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

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