The expression for entropy change given by, $$\Delta {\text{S}} = {\text{nR}}l{\text{n}}\left( {\frac{{{{\text{V}}_2}}}{{{{\text{V}}_1}}}} \right) + {\text{n}}{{\text{C}}_{\text{v}}}l{\text{n}}\left( {\frac{{{{\text{T}}_2}}}{{{{\text{T}}_1}}}} \right)$$      is valid for

Solution(By Examveda Team)

For a pure substance in a closed system has a degree of freedom = 2
So, entropy(s) can be expressed as a function of two variables (properties) $$S = fn(T,V)$$

\[\begin{align} & \Rightarrow dS={{\left( \frac{\partial S}{\partial T} \right)}_{V}}dT+{{\left( \frac{\partial S}{\partial v} \right)}_{T}}dT \\ & \Rightarrow dS=\frac{{{C}_{V}}dT}{T}+{{\left( \frac{\partial P}{\partial T} \right)}_{V}}dV\left( \text{for one mole of ideal gas}{{\left( \frac{\partial P}{\partial T} \right)}_{V}}=\frac{R}{V} \right) \\ & \Rightarrow dS=\frac{{{C}_{V}}}{T}dT+\frac{nR}{V}dV \\ \end{align}\]
So, on integrating this equation from temperature T₂ to T₁ expanding volume from V₁ to V₂ GIVES the equation

$$\Delta S = nRln\left( {\frac{{{V_2}}}{{{V_1}}}} \right)\, + nCvln\left( {\frac{{{T_2}}}{{{T_1}}}} \right)$$

for an ideal gas of n moles.