Question

If one mole of a van der Waals’ gas undergoes reversible isothermal transformation from an initial volume $V_{1}$ to a final volume $V_{2}$ , the expression for the work done is

• A. $RT\, ln\, \frac{V_{2}}{V_{1}} +a (V_{2}-V_{1})$
• B. $-nRT\, ln \, \frac{V_{2}-b}{V_{1}-b}+a \left(\frac{1}{V_{1}}-\frac{1}{V_{2}}\right)$
• C. $RT\, ln \, \frac{P_{2}}{P_{1}}$
• D. $RT\, ln \,\frac{V_{2}-b}{V_{1}-b}-a \left(\frac{1}{V_{1}}-\frac{1}{V_{2}}\right)$

Answer 1

Answer: (B)

For the van der Waals’ gas
$w=-\int\limits_{V_1}^{V_{2}} p_{\text{external}} dV =-\int\limits_{V_1}^{V_{2}} \left(\frac{RT}{V_{m}-b}-\frac{a}{V_{m}^{2}}\right)dV$
$=-\int\limits_{V_{1}}^{V_{2}}\left(\frac{R T}{V_{m}-b}\right) d V+\int\limits_{V_{1}}^{V_{2}}\left(\frac{a}{V_{m}^{2}}\right) d V$
The first integral is solved by making the substit
$y=V_{m}-b$
$=-\int\limits_{V_{1}}^{V_{2}}\left(\frac{R T}{V_{m}-b}\right) d V=-\int\limits_{V_{1}}^{V_{2}}\left(\frac{R T}{y^{*}}\right) d y$
$=-R T\left[\ln \left(V_{2}+b\right)-\ln \left(V_{1}+b\right)\right]$
Therefore, the work is given by $w=-nRT \ln \left(\frac{V_{2}-b}{V_{1}-b}\right)+a\left(\frac{1}{V_{1}}-\frac{1}{V_{2}}\right)$