UPSEE Solved Model Paper 6

Here, we have to find out the work done in an isothermal expansion of gas. We have, The work done by a gas when it expands by a very small volume, dV under a pressure, P. dW = PdV Hence, the total work done by a gas when the volume of one mole of gas increases from $V_1$ to $V_2$. W = $\int\limits_{V_2}^{V_2} P \, dV$ = $\int\limits_{V_1}^{V_2} \frac{RT}{V} \, dV$ [Since PV = RT ⇒ P = $\frac{RT}{V}$] Since, the temperature, T is a constant for an isothermal change, W = $RT \int\limits_{V_1}^{V_2} \frac{dV}{V}$ = $RT \left[ \log_e V \right]_{V_1}^{V_2}$ = $RT(\log_e V_2 - \log_e V_1)$ = $RT \log_e \frac{V_2}{V_1}$ But, $P_1V_1$ = $P_2V_2$ = RT ∴ W = $P_1V_1 \log_e \left( \frac{V_2}{V_1} \right)$ That is, the work done in an isothermal expansion of gas is given by W = $P_1V_1 \log_e \frac{V_2}{V_1}$