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According to Finn's Thermal Physics Third Edition, the First Law of Thermodynamics is described as:

If a thermally isolated system is brought from one equilibrium state to another, the work necessary to achieve this change is independent of the process used.

Which I think must be a rephrasing of the law of energy conservation. However, work is an inexact differential, in that the work done in a system transitioning reversibly from $(P_1,V_1)$ to $(P_2,V_2)$ is entirely dependent on the path taken. But here the quote sounds like its contradicting the previous sentence. The textbook goes on to say

The statement above is equivalent to saying that the adiabatic work $W_{\text{adiabatic}}$ expended in a process is path independent, ...

which makes some sense to me, I think. If the walls of the system are insulating, then no work on the gas can go to heat. However, how does this fit in with the general concept that work is path dependent? Are they saying there is always a minimum amount of work necessary? Because a bunch of paths can be made from $(P_1,V_1)$ to $(P_2,V_2)$ which don't necessarily expend the same amount of work.

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  • $\begingroup$ In general but not always work is path dependent but specifically adiabatic work is path independent. There is no contradiction and has nothing to do with being minimum or some such. $\endgroup$ – hyportnex Feb 3 '19 at 17:31
  • $\begingroup$ Just emphasising hyportnex's comment, the adiabatic constraint is a powerful one! $\endgroup$ – Philip Wood Feb 3 '19 at 18:28
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For a closed system (no mass entering or leaving), the first law tells us that $$\Delta U=Q-W$$ Since the internal energy U is a function of state, once the initial and final end states are specified, the internal energy change is fixed; there can be multiple process paths between these two end state (involving different Q and W), but for all such process paths, the difference between Q and W is fixed. A subset of these multiple paths are those for which the system is thermally isolated: Q = 0. That means that, for the subset of thermally isolated paths, $W=-\Delta U$.

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If a thermally isolated system is brought from one equilibrium state to another, the work necessary to achieve this change is independent of the process used.

The first law for a closed system (no mass transfer between system and surroundings) is

$$\Delta U=Q-W$$

To say a system is “thermally isolated” means no heat transfer, so the first law becomes

$$\Delta U=-W$$

Now, internal energy $U$ is a state function, meaning it is independent of the path taken between initial and final states. But if heat transfer is not possible, the only path possible is an adiabatic work path. Any different paths connecting the same states would need to involve heat transfer, as discussed below.,

Because a bunch of paths can be made from $(P_{1}V_{1}$ to $ P_{2}V_{2})$ which don't necessarily expend the same amount of work.

Yes, but in order for that to happen heat transfer needs to be involved.

For example, you could connect states 1 and 2 with the following paths:

A. Constant pressure path from 1 to 1a followed by a constant volume path from 1a to 2.

B. Constant volume path from 1 to 1a followed by a constant pressure path from 1a to 2.

For path A the process from 1 to 1a involves both heat and work transfers while the process from 1a to 2 involves heat transfer only.

For path B the process from 1 to 1a involves heat transfer only while the process from 1a to 2 involves both heat and work transfer.

If you look at the PV graphs for both paths, the area under path A is greater than path B; the path A work is greater than path B. But this was only because both paths involved heat transfer in addition to work.

In order to convince yourself, try and find multiple paths between two states where none of the paths involves heat transfer.

Hope this helps.

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