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## Q. 2.9

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

## Verified Solution

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

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.