Question

The van der Waals equation of state represents an approximate behaviour of gases at high pressure and is given by $(P+\frac{a{n}^2}{V^2})(V-nb)=nRT$ where a and b are constants having different values for different gases. In a special case if $a=0,b=0$ van der Waals equation will take the shape of ideal gas equation. For ethane $\left(C_2{H}_6\right)$ gas $a=0.554J{m}^{3}mo{l}^{-2}$ and $b=6.38\times {10}^{-5}{m}^{3}mo{l}^{-1}$. 1.8 mole of ethane allowed to expand from $2\times {10}^{-3}{m}^3$ to $4\times {10}^{-3}{m}^3$ isothermally at $300K$.

Find the work done using van der Waal's equation in an isothermal condition.

• A. $nRT{\log }_e\frac{V_2-nb}{V_1-nb}+a{n}^2[\frac{1}{V_2}-\frac{1}{V_1}]$
• B. $nRT{\log }_e\frac{V_2-nb}{V_1-nb}+a{n}^2[\frac{1}{V_1}-\frac{1}{V_2}]$
• C. $nRT{\log }_e\frac{V_1-nb}{V_2-nb}+a{n}^2[\frac{1}{V_1}-\frac{1}{V_2}]$
• D. None of these

Answer 1

The correct option is A $nRT{\log }_e\frac{V_2-nb}{V_1-nb}+a{n}^2[\frac{1}{V_2}-\frac{1}{V_1}]$

$dW=PdV$
$P=\frac{nRT}{V-nb}-\frac{a{n}^2}{V^2}$

$W=\int dW={\int }_{V_1}^{V_2}(\frac{nRT}{V-nb}-\frac{a{n}^2}{V^2})$
$=nRT{\log }_e\frac{V_2-nb}{V_1-nb}-a{n}^2[\frac{1}{V_1}-\frac{1}{V_2}]$
or
$=nRT{\log }_e\frac{V_2-nb}{V_1-nb}+a{n}^2[\frac{1}{V_2}-\frac{1}{V_1}]$