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

Correct Answer: $${\left( {\frac{{\partial {\text{S}}}}{{\partial {\text{V}}}}} \right)_{\text{T}}} = {\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{T}}}}} \right)_{\text{V}}}$$

Helmholtz function :
\
So, we can derive the Maxwell function : \
Bissoy MCQ
#Chemical Engineering #Chemical Engineering Thermodynamics

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