Question
Derive an expression for the work done in an isothermal process.
✓
Answer
For a small change in volume, work done is given by, DW =P dV We, know, PV = nRT $\Rightarrow\text{P}=\frac{\text{nRT}}{\text{V}}$ For T = costant, $\text{dW}=\text{nRT}=\frac{\text{dV}}{\text{V}}$ Net work done under isothermal condition to change the valume from $V_i$ to $V_f$ is, $\text{W}=\int\limits^{\text{V}_\text{f}}_{\text{V}_\text{i}}\text{dW}=\text{nRT}\int\limits^{\text{V}_\text{f}}_{\text{V}_\text{i}}\frac{\text{dV}}{\text{V}}$
$=\text{nRT}\Big|\log_\text{e}\text{V}\Big|^{\text{V}_\text{f}}_{\text{V}_\text{i}}$
$\text{W}=\text{nRT}\log_\text{e}\Big(\frac{\text{V}_\text{f}}{\text{V}_\text{i}}\Big)$
$\therefore\text{W}=2.3026\text{ nRT }\log_{10}\Big(\frac{\text{V}_\text{f}}{\text{v}_i}\Big)$ Where n is the number of moles. If $P_f$ and $P_i$ are the pressures, we can also write, $\text{W}=2.3026\text{ nRT }\log_{10}\Big(\frac{\text{P}_\text{i}}{\text{P}_\text{i}}\Big)$
$=\text{nRT}\Big|\log_\text{e}\text{V}\Big|^{\text{V}_\text{f}}_{\text{V}_\text{i}}$
$\text{W}=\text{nRT}\log_\text{e}\Big(\frac{\text{V}_\text{f}}{\text{V}_\text{i}}\Big)$
$\therefore\text{W}=2.3026\text{ nRT }\log_{10}\Big(\frac{\text{V}_\text{f}}{\text{v}_i}\Big)$ Where n is the number of moles. If $P_f$ and $P_i$ are the pressures, we can also write, $\text{W}=2.3026\text{ nRT }\log_{10}\Big(\frac{\text{P}_\text{i}}{\text{P}_\text{i}}\Big)$
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