Using an appropriate cyclic rule and an appropriate Maxwell relation, show that (∂P / ∂V)S = γ (∂P / ∂V)T where γ = CP / CV

We know that cyclic is given as:

(∂P / ∂V)S (∂V / ∂S)P (∂S / ∂P)V = -1

(∂P / ∂V)S = – [(∂S / ∂V) P]/[(∂S / ∂P)V] = [- (∂S / ∂T)P (∂T / ∂V)P]/[(∂S / ∂T) V (∂T / ∂P)V]

= -[ (CP / T) (∂T / ∂V) P]/[ (CV / T) (∂T / ∂P)V ]

= – γ/(∂V / ∂T) P (∂T / ∂P)V

Also, (∂V / ∂T) P (∂T / ∂P) V (∂P / ∂V) T = -1

Therefore, (∂V / ∂T) P (∂T / ∂P)V = – 1 / (∂P / ∂V) T

Hence, (∂P / ∂V) S = γ((∂P / ∂V) T)

Category: Second Law of Thermodynamics

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