Question

The equation of state for one mole of a non-ideal gas is given by $PV = A\left(1 + \dfrac{B}{V}\right)$, where the coefficients $A$ and $B$ are temperature dependent. If the volume changes from $V_1$ to $V_2$ in an isothermal process, the work done by the gas is

• A. $AB\left(\dfrac{1}{V_1} - \dfrac{1}{V_2}\right)$
• B. $AB\,\ln\left(\dfrac{V_2}{V_1}\right)$
• C. $A\ln\left(\dfrac{V_2}{V_1}\right) + AB\left(\dfrac{1}{V_1} - \dfrac{1}{V_2}\right)$
• D. $A\ln\left(\dfrac{V_2 - V_1}{V_1}\right) + B$

Hint:

Always rewrite non-ideal equations of state in terms of $P(V)$ to simplify $\int P\, dV$.

Answer 1

The Correct Option is C

Step 1: Start from the definition of work.
For expansion from $V_1$ to $V_2$: $W = \displaystyle \int_{V_1}^{V_2} P\, dV$.

Step 2: Substitute the equation of state.
$P = \dfrac{A}{V} + \dfrac{AB}{V^2}$.
Thus, $W = \displaystyle \int_{V_1}^{V_2} \left( \dfrac{A}{V} + \dfrac{AB}{V^2} \right) dV$.

Step 3: Integrate term-by-term.
$\int \dfrac{A}{V}\, dV = A \ln\left(\dfrac{V_2}{V_1}\right)$,
$\int \dfrac{AB}{V^2}\, dV = AB\left(\dfrac{1}{V_1} - \dfrac{1}{V_2}\right)$.

Step 4: Final expression.
$W = A \ln\left(\dfrac{V_2}{V_1}\right) + AB\left(\dfrac{1}{V_1} - \dfrac{1}{V_2}\right)$.

Step 5: Conclusion.
This matches option (C).