The Maxwell relation derived from the differential expression for the Helmholtz free energy (dA) is

A. $${\left( {\frac{{\partial {\text{T}}}}{{\partial {\text{V}}}}} \right)_{\text{S}}} = - {\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{S}}}}} \right)_{\text{V}}}$$

B. $${\left( {\frac{{\partial {\text{S}}}}{{\partial {\text{P}}}}} \right)_{\text{T}}} = - {\left( {\frac{{\partial {\text{V}}}}{{\partial {\text{T}}}}} \right)_{\text{P}}}$$

C. $${\left( {\frac{{\partial {\text{V}}}}{{\partial {\text{S}}}}} \right)_{\text{P}}} = {\left( {\frac{{\partial {\text{T}}}}{{\partial {\text{P}}}}} \right)_{\text{S}}}$$

D. $${\left( {\frac{{\partial {\text{S}}}}{{\partial {\text{V}}}}} \right)_{\text{T}}} = {\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{T}}}}} \right)_{\text{V}}}$$

Answer: Option D

Solution(By Examveda Team)

Helmholtz function :
\[\begin{array}{l} A = U - TS\\ \Rightarrow dA = - PdV - SdT \end{array}\]
So, we can derive the Maxwell function : \[{\left( {\frac{{\partial P}}{{\partial T}}} \right)_V} = {\left( {\frac{{\partial S}}{{\partial V}}} \right)_T}\]