# On the meaning of $dU = \delta w$ for adiabatic processes

For an adiabatic transformation between state A and B $$\delta q = 0$$ and consequently from the first law of thermodynamics $$dU = \delta w$$, since $$U$$ is a state function its variation should be the same whether the process is reversible or irreversible.

The possibility to go via an irreversible or irreversible path between the two states seems feasible by what I'm reading on "Chemical Thermodynamics: Classical, Statistical and Irreversible By J. Rajaram page 66"

For the adiabatic expansion of an ideal gas the work done by the gas is equal to the decrease in internal energy, -ΔU = —w. However, if an ideal gas is taken from state A to state B by a reversible path as well as an irreversible path, while the change in internal energy is the same because the initial and final states are the same, the work done against external pressure will not be the same. The work done in the irreversible process must be less than that done in the reversible process. The decrease in internal energy in either case will be ΔU =n C_v(T_2 - T_1).

$$dU = \Delta w$$ seems to indicate that also work to be the same for the reversible and irreversible path. But how can work for the irreversible and reversible process be the same? We all know that the maximum work can be extracted by the reversible path. Hence even if $$\Delta U$$ is equal for the two processes $$w$$ should not. So Does the equal sign in $$dU = \delta w$$, hold only for reversible processes? if Yes why? If No, how should I read $$dU = \delta w$$?

You are assuming that you can take either a reversible or irreversible adiabatic path and end up in the same final state. On the irreversible you will generate entropy and since the path is adiabatic you cannot pass that entropy to the surroundings as heat, so the final state of the irreversible path will have a higher entropy but the same energy.

We all know that the maximum work can be extracted by the reversible path

This is true of cycles rather than paths. Making an analogous statement about paths is difficult for exactly the same reason as happened here; it is difficult to define a "irreversible version" of a path in a general way.

• what does it mean that $U$ is a state function and its delta does not depend on the path? isn't it implicit that there can be different paths between the same states? or does it hold only for reversible paths? If so what does forbid an irreversible path? Commented Sep 3, 2020 at 17:43
• In general yes, $\delta w$ depends on the path. If you then say, however, that you are going to limit yourself to adiabatic paths you reduce the possibilities considerably. There generally will be other paths with the same start and end points with different amounts of work done. It is just that the others will not be adiabatic. Commented Sep 3, 2020 at 17:46
• the manual I'm reading Chemical Thermodynamics: Classical, Statistical and Irreversible By J. Rajaram page 66, states that such irreversible and reversible processes (in particular it treats expansions of an ideal gas) are possible but in the irreversible you end up with less work been done by the system. Commented Sep 3, 2020 at 17:58
• Then the final end state and the change in internal energy must be different (for example, in a case where the final volumes are the same). Commented Sep 3, 2020 at 19:38

If you carry out an adiabatic irreversible expansion or compression between two end states, you will find that there is no adiabatic reversible path between the same two end states. If you carry out an irreversible and a reversible process such that the work and the change in internal energy is the same for the two paths, you will find that the final end states will be different. For an ideal gas, since the change in internal energy depends only on the temperature change, you will find that the final pressures and volumes will be different for the reversible and irreversible paths.

• Only for ad. processes or can we generalize it? And for what thermodynamic principle? For example for an irreversible process, we cannot integrate $dS = dq /T$ to obtain $\Delta S$ because $dS$ equals $dq/T$ only for reversible processes. However we can conceive a reversible path between the same states and $\Delta S$ should be the same. Commented Sep 5, 2020 at 8:52
• I don't understand this question. Commented Sep 5, 2020 at 12:44

$$dU=\delta W+\delta Q$$ is always true. Of course, for an adiabatic process $$dU=\delta W$$ is always true. As far as I understand, the confusion is the following: reversible/not reversible is not about different paths between two same initial and final states. They are different paths with different final states. They can't be made to meet again even after another (adiabatic) transform.

A state is made of external variable(s) (choose volume $$V$$ for a gas) and one internal variable (say energy $$U$$). The gas is just an example, these variables are just one possible choice. Imagine the starting point is $$A=(V_A,U_A)$$. Acting on the system by a change in external variables (adiabatically), you end in different states. Assume you always decide to end with a certain fixed volume $$V_B$$. Depending on whether you do it reversibly or not, you will end in different energies. It can be $$B=(V_B,U_B)$$ or $$B'=(V_B,U'_B)$$. Similarily, you can decide to spend/extract the same amount of work, and then you will end with the same energy, but different volumes.

The paragraph you quote does not mention adiabaticity as a constraint. It is possible to have the same start/end states for a reversible process and a non reversible proces. Typically, if one theses paths is adiabatic, the other is not.

Because I do not have enough reputation points to comment, I will submit this as an answer, which is, $$dU = dw_{rev}$$.

If we want to calculate the internal energy change from an adiabatic process, we should do so reversibly, in order to guarantee that the final state of interest has been reached.

Source: Consider 36:06 in Lecture 3 of the MIT 5.60 "Thermodynamics & Kinetics", 2008 Lecture series.