Question

Explain the work done in an isothermal process.

Answer 1

Essential conditions for an isothermal process.
(i) The process of compression of expansion should be infinitely slow. So as to provide a sufficient time for the exchange of heat.
(ii) The walls of the container must be perfectly conducting to allow free exchange of heat between container and surroundings.
Consider a cylinder with $\mu$, moles of an ideal gas at isothermal expansion, then it reaches its final state $(P_2,V_2)$ from initial state $(P_1,V_1)$ Meanwhile, in the state when pressure is $P$ and change in volume is $V+dV$ from $V$, then pressure $\left(P\right)$.
$dW=PdV$
Thus, total work done:
$E_{isothermal}={\int }_{V_1}^{V_2}PdV$ $...\left(13.20\right)$
From ideal gas equation for $\mu$ moles of a gas, we know that,
$PV=\mu RT$
$P=\frac{\mu RT}{V}$
From equation $\left(13.20\right),W_{isothermal}={\int }_{V_1}^{V^2}\frac{\mu RT}{V}dV$
$=\mu RT[{\log }_{e}V{]}_{V_1}^{V_2}$
$=\mu RT[{\log }_e{V}_2-{\log }_e{V}_1]$
$=\mu RT{\log }_e\frac{V_2}{V_1}$
$W_{isothermal}=2.303\mu RT{\log }_{10}\frac{V_2}{V_1}$ $...\left(13.20a\right)$
Since pressure and volume are inversely proportional to each other in an isothermal process, then from equation $\left(13.20a\right)$, from Boyle's law for ideal gas
$P_1{V}_1=P_2{V}_2$
$\frac{P_1}{P_2}=\frac{V_2}{V_1}$
$W_{isothermal}=2.303\mu RT{\log }_{10}\frac{P_1}{P_2}$ $...\left(13.20b\right)$
The above expression shows the work done for $p$ moles of an ideal gas in an isothermal process, State equation for isothermal process,
$PV=$ constant
$\therefore PdV+VdP=0$
$VdP=-PdV$
or, $\frac{dP}{dV}=\frac{-P}{V}$ $...\left(13.21\right)$
The slope for $P.V$ curve in an isothermal process is negative.