thermodynamics - Generalized work to increase both entropy and energy of a system

Suppose we have an adiabatic container with $N$ ideal gas particles, and each particle consists of two identical atoms so that each possesses a vibrational mode. For simplicity, let us assume that the vibrational modes are approximated by harmonic oscillators. In other words, we have $N$ ideal gases and, simultaneously, $N$ harmonic oscillators in the container.

If we exert an external force to compress the container, some amount of work will be transferred into the container, increasing total kinetic energy of ideal gases. In other words, the number of allowable microstates in a momentum phase space of the gases increases. On the other hand, the shirinkage of a coordinate phase space (= reduced volume) cancels out the expansion of the momentum phase space. Thus, the total number of microstates of the container does not change, resulting in no change in entropy of the container, i.e.

$$dS =k_B(dln\Omega)= \frac{1}{T}(dE + PdV - \mu dN - fdX) = 0$$ since $dE = -PdV$ and $\mu dN = fdX = 0$ ($f$ and $X$ stand for generalized forces and generalized coordinates respectively). This result agrees with what we already know: $dS=0$ since $dQ_{rev}/T=0$ in an adiabatic compression of an ideal container.

However, suppose that we radiate an electric field to the container. The field will transfer some work to the container, assuming that the field interacts with electron density polarizations of the gases, which results in excitation of the vibrational modes. As a result, the number of allowed modes inside the container will increase and subsequently, the entropy of the container will also increase (the number of microstates rise), i.e.

$$dS=k_B(dln\Omega) >0 \tag{1} \label{1}$$ . However, since all work done by the electric field is equivalent to the change of the energy of the container, $$dE=fdX$$ and thus, $$dS=\frac{1}{T}(dE + PdV - \mu dN - fdX) =0 .\tag{2} \label{2}$$ Obviously, equations $\eqref{1}$ and $\eqref{2}$ are contradictory. If the generalized work has a form of $-PdV$, I do not have any problem (as illustrated above). The origin of such contradiction may result from the fact that the generalized work done by the electric field increase both the energy and the total number of allowed microstates inside the container.

How can I solve this contradiction? Shouldn't I interpret the energy transferred by the field as a work? If this is true, why? I mean, how can I judge whether an energy transferred to a system is $TdS$ (~ heat) or $fdX$ (~ work)?