\(\triangle S = C_vdn\frac {T_2}{T_1} + nR\,ln\frac{V_2}{V_1}\)
\(\because \triangle T = 0 \) (isothermally)
\(\therefore \triangle S = nR\, ln \frac {V_2}{V_1}\) ......(i)
Using boyle's law PV = constant
\(P_1V_1 = P_2V_2 = constant\)
\(\frac{V_2}{V_1} = \frac {P_1}{P_2}\) ......(ii)
From equation (i) & (ii)
\(\triangle S = nR\, ln \frac {P_1}{P_2}\)
\(\triangle S =- nR\, ln \frac {P_2}{P_1}\)
Using ideal gas equation-
\(PV = nRT\)
\(P_1 \times 20 = 2 \times 0.0821 \times 243.6\)
\(P_1 = 2atm\)
given, \(P_2 = 1atm\)
\(\triangle S =- 2\times 8.314\; ln\frac 12\)
\(= 2 \times 8.34 \times ln2\)
\(= 11.53 J/K\)
\(\triangle S = 2.76 \,cal/K\)
