Question

The equation of state of a gas is given by $P(V-b)=nRT$. If $1\text{mole}$ of a gas is isothermally expanded from volume $V$ and $2V$, the work done during the process is

• A. $\text{RT ln}\left(\frac{2V-b}{V-b}\right)$
• B. $\text{RT ln}\left(\frac{V-b}{V}\right)$
• C. $\text{RT ln}\left(\frac{V-b}{2V-b}\right)$
• D. $\text{RT ln}\left(\frac{V}{V-b}\right)$

Answer 1

The correct option is A $\text{RT ln}\left(\frac{2V-b}{V-b}\right)$

Given, equation of state of gas
$P(V-b)=nRT$
Initial volume $\left(V_i\right)=V$
Final volume $\left(V_f\right)=2V$
Work done in an isothermal process,
$W={\int }_{V_i}^{V_f}PdV$
$W={\int }_V^{2V}\frac{nRT}{V-b}dV=nRT\ln (V-b){|}_V^{2V}$
For $1\text{mole}$,
$W=\text{RT ln}\left(\frac{2V-b}{V-b}\right)$