Obtain an expression for work done by an ideal gas in an isothermal expansion.

(N/A) An isothermal process is a thermodynamic process in which the temperature of the system remains constant throughout.
For an ideal gas,the equation of state is $PV = \mu RT$. Since $T$ is constant,$PV = \text{constant}$.
Work done by a gas during a small change in volume $dV$ is given by $dW = P dV$.
To find the total work done during expansion from initial volume $V_1$ to final volume $V_2$,we integrate the expression:
$W = \int_{V_1}^{V_2} P dV$
Substituting $P = \frac{\mu RT}{V}$ into the integral:
$W = \int_{V_1}^{V_2} \frac{\mu RT}{V} dV$
Since $\mu, R,$ and $T$ are constants for an isothermal process:
$W = \mu RT \int_{V_1}^{V_2} \frac{1}{V} dV$
$W = \mu RT [\ln V]_{V_1}^{V_2}$
$W = \mu RT (\ln V_2 - \ln V_1)$
$W = \mu RT \ln \left( \frac{V_2}{V_1} \right)$
Since $P_1 V_1 = P_2 V_2$,we can also write this as $W = \mu RT \ln \left( \frac{P_1}{P_2} \right)$.