Van der Waals Equation
In the examples below, we focus on the investigation of the Van der Waals equation with the help of the derivative.
Solved Problems
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
\[D = 4{a^2}{b^2} - 4a{b^2}\left( {a - bRT} \right) = \cancel{4{a^2}{b^2}} - \cancel{4{a^2}{b^2}} + 4a{b^3}RT = 4a{b^3}RT \gt 0.\]
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 \(1 - {\frac{b}{a}} RT \gt 0\), 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 \(V.\)
