Entropy $dQ=TdS$ and Work $dW = -pdV$ conditions?

Question

What are the conditions in order for the equations: Entropy $dQ = TdS$ and Work $dW = -p dV$ to work?

I think for $dQ = T dS$, it must be a reversible process?

But for $dW = -p dV$, shouldn't it always hold?

Answer 1

I think for $dQ = T dS$, it must be a reversible process?

Not necessarily. It has to be quasistatic process in which the system passes through equilibrium states. Quasistatic process in theory may be reversible, but need not be. For example, when heat passes from colder body to warmer body through separating insertion (walls) slowly enough so that the bodies remain near equilibrium state, for any one body the relation $dQ=TdS$ holds, where $T$ is temperature of this body. This process is irreversible, since heat is going from warmer to colder body, but the formula applies for both bodies separately. Only for the combined system, the formula does not apply; the combined system does not have one temperature.

But for $dW = -p dV$, shouldn't it always hold?

Again, it holds if the process is quasistatic and the system has the same pressure throughout.

Answer 2

I think for $dQ = T dS$, it must be a reversible process?

Yes.

But for $dW = -p dV$, shouldn't it always hold?

No, it is only for reversible Pressure-Volume work. It doesn't apply to irreversible processes. It does not apply to situations with other types of work like electrical, surface tension, etc.

According to Physical Chemistry by Levine, "work in a mechanical irreversible volume change sometimes cannot be calculated by thermodynamics". Non-uniform pressure, turbulence and kinetic energy are then discussed.