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Determine whether d q=f(T) d T+\left[\frac{R T}{v}\right] d v is a thermodynamic property or not.

\frac{\partial f(T)}{\partial v}=0

and \left(\frac{\partial}{\partial T}\right)\left(\frac{R T}{v}\right)=\frac{R}{v}

Hence \left(\frac{\partial}{\partial v}\right) f(T) \neq\left(\frac{\partial}{\partial T}\right)\left(\frac{R T}{v}\right)

Thus dq is not an exact differential, and hence not a thermodynamic property.