Chapter 2, Problem 2D.5P

Interpretation Introduction

Interpretation: The values of the isothermal compressibility and the expansion coefficient of a van der Waals gas have to be calculated.  Also, it has to be shown that ${\kappa }_{T}R=\alpha \left(V_{\text{m}}-b\right)$ with the help of Euler’s chain relation.

Concept introduction: The van der Waals equation defined the behavior of real gases.  This includes the forces of attraction and repulsion in the real gases.  This is represented by the formula given below as,

    $p=\frac{nRT}{V-nb}-\frac{n^{2}a}{V^2}$

Answer to Problem 2D.5P

The expansion coefficient of a van der Waals gas is,

    $\alpha =\frac{R}{P{V}_{\text{m}}-\frac{a}{V_{\text{m}}}+\frac{2ab}{V_{\text{m}}^2}}$

The isothermal compressibility of a van der Waals gas is,

    ${\kappa }_T=\frac{\left(V_{\text{m}}-b\right)}{p{V}_{\text{m}}-\frac{a}{V_{\text{m}}}+\frac{2ab}{V_{\text{m}}^2}}$

It is proved that ${\kappa }_{T}R=\alpha \left(V_{\text{m}}-b\right)$.

Explanation of Solution

The expression for one mole of a van der Waals gas is,

    $\left(p+\frac{a}{V_{\text{m}}^2}\right)\left(V_{\text{m}}-b\right)=RT$                                                                                 (1)

The above expression is also written as,

    $\begin{array}{c}p{V}_{\text{m}}+\frac{a}{V_{\text{m}}}-pb-\frac{ab}{V_{\text{m}}^2}=RT\\ p{V}_{\text{m}}^3+a{V}_{\text{m}}-p{V}_{\text{m}}^{2}b-ab=V_{\text{m}}^{2}RT\end{array}$

Differentiate the above equation as follows,

    $V_{\text{m}}^{3}dp+3V_{\text{m}}^{2}pd{V}_{\text{m}}+ad{V}_{\text{m}}-V_{\text{m}}^{2}bdp-2p{V}_{\text{m}}bd{V}_{\text{m}}=2V_{\text{m}}RTd{V}_{\text{m}}+V_{\text{m}}^{2}RdT$        (2)

At constant pressure, $dp=0$.  Divide the above equation by $dT$, it is simplified as,

    $\begin{array}{c}3V_{\text{m}}^{2}p{\left(\frac{\partial V_{\text{m}}}{\partial T}\right)}_p+a{\left(\frac{\partial V_{\text{m}}}{\partial T}\right)}_p-2p{V}_{\text{m}}b{\left(\frac{\partial V_{\text{m}}}{\partial T}\right)}_p=2V_{\text{m}}RT{\left(\frac{\partial V_{\text{m}}}{\partial T}\right)}_p+V_{\text{m}}^{2}R\\ {\left(\frac{\partial V_{\text{m}}}{\partial T}\right)}_p=\frac{V_{\text{m}}^{2}R}{3V_{\text{m}}^{2}p+a-2p{V}_{\text{m}}b-2V_{\text{m}}RT}\end{array}$ (3)

Now, the value of the expansion coefficient of a van der Waals gas is,

    $\alpha =\frac{1}{V}{\left(\frac{\partial V}{\partial T}\right)}_p$                                                                                                 (4)

Substitute the value of ${\left(\frac{\partial V_{\text{m}}}{\partial T}\right)}_p$ from equation (3) to equation (4).

    $\alpha =\frac{R}{3V_{\text{m}}p+\frac{a}{V_{\text{m}}}-2pb-2RT}$

As, it is known that $\left(p+\frac{a}{V_{\text{m}}^2}\right)\left(V_{\text{m}}-b\right)=RT$.  Therefore, the above expression becomes,

    $\begin{array}{c}\alpha =\frac{R}{3V_{\text{m}}p+\frac{a}{V_{\text{m}}}-2pb-2\left(p+\frac{a}{V_{\text{m}}^2}\right)\left(V_{\text{m}}-b\right)}\\ =\frac{R}{3V_{\text{m}}p+\frac{a}{V_{\text{m}}}-2pb-2P{V}_{\text{m}}+2pb-\frac{2a{V}_{\text{m}}}{V_{\text{m}}^2}+\frac{2ab}{V_{\text{m}}^2}}\\ =\frac{R}{P{V}_{\text{m}}-\frac{a}{V_{\text{m}}}+\frac{2ab}{V_{\text{m}}^2}}\end{array}$

Hence, the expansion coefficient of a van der Waals gas is,

    $\alpha =\frac{R}{P{V}_{\text{m}}-\frac{a}{V_{\text{m}}}+\frac{2ab}{V_{\text{m}}^2}}$

Now, At constant temperature, $dT=0$.  Divide the above equation (2) by $dp$, it is simplified as,

    $V_{\text{m}}^3+3V_{\text{m}}^{2}p{\left(\frac{\partial V_{\text{m}}}{\partial p}\right)}_T+a{\left(\frac{\partial V_{\text{m}}}{\partial p}\right)}_T-V_{\text{m}}^{2}b{\left(\frac{\partial V_{\text{m}}}{\partial p}\right)}_T-2p{V}_{\text{m}}b{\left(\frac{\partial V_{\text{m}}}{\partial p}\right)}_T=2V_{\text{m}}RT{\left(\frac{\partial V_{\text{m}}}{\partial p}\right)}_T$

