphi {V_K} – b} \right) }={ R\tau {T_K},\;\;}\Rightarrow {\left( {\frac{{\pi a}}{{27{b^2}}} + \frac{a}{{{{\left( {3b} \right)}^2}{\varphi ^2}}}} \right)\left( {3b\varphi – b} \right) }={ R\tau \cdot \frac{{8a}}{{27bR}},\;\;}\Rightarrow {\left( {\pi + \frac{3}{{{\varphi ^2}}}} \right)\left( {3\varphi – 1} \right) = 8\tau .}$

This equation is more versatile than the original Van der Waals equation. The isotherms of various substances built in accordance with the above equation will coincide at the same values of $$\tau.$$ The states of different substances, which are described by the same reduced parameters $$\pi,$$ $$\varphi,$$ $$\tau$$ are called the corresponding states. If two bodies or substances have equal values of any two of the three reduced quantities, then they have the same third quantity as well. This universal property is called the law of corresponding states.

Consider another interesting invariant. Since the critical quantitites $${V_K}$$, $${p_K}$$, $${T_K}$$ in the Van der Waals model are expressed only in terms of the two parameters $$a$$ and $$b,$$ we can construct a combination of these quantities, which is independent of $$a$$ and $$b.$$ One can write the following relationship:

${\frac{{{p_K}{V_K}}}{{{T_K}}} } = {\frac{{3b \cdot \frac{a}{{27{b^2}}}}}{{\frac{{8a}}{{27bR}}}} } = {\frac{{3\cancel{a}\cancel{b} \cdot \cancel{27}\cancel{b}R}}{{\cancel{27}\cancel{b^2} \cdot 8\cancel{a}}} } = {\frac{{3R}}{8}}\;\;\;\kern-0.3pt {\text{or}\;\;\;\frac{{{p_K}{V_K}}}{{{T_K}R}} } = {\frac{3}{8} = 0,375.}$

This ratio is called the universal compressibility factor and does not depend on any quantities characterizing the substance, that is this ratio is an invariant.

### Liquid-Gas Phase Transition and Metastable States

Consider again the Van der Waals equation at temperatures above the critical point: $$T \lt {T_K}.$$ In this case, the isotherms have an undulating shape. A portion of such isotherm between the minimum and maximum is characterized by a positive derivative $${\large\frac{{dp}}{{dV}}\normalsize} \gt 0,$$ which corresponds to an unstable state of the substance. Any small positive perturbation of the volume $$dV \gt 0$$ causes an increase in the gas pressure (since $$dp \gt 0$$), which will lead to an explosive expansion of the gas. In the case of a negative perturbation $$dV \lt 0,$$ on the contrary, we would observe a quick collapse of the gas.

Actually, the liquid-gas phase transition occurs in this area of the $$PV$$-diagram. This transition is accompanied by a significant change in the volume $$V$$ at constant pressure $$p$$ and constant temperature $$T.$$ Such a transition is represented by a straight horizontal line in the $$PV$$-diagram (the segment line $$FJ$$ in Figure $$3$$).

The two phases coexist in the container at this moment. As the volume of the gas decreases when moving to the left along the segment line $$JF,$$ the volume of the liquid correspondingly increases.

The location of the horizontal section $$FJ$$ is determined from thermodynamic considerations − on the basis of the so-called Maxwell equal area rule, according to which the areas of the curved shapes $$FGH$$ and $$HIJ$$ should be equal (see Figure $$3$$ above).

The set of start and end points of the horizontal segment lines at different temperatures $$T$$ (where $$T \le {T_K}$$) forms a curve, which is called the binodal curve. The curve passing through the minimum and maximum points of all isotherms at $$T \le {T_K}$$ is called the spinodal curve.

The unstable states (for which $${\large\frac{{dp}}{{dV}}\normalsize} > 0$$) are within the region bounded by the spinodal curve. A part of the diagram between the spinodal and binodal curves in principle satisfies the stability criterion $${\large\frac{{dp}}{{dV}}\normalsize} \lt 0.$$ Therefore, the states in this region are called metastable states. The left half of the indicated region represents the superheated liquid, and the right half corresponds to the supercooled vapor.

It is surprising that such a simple model of a real gas, which takes into account the intermolecular forces only very roughly is capable to reflect the actual set of different states of gas and liquid. Besides the regular gas and liquid, the “liquid-gas” phase transition, and the metastable states considered above, the Van der Waals model also captures such an exotic phenomenon as stretched liquid (the curve $$MNS$$ in Figure $$3\text{).}$$ This state of the liquid is unstable.

