First law of thermodynamics: state and path functions

Question

In the first law of thermodynamics

$$\Delta U=Q+W=\int TdS-\int PdV$$

it is known that $U$ is a state function while $Q$ and $W$ are path functions. However consider the case when there is no work done on a system

$$\Delta U=\int TdS$$

and the case when there is no heat flow

$$\Delta U=-\int PdV$$

In these two cases, I can't tell whether $U$ is a state function or a path function. If $U$ is a state function, then can anyone give me examples when $\Delta U$ is independent of the path used when calculating $Q$ or $W$?

Answer 1

A useful analogy I've heard involves a pond that can gain or lose water by two mechanisms: (1) liquid inflow/outflow and (2) condensation/evaporation. Any change in the pond level—from yesterday to today—tells you nothing on its own about how much each mechanism participated.

That is, there might have been a lot of net inflow and moderate net evaporation, say, or moderate net inflow and minimal net evaporation, and so on.

Here, the water level is analogous to the system internal energy, and the net liquid flow and net phase change are analogous to the work and heat (it doesn't matter which) through which the system interacts with the surroundings. The water level is a state function; the water transport mechanisms are path functions.

Further, there is no inflow "contained" in the pond; any incremental amount of water that flowed in might have remained in the pond, or flowed out, or evaporated; we just don't know.

You may find the analogy well suited to this question because if you conceptually cover the pond to eliminate condensation/evaporation, you know that the water level change is due solely to liquid inflow/outflow. Similarly, if you block the inflow/outflow channels, then the water level change is due solely to condensation/evaporation. (Note that you still know only the net amounts for each mechanism.)

(A caution to avoid confusion: The original system is assumed to be closed; the pond in the analogy is not closed because we're using water transfer as a surrogate for heat and work.)

To address the original question, if heat or work is blocked for a closed system, then the fact that internal energy is a state function implies that the contribution of the nonblocked mode of energy transport can now be determined solely from the system state—which is not the case if both heat and work interactions are allowed.

Answer 2

For any state function $f(x,y)$: $$ df = f_x dx + f_y dy $$ where $f_x$ and $f_y$ ar the partial derivatives. If we restrict the function to constant $y=y^*$, then along that line –and that only– we have $$ df = f_x dx\quad (y=y^*) $$ This does not mean that $f$ is no longer a function of two variables. it only means that we are focusing on a region of the $(x,y)$ plain where $y$ is held constant and only $x$ is allowed to vary.

In your example, no work means $V = V^* = \text{const}$. Along this path $U=U(T,V^*)$, still a function of two variables but you choose to keep one constant and change only the other one.

Answer 3

If you accept that the internal energy is a state function then if you break up its change into an arbitrary sum of component terms, say, $\Delta U = b_1 + b_2+...b_J$ and then you assume that $J-1$ of those terms are zero in the process except for one, say, $b_j$, $1\le j \le J$ so that $\Delta U= b_j$, then the change of that $j^{th}$ term $b_j$ in that process is independent of the path and depends on the end states only.