Chapter 2
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.
Step-by-Step
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.