Why $dQ$ and $dL$ are not exact differentials, while instead $dU$ is exact?

Question

Why $dQ$ and $dL$ are not exact differentials, while instead $dU$ is exact? Is there a way to see it analytically?

Answer 1

If you consider the generic system in the figure above, the balance of energy is given by: $$\int_{t_{0}}^t{\dot{Q(s)} ds}-\int_{t_{0}}^t{\dot{L(s)} ds}+U(t_{0})=U(t)$$ $U(t)$ represents the energy which is stored by the system.

(It is the first tdm principle: $Q-L=\Delta U$)
Also you have: $$[\int{\dot{Q(s)}-\dot{L(s)} ds}](t)-[\int{\dot{Q(s)}-\dot{L(s)} ds}](t_{0})+U(t_{0})=U(t)$$ $$[\int{\dot{Q(t)}-\dot{L(t)} dt}]+cost=U(t)$$ Deriving you obtain: $$\dot{Q(t)}-\dot{L(t)}=\dot{U(t)}$$ and also: $$\delta Q-\delta L=dU$$ Now you integer this expression: $$\int_{t_0}^{t}{\delta Q}-\int_{t_0}^{t}{\delta L}=\int_{t_0}^{t}{dU}$$ And you compare it with the first expression: $$\int_{t_{0}}^t{\delta Q}-\int_{t_{0}}^t{\delta L}+U(t_{0})=U(t)$$ You see that: $$\int_{t_0}^{t}{dU}=U(t)-U(t_{0})$$ Which is the definition of exact differential. While: $$\int_{t_0}^{t}{\delta Q}\neq Q(t)-Q(t_{0}) $$ $$\int_{t_0}^{t}{\delta L}\neq L(t)-L(t_{0}) $$

Answer 2

U is a physical property of the material that depends only on its present thermodynamic state. Q and L are quantities that not only depend on the present state, but also the path by which it arrived at its present state.