In the examples below, we focus on the investigation of the Van der Waals equation with the help of the derivative.

• ## Solved Problems

Click a problem to see the solution.

### Example 1

Show that at a critical point, the first and second derivatives of the pressure with respect to the volume at constant temperature are equal to zero.

### Example 2

Investigate the dependence of $$pV$$ on $$V$$ for real gases.

### Example 1.

Show that at a critical point, the first and second derivatives of the pressure with respect to the volume at constant temperature are equal to zero.

Solution.

The equation of an isotherm in the Van der Waals model can be written as

$p\left( V \right) = \frac{{RT}}{{V – b}} – \frac{a}{{{V^2}}}.$

We write the expressions for the first and second derivative assuming that the temperature $$T$$ is constant:

${{\left( {\frac{{\partial p}}{{\partial V}}} \right)_T} } = {p’\left( V \right) } = {{\left( {\frac{{RT}}{{V – b}} – \frac{a}{{{V^2}}}} \right)^\prime } } = { – \frac{{RT}}{{{{\left( {V – b} \right)}^2}}} + \frac{{2a}}{{{V^3}}};}$

${{\left( {\frac{{{\partial ^2}p}}{{\partial {V^2}}}} \right)_T} } = {p^{\prime\prime}\left( V \right) } = {{\left( { – \frac{{RT}}{{{{\left( {V – b} \right)}^2}}} + \frac{{2a}}{{{V^3}}}} \right)^\prime } } = {\frac{{2RT}}{{{{\left( {V – b} \right)}^3}}} – \frac{{6a}}{{{V^4}}}.}$

Find the values of $$V$$ and $$T,$$ at which both derivatives are equal to zero:

${\left\{ \begin{array}{l} p’\left( V \right) = 0\\ p^{\prime\prime}\left( V \right) = 0 \end{array} \right.,\;\;}\Rightarrow {\left\{ \begin{array}{l} – \frac{{RT}}{{{{\left( {V – b} \right)}^2}}} + \frac{{2a}}{{{V^3}}} = 0\\ \frac{{2RT}}{{{{\left( {V – b} \right)}^3}}} – \frac{{6a}}{{{V^4}}} = 0 \end{array} \right.,\;\;}\Rightarrow {\left\{ \begin{array}{l} \frac{{RT}}{{{{\left( {V – b} \right)}^2}}} = \frac{{2a}}{{{V^3}}}\\ \frac{{2RT}}{{{{\left( {V – b} \right)}^3}}} = \frac{{6a}}{{{V^4}}} \end{array} \right..}$

Divide the second equation by the first:

${\frac{{2RT}}{{{{\left( {V – b} \right)}^3}}}:\frac{{RT}}{{{{\left( {V – b} \right)}^2}}} = \frac{{6a}}{{{V^4}}}:\frac{{2a}}{{{V^3}}},\;\;}\Rightarrow {\frac{{2\cancel{RT} \cdot \cancel{{\left( {V – b} \right)}^2}}}{{{{\left( {V – b} \right)}^{\cancel{3}}} \cdot \cancel{RT}}} = \frac{{6\cancel{a} \cdot \cancel{V^3}}}{{{V^{\cancel{4}}} \cdot 2\cancel{a}}},\;\;}\Rightarrow {\frac{2}{{V – b}} = \frac{3}{V},\;\;}\Rightarrow {2V = 3\left( {V – b} \right),\;\;}\Rightarrow {2V = 3V – 3b,\;\;}\Rightarrow {V = 3b.}$

We find $$T$$ from the first equation:

${\frac{{RT}}{{{{\left( {V – b} \right)}^2}}} = \frac{{2a}}{{{V^3}}},\;\;}\Rightarrow {T = \frac{{2a{{\left( {V – b} \right)}^2}}}{{R{V^3}}} } = {\frac{{2a{{\left( {3b – b} \right)}^2}}}{{R{{\left( {3b} \right)}^3}}} } = {\frac{{2a \cdot 4{b^2}}}{{R \cdot 27{b^3}}} = \frac{{8a}}{{27bR}}.}$

Finally, we compute $$p$$ from the original Van der Waals equation:

${p = \frac{{RT}}{{V – b}} – \frac{a}{{{V^2}}} } = {\frac{{R \cdot \frac{{8a}}{{27bR}}}}{{3b – b}} – \frac{a}{{{{\left( {3b} \right)}^2}}} } = {\frac{{4a}}{{27{b^2}}} – \frac{2}{{9{b^2}}} } = {\frac{a}{{27{b^2}}}.}$

As it can be seen, the obtained values are completely consistent with the critical values of $${V_K}$$, $${p_K}$$ and $${T_K}$$ derived above.

