Define isothermal process. Obtain an expression for work done during isothermal process. Draw P–V diagram.

Answer: \(W = nRT\ln\dfrac{V_f}{V_i} = 2.303\,nRT\log_{10}\dfrac{V_f}{V_i}\)

Isothermal process: A thermodynamic process in which the temperature of the system remains constant throughout is called an isothermal process (\(\Delta T = 0\)). For an ideal gas, \(PV\) = constant and \(\Delta U = 0\).

Work done: Consider \(n\) moles of an ideal gas in a cylinder with a frictionless piston, kept in contact with a heat reservoir at temperature \(T\). Let the gas expand slowly (quasi-statically) from \((P_i, V_i)\) to \((P_f, V_f)\). For a small change in volume \(dV\), \(dW = P\,dV\). Total work

Since \(PV = nRT\), \(P = \dfrac{nRT}{V}\) with \(T\) constant:

Since \(P_iV_i = P_fV_f\), also \(W = nRT\ln\dfrac{P_i}{P_f} = 2.303\,nRT\log_{10}\dfrac{V_f}{V_i}\). For expansion \(W\) is positive; for compression it is negative. As \(\Delta U = 0\), the heat absorbed equals the work done, \(Q = W\).

P–V diagram: [Figure: pressure P on the vertical axis and volume V on the horizontal axis. A smooth curve (rectangular hyperbola, \(PV\) = constant) falls from point i \((P_i, V_i)\) to point f \((P_f, V_f)\), marked with an arrow and labelled “isotherm, T = constant”. The area under the curve between \(V_i\) and \(V_f\) is shaded and represents the work done \(W\).]