The expression for entropy change given by, S = nRl n ( V_2 V_1 ) + n C_ vl n ( T_2 T_1 ) is valid for

Answer: Option D

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) $$S=fn(T,V)$$

$$\begin{array}{rl}& \Rightarrow dS={\left(\frac{\partial S}{\partial T}\right)}_{V}dT+{\left(\frac{\partial S}{\partial v}\right)}_{T}dT\\ & \Rightarrow dS=\frac{C_{V}dT}{T}+{\left(\frac{\partial P}{\partial T}\right)}_{V}dV\left(\text{for one mole of ideal gas}{\left(\frac{\partial P}{\partial T}\right)}_V=\frac{R}{V}\right)\\ & \Rightarrow dS=\frac{C_V}{T}dT+\frac{nR}{V}dV\end{array}$$
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.