Why is work entropy-free? These notes I found online state that "work is entropy‐free, and no entropy is transferred with work." I take this to mean that entropy is not generated in a work process. Why is this? Why is no entropy created when work is done on a system?
 A: Entropy is generated in work processes involving viscous dissipation of mechanical energy such as rapid expansion or compression, stirring, etc.
A: Because the entropy in this notes is defined in terms of the heat transfer:
$$
dS=\left(\frac{\delta Q}{T}\right)_{int, rev},
$$
which is the increase of the internal energy not associated with work.
Remarks

*

*As pointed by @ChetMiller in their answer and in the comments, the statement applies to the entropy transfer, but not to the entropy generation, which can result from work in an irreversible process (note that the entropy above is defined for a reversible process).

*There are different ways to define entropy, notably the axiomatic definition in thermodynamics (the one above), and the statistical physics definition as the logarithm of the number of microstates (Boltzmann entropy). See, e.g., this answer.

A: The notes should read “reversible work”, which is a useful idealization because it lets us ignore entropy generation. (All real processes generate entropy, often with a rate dependence that overly complicates introductory problems.)
Work doesn’t transfer entropy because it raises the energy of all particles in a system equally, with no dispersion. In this way, it can be contrasted with heating, which widens the energy spectrum (and does transfer entropy). Reversible work further doesn’t generate entropy because it’s performed at the limit of slowness, with no gradients (in force, pressure, voltage, magnetic field, chemical potential, or surface tension, for example).
A: From an atomistic and quantum perspective,

*

*the entropy of a system is defined (the "Gibbs definition") in terms of the probability $P_i$ that a randomly selected microsystem from that system is found in the $i$th of the discrete quantum stationary states available for a microsystem to be in, as $$S = -Nk_{\textrm{B}}\sum_{\textrm{All }i}\left(P_i\ln\left(P_i\right)\right)$$ where $N$ is the total number of microsystems in the system;

*heat is defined as a transfer of energy into or out of the system which proceeds by some of the microsystems making transitions between different quantum stationary states: it's possible for this to change (some of) the $P_i$ values, and therefore to change the entropy of the system;

*work is defined as a transfer of energy into or out of the system which proceeds by leaving the microsystems in the same quantum stationary states, but altering the potential field in such a way that the characteristic energy of each quantum stationary state changes: since the microsystems all stay in the same quantum stationary states, the $P_i$ values don't change, and neither does the entropy of the system.

This atomistic definition of entropy is equivalent to the macroscopic definition of entropy given by @RogerVadim if the probability distribution over the quantum stationary states is given by the Boltzmann distribution.
