Question

Starting from $C_v=T\left(\frac{\partial S}{\partial T}\right)_V$ and using the appropriate Maxwell relation, show that for a joule free expansion

$\frac{dT}{dV}=\frac{P-T\left(\frac{\partial P}{\partial \:T}\right)_V}{C_v\:}$

Answer 1

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Since, C_v = T(partial differential S/partial differential T)_V = partial differential V/partial differential T)_V rightdoublearrow Now (partial differential T/partial differential V)_V = -(partial differential T/partial differential U)_V (partial differential V/partial differential V)_T = -(partial differential U/partial differential V)_T/C_v .. (1) Now partial differential U = T partial differential S - P partial differential V so (partial differential U/partial differential V)_T = T(partial differential S/partial differential V)_T - P from redaction, it since (partial differential S/partial differential V)_T = (partial differential P/partial differential T)_T so (partial differential U/partial differential V)_T = T (partial differential P/partial differential T)_T - P Now, (1) rightdoublearrow (partial differential T/partial differential V)_V = (1 -)/C_v (T (partial differential P/partial differential T)_T - P) on (partial differential T/partial differential V)_v = 1/C_v [P - T (partial differential P/partial differential T)_T]