Using an appropriate cyclic rule and an appropriate Maxwell relation, show that (∂P / ∂V)_S = γ (∂P / ∂V)_T where γ = C_{P / 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]

= -[ (C_P / T) (∂T / ∂V) _P]/[ (C_V / 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)