To investigate the nature of the critical point in more detail, we calculate the third derivative in it:

${{\left( {\frac{{{\partial ^3}p}}{{\partial {V^3}}}} \right)_T} } = {p^{\prime\prime\prime}\left( V \right) } = {{\left( {\frac{{2RT}}{{{{\left( {V – b} \right)}^3}}} – \frac{{6a}}{{{V^4}}}} \right)^\prime } } = { – \frac{{6RT}}{{{{\left( {V – b} \right)}^4}}} + \frac{{24a}}{{{V^5}}};}$

${\Rightarrow {\left( {\frac{{{\partial ^3}p}}{{\partial {V^3}}}} \right)_{T = {T_K}}} } = {p^{\prime\prime\prime}{\left( V \right)_{T = {T_K}}} } = { – \frac{{6R \cdot \frac{{8a}}{{27bR}}}}{{{{\left( {3b – b} \right)}^4}}} + \frac{{24a}}{{{{\left( {3b} \right)}^5}}} } = { – \frac{{3a}}{{27{b^5}}} + \frac{{8a}}{{81{b^5}}} } = {\frac{{ – 9a + 8a}}{{81{b^5}}} } = { – \frac{a}{{81{b^5}}} \lt 0.}$

Thus, the third derivative at the critical point is negative. Therefore, the critical point is the point of inflection of the function $$p\left( V \right).$$ However, according to the third derivative test there is no extremum at this point.

### Example 2.

Investigate the dependence of $$pV$$ on $$V$$ for real gases.

Solution.

For an ideal gas, the product of the pressure and volume in an isothermal process remains constant:

$pV = RT = \text{const}.$

(The equation is written for $$1$$ mole of gas.)

For a real gas, the situation may be quite different. In order to investigate this topic, we write the Van der Waals equation in the form

$p = \frac{{RT}}{{V – b}} – \frac{a}{{{V^2}}}.$

Consequently,

$pV = \frac{{RTV}}{{V – b}} – \frac{a}{V}.$

Examine the product $$pV$$ for relative extrema. The derivative of $$pV$$ with respect to $$V$$ is given by

${{\left( {pV} \right)^\prime } } = {{\left( {\frac{{RTV}}{{V – b}} – \frac{a}{V}} \right)^\prime } } = {RT \cdot \frac{{1 \cdot \left( {V – b} \right) – V \cdot 1}}{{{{\left( {V – b} \right)}^2}}} + \frac{a}{{{V^2}}} } = { – \frac{{bRT}}{{{{\left( {V – b} \right)}^2}}} + \frac{a}{{{V^2}}}.}$

Equating the derivative to zero, we find the critical points:

${{\left( {pV} \right)^\prime } = 0,\;\;}\Rightarrow { – \frac{{bRT}}{{{{\left( {V – b} \right)}^2}}} + \frac{a}{{{V^2}}} = 0,\;\;}\Rightarrow {\frac{{ – bRT{V^2} + a{{\left( {V – b} \right)}^2}}}{{{V^2}{{\left( {V – b} \right)}^2}}} = 0,\;\;}\Rightarrow { – bRT{V^2} + a{\left( {V – b} \right)^2} = 0,\;\;}\Rightarrow { – bRT{V^2} + a\left( {{V^2} – 2bV + {b^2}} \right) = 0,\;\;}\Rightarrow { – bRT{V^2} + a{V^2} – 2abV + a{b^2} = 0,\;\;}\Rightarrow {\left( {a – bRT} \right){V^2} – 2abV + a{b^2} = 0.}$

We obtained a quadratic equation. Its discriminant is equal to

${D = 4{a^2}{b^2} – 4a{b^2}\left( {a – bRT} \right) } = {\cancel{4{a^2}{b^2}} – \cancel{4{a^2}{b^2}} + 4a{b^3}RT } = {4a{b^3}RT > 0.}$