The above expression is simplified as,

    $\begin{array}{c}{\left(\frac{\partial V_{\text{m}}}{\partial p}\right)}_T=\frac{V_{\text{m}}^{2}b-V_{\text{m}}^3}{3V_{\text{m}}^{2}p+a-2p{V}_{\text{m}}b-2V_{\text{m}}RT}\\ =\frac{V_{\text{m}}\left(b-V_{\text{m}}\right)}{3V_{\text{m}}^{2}p+a-2p{V}_{\text{m}}b-2V_{\text{m}}RT}\end{array}$

As, it is known that $\left(p+\frac{a}{V_{\text{m}}^2}\right)\left(V_{\text{m}}-b\right)=RT$.  Therefore, the above expression becomes,

    $\begin{array}{c}{\left(\frac{\partial V_{\text{m}}}{\partial p}\right)}_T=\frac{V_{\text{m}}\left(b-V_{\text{m}}\right)}{3V_{\text{m}}^{2}p+a-2p{V}_{\text{m}}b-2V_{\text{m}}\left(p+\frac{a}{V_{\text{m}}^2}\right)\left(V_{\text{m}}-b\right)}\\ =\frac{V_{\text{m}}\left(b-V_{\text{m}}\right)}{3V_{\text{m}}^{2}p+a-2p{V}_{\text{m}}b-2V_{\text{m}}\left(p+\frac{a}{V_{\text{m}}^2}\right)\left(V_{\text{m}}-b\right)}\\ =\frac{V_{\text{m}}\left(b-V_{\text{m}}\right)}{p{V}_{\text{m}}-\frac{a}{V_{\text{m}}}+\frac{2ab}{V_{\text{m}}^2}}\end{array}$                                 (5)

Now, the value of isothermal compressibility is given by the expression,

    ${\kappa }_T=-\frac{1}{V_{\text{m}}}{\left(\frac{\partial V_{\text{m}}}{\partial p}\right)}_T$                                                                                          (6)

Substitute the value of ${\left(\frac{\partial V_{\text{m}}}{\partial p}\right)}_T$ from equation (5) to equation (6).

    $\begin{array}{c}{\kappa }_T=-\frac{1}{V_{\text{m}}}\times \frac{V_{\text{m}}\left(b-V_{\text{m}}\right)}{p{V}_{\text{m}}-\frac{a}{V_{\text{m}}}+\frac{2ab}{V_{\text{m}}^2}}\\ =\frac{\left(V_{\text{m}}-b\right)}{p{V}_{\text{m}}-\frac{a}{V_{\text{m}}}+\frac{2ab}{V_{\text{m}}^2}}\end{array}$

Hence, the isothermal compressibility of a van der Waals gas is,

    ${\kappa }_T=\frac{\left(V_{\text{m}}-b\right)}{p{V}_{\text{m}}-\frac{a}{V_{\text{m}}}+\frac{2ab}{V_{\text{m}}^2}}$

According to Euler’s chain relation, the partial derivates are expressed as,

    ${\left(\frac{\partial V_{\text{m}}}{\partial p}\right)}_T{\left(\frac{\partial T}{\partial V_{\text{m}}}\right)}_p{\left(\frac{\partial p}{\partial T}\right)}_{V_{\text{m}}}=-1$                                                                      (7)

Since, ${\left(\frac{\partial V_{\text{m}}}{\partial p}\right)}_T=-{\kappa }_T{V}_{\text{m}}$                                                                                          (8)

And, ${\left(\frac{\partial T}{\partial V_{\text{m}}}\right)}_p=\frac{1}{\alpha V_{\text{m}}}$                                                                                                (9)

Substitute the values of ${\left(\frac{\partial V_{\text{m}}}{\partial p}\right)}_T$ and ${\left(\frac{\partial T}{\partial V_{\text{m}}}\right)}_p$ from equations (8) and (9) to equation (7).

    $\begin{array}{c}-{\kappa }_T{V}_{\text{m}}\times \frac{1}{\alpha V_{\text{m}}}{\left(\frac{\partial p}{\partial T}\right)}_{V_{\text{m}}}=-1\\ {\kappa }_T{\left(\frac{\partial p}{\partial T}\right)}_{V_{\text{m}}}=\alpha \end{array}$                                                                          (10)

At constant volume, $dV=0$.  Divide the above equation (2) by $dT$, it is simplified as,

    $\begin{array}{c}{V}_{\text{m}}^3{\left(\frac{\partial p}{\partial T}\right)}_{V_{\text{m}}}-V_{\text{m}}^{2}b{\left(\frac{\partial p}{\partial T}\right)}_{V_{\text{m}}}=V_{\text{m}}^{2}R\\ {\left(\frac{\partial p}{\partial T}\right)}_{V_{\text{m}}}=\frac{R}{V_{\text{m}}-b}\end{array}$                                                              (11)

Now, substitute the value of ${\left(\frac{\partial p}{\partial T}\right)}_{V_{\text{m}}}$ from equation (11) to equation (10).

    $\begin{array}{c}{\kappa }_T\times \frac{R}{V_{\text{m}}-b}=\alpha \\ {\kappa }_{T}R=\alpha \left(V_{\text{m}}-b\right)\end{array}$

Hence, it is proved that ${\kappa }_{T}R=\alpha \left(V_{\text{m}}-b\right)$.