Clearing up concepts regarding quasi-static processes I have two main questions/problems:

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*Most definitions I've seen about quasi-static processes talk about them being "infinitely slow processes", or "slow enough as for the system to remain infinitesimally close to internal equilibrium". My confusion comes in that my notes talk about extremely fast internal redistribution of the system and don't really mention the "slowness" of the process. I suppose, however, that the "slowness" that is often talked about is in regard to external changes to the system. That effect is then redistributed very quickly across the system such that it is in equilibrium as a whole. Essentially, the external change is very slow in comparison with the redistribution. Is this correct?


*As mentioned in questions such as: 'Equations of State' - is it wrong to derive dynamic relationships from these equations?, it seems strange to take derivatives of variables that come from an equation that talks about stationary, equilibrium states. However this is done all the time, for example: $\delta Q_X=C_X \left(\frac{\partial T}{\partial V} \right)_XdV$. Here,$\left(\frac{\partial T}{\partial V} \right)_X $ is a dynamical relationship between temperature and volume. But the equations of state that relate temperature and volume assume equilibrium, such that no dynamical relationship exists. Is it simply the case that quasi-static processes "save the day" here, again, working as an idealization that yields reasonable approximations?
 A: *

*This is an annoying problem, I agree. I find that the most useful definition of a quasi-static process is: a process during which the system reaches internal equilibrium at every instant.

Internal equilibrium means that all intensive parameters are defined, which means that their standard deviation is negligible (you can always compute an average value, but it's only representative of the system if standard deviation is small).
A system in internal equilibrium doesn't have to be in equilibrium with the outside environment, however.
While being slow helps a lot to realize this situation, it isn't strictly part of the definition. It's just the most common way to approach a quasi-static transformation in the real world.
This being said, in practice, there's one kind of equilibrium that's easily realized between a system and the outside: mechanical equilibrium. This explains why $P=P_\text{ext}$ is usually assumed for a quasi-static process, while $T=T_\text{ext}$ isn't: mechanical equilibrium is very quickly reached, while thermal equilibrium is dog slow.


*I see this the following way. "Traditional" thermodynamics is the theory of equilibrium. It doesn't say anything about system outside equibrium.

However, the trick is to replace the real transformation by an imaginary one that has the same initial and final states, but that's quasi-static in between.
You can write $\Delta U=W+Q$ for the real transformation. Since $U$ is a state parameter, $\Delta U$ has the same value for all transformations that share the same initial and final states. $Q$ and $W$, on the other hand, don't have that property.
So you want to compute $\Delta U$ for the real transformation. You can compute it for the quasi-static transformation. And since, it's quasi-static, you have internal equilibrum at all times, between two close instants $t$ and $t+dt$, you can write $dU=\delta W+\delta Q$.
That's how you can get $dU$ and, from there, derivatives. For example, if you study the evolution of an ideal gas between $t$ and $t+dt$, you might want to use $dU=C\,dT$. In principe, you can't because $T$ isn't defined during the transformation. But if you switch to the related quasi-static process, then $T$ becomes defined, and you can write that. From there you can compute $\Delta U$, which has the same value for both transformations.