We see that the discriminant is always positive, i.e. the quadratic equation has two real roots. They are given by the following formula:

${{V_{1,2}} = \frac{{2ab \pm \sqrt {4a{b^3}RT} }}{{2\left( {a – bRT} \right)}} } = {\frac{{ab \pm ab\sqrt {\frac{b}{a}RT} }}{{a\left( {1 – \frac{b}{a}RT} \right)}} } = {\frac{{b\left( {1 \pm \sqrt {\frac{b}{a}RT} } \right)}}{{1 – \frac{b}{a}RT}}.}$

The first root $${V_1}$$ with a minus sign in the numerator has no physical meaning since $${V_1} \lt b.$$ Therefore, the product $$pV$$ has an extremum at

${{V_2} = {V_{\min}} = \frac{{b\left( {1 + \sqrt {\frac{b}{a}RT} } \right)}}{{1 – \frac{b}{a}RT}} } = {\frac{{b\left( {1 + \sqrt {\frac{b}{a}RT} } \right)}}{{1 – {{\left( {\sqrt {\frac{b}{a}RT} } \right)}^2}}} } = {\frac{{b\cancel{\left( {1 + \sqrt {\frac{b}{a}RT} } \right)}}}{{\left( {1 – \sqrt {\frac{b}{a}RT} } \right)\cancel{\left( {1 + \sqrt {\frac{b}{a}RT} } \right)}}} } = {\frac{b}{{1 – \sqrt {\frac{b}{a}RT} }}.}$

Provided $$1 – {\large\frac{b}{a}\normalsize} RT > 0$$, the branches of the parabola corresponding to the numerator of the derivative will be directed upwards. Therefore, the derivative changes sign from minus to plus when passing through the right root $${V_2}.$$ This means that $${V_2}$$ is the minimum point of the function $$pV:$$ $${V_2} = {V_{\min}}.$$

Thus, when a rarefied Van der Waals gas is compressed at a constant temperature, the product $$pV$$ initially decreases, reaching a minimum value $${\left( {pV} \right)_{\min }},$$ and then begins to increase. The first stage is explained by the dominating attractive forces between the molecules, and the second stage is associated with increased contribution of repulsive forces.

The minimum value $${\left( {pV} \right)_{\min }}$$ is determined by the following expression:

${{\left( {pV} \right)_{\min }} } = {\frac{{RT{V_{\min }}}}{{{V_{\min }} – b}} – \frac{a}{{{V_{\min }}}} } = {\frac{{RT \cdot \frac{b}{{1 – \sqrt {\frac{b}{a}RT} }}}}{{\frac{b}{{1 – \sqrt {\frac{b}{a}RT} }} – b}} – \frac{a}{{\frac{b}{{1 – \sqrt {\frac{b}{a}RT} }}}} } = {\frac{{\frac{{bRT}}{\cancel{1 – \sqrt {\frac{b}{a}RT} }}}}{{\frac{{b – b\left( {1 – \sqrt {\frac{b}{a}RT} } \right)}}{\cancel{1 – \sqrt {\frac{b}{a}RT} }}}} – \frac{{a\left( {1 – \sqrt {\frac{b}{a}RT} } \right)}}{b} } = {\frac{{bRT}}{{b\left( {\cancel{1} – \cancel{1} + \sqrt {\frac{b}{a}RT} } \right)}} – \frac{a}{b} }+{ \frac{a}{b}\sqrt {\frac{b}{a}RT} } = {\sqrt {\frac{a}{b}RT} – \frac{a}{b} + \sqrt {\frac{a}{b}RT} } = {2\sqrt {\frac{a}{b}RT} – \frac{a}{b}.}$

Note that the value $${V_{\min}}$$ increases as the temperature rises and becomes equal to infinity when the temperature is

$T = {T_B} = \frac{a}{{2R}}.$

This transition point $${T_B}$$ is called the Boyle temperature. Below this temperature, the product $$pV$$ has a minimum in the isothermal compression. At the temperatures higher than $${T_B},$$ the product $$pV$$ will increase monotonically in the compression process.

Some gases have a rather low value of the Boyle temperature. For example, for Helium $$He$$ it is $${T_B} = 22,6\;\text{K},$$ for Neon $$Ne$$ it is equal to $${T_B} = 122,1\;\text{K}.$$ Consequently at room temperatures, such gases exhibit monotonic dependence of $$pV$$ on $$V.$$