Q.19 What is an isothermal process? Obtain an expression for work done by a gas in an isothermal process.
Best answer
Answer: A process in which the temperature of the system remains constant throughout is an isothermal process. Work done: \(W = nRT\ln\dfrac{V_f}{V_i} = 2.303\,nRT\log_{10}\dfrac{V_f}{V_i}\).
For an isothermal change the gas is kept in contact with a large heat reservoir and the change is carried out very slowly, so \(\Delta T = 0\) and, for an ideal gas, \(\Delta U = 0\).
Derivation: Consider \(n\) moles of an ideal gas at temperature \(T\) expanding isothermally from volume \(V_i\) to \(V_f\). When the volume increases by \(dV\) at pressure \(P\), the work done is \(dW = P\,dV\).
Total work: \(W = \displaystyle\int_{V_i}^{V_f} P\,dV\). For an ideal gas \(PV = nRT\), so \(P = \dfrac{nRT}{V}\).
\(W = \displaystyle\int_{V_i}^{V_f}\dfrac{nRT}{V}dV = nRT\displaystyle\int_{V_i}^{V_f}\dfrac{dV}{V} = nRT\,[\ln V]_{V_i}^{V_f}\) (T is constant)
\(W = nRT\ln\dfrac{V_f}{V_i} = 2.303\,nRT\log_{10}\dfrac{V_f}{V_i}\).
Since \(P_iV_i = P_fV_f\), this can also be written \(W = 2.303\,nRT\log_{10}\dfrac{P_i}{P_f}\). In an expansion \(W \gt 0\); as \(\Delta U = 0\), \(Q = W\).
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