Why can pressure be identified as partial of energy with respect to volume?

Question

There's something I'm still struggling to grasp about the fundamental relation of thermodynamics. For a closed system, we have

$$dU = \left.\frac{\partial U}{\partial S}\right|_V dS + \left.\frac{\partial U}{\partial V}\right|_S dV$$

We identify the first partial with temperature, and the second with (negative) pressure. For a reversible process ($p_{ext} = p_{sys}$), this of course makes the second term equal to the work done on the system. But that implies that when work is done with no heat flow (i.e. reversible adiabatic expansion), the entropy of the system doesn't change. To me that seems like a bold claim. We have a gas, let's say, that is expanding in such a way that the pressure against it always matches its own pressure. It therefore loses energy as it expands, which tends to decrease its entropy, but it gains entropy due to the increase in volume, and somehow these two effects perfectly cancel out! Not only that, but they cancel out for any thermodynamic system, not just a gas!

So as I see it, there must be a deep reason that a system expanding against its own pressure doesn't change its entropy, and to be fair, it is only due to this reason that we can even identify the first term as the reversible heat in the first place - for otherwise the second term would include some heat as well!

So what is the reason?

Answer 1

But that implies that when work is done with no heat flow (i.e. reversible adiabatic expansion), the entropy of the system doesn't change.

Reversible processes are by definition the processes which doesn't generate entropy - otherwise they would not be reversible. In fact, a process occurring at finite speed necessarily involves generation of entropy, but from the point of view of thermodynamics a process can be always made slow enough for this change to be negligible.

We identify the first partial with temperature, and the second with (negative) pressure.

The logic goes in the opposite way: the energy of a closed system is conserved. In thermo/stat. physics we assume that this energy can be changed in two ways: via macroscopic changes, which we call work and on a microscopic level, which we call heat, that is $$ dU=dQ - dW. $$

Work can be performed in many different ways - by changing the system volume, by changing magnetic field, etc. So generally we write: $$ dW = \sum_i x_i dX_i $$

$pdV$ is just the most common type of work encountered when dealing with gases, which is used throughout statistics physics textbooks for illustrative purposes, as well as for historical reasons (when Sadi Carnot had in mind real steam engines, even though his results are applicable to nuclear reactors.)

Entropy as an adjoint variable to temperature was defined by Clausius as (see this answer for alternative entropy definitions): $$ dS = \frac{\delta Q_{rev}}{T} $$

Now we have $$ dU = TdS - pdV, $$ and, since $U$ is a function of state, i.e., it is unique for any combination of variables $S, V$, we have to identify the pre-factors as the partial derivatives: $$ T=\frac{\partial U}{\partial T},p=-\frac{\partial U}{\partial V}. $$ (For more variables we would also have $x_i=\frac{\partial U}{\partial X_i}$, see also this answer for a crash course on differentials.)

We have a gas, let's say, that is expanding in such a way that the pressure against it always matches its own pressure. It therefore loses energy as it expands, which tends to decrease its entropy, but it gains entropy due to the increase in volume, and somehow these two effects perfectly cancel out!

(emphasis is mine)

It is not clear where the intuition behind the statements in bold comes from (that decrease in energy should cause change in entropy or that change in volume should increase entropy) - this does not follow from the equations discussed above. Rather, such an expansion corresponds to changing the energy along the path with $dS=0$: as energy is a function of multiple variables there are different paths connecting two energy values (imagine the image below in variables $S,T$, image source):

Answer 2

Usually you start with the first law of thermodynamics, which says that, $$dE=\delta q+ \delta w$$ and then define the $PV$ work as, $$\delta w= -P_{ex} dV$$ $$\implies \delta w_{rev} = -PdV$$ With some lengthy additional work to show the Clausius entropy is related to reversible heat flow, $$ dS = \frac{\delta q_{rev}}{T} $$ you can then give the total differential of internal energy for a reversible process where $P=P_{ex}$ in terms of entropy and volume dependence, $$dE=TdS-PdV$$ Since the heat and work are path functions but the internal energy is a state function, we have effectively found a general statement using a specific one. The variable next to the volume differential in this case is the pressure, and you have shown that the partial of energy with respect to volume is the pressure, $$P= -\frac{\partial E}{\partial V}$$

Answer 3

But that implies that when work is done with no heat flow (i.e. reversible adiabatic expansion), the entropy of the system doesn't change.

And that would be correct because there are only two ways that the entropy of the system can change. One is if there is a transfer of entropy, which requires heat transfer with the surroundings. The other is if entropy is generated within the system.

Entropy can be generated within the system by, for example, a non-quasistatic expansion ($p_{gas}\ne p_{ext}$), or by a quasistatic expansion ($p_{gas}=p_{ext}$) involving mechanical friction. The latter causes internal temperature gradients with irreversible internal heating effects. So for a reversible adiabatic process, a.k.a an “isentropic” process, the change in system entropy must be zero.

and somehow these two effects perfectly cancel out! Not only that, but they cancel out for any thermodynamic system, not just a gas!

They have to to in order to satisfy the above non statistical requirements for entropy change of the system discussed above.

Regarding your last paragraph, as previously stated mechanical friction causing internal temperature gradients can result in irreversible internal heating generating entropy within the system. Note that such internal heating effects are not covered in the first term of your equation. It is a consequence of the irreversible friction work in the second term. The first term only covers heat between the system and its surroundings.

Hope this helps.