Kinetic theory of gases In the kinetic theory of gases I know that the temperature of an ideal monatomic gas is proportional to the average kinetic energy of its atoms. If I give energy to the system through irradiation, atoms absorb this energy and increase their kinetic energy. The question is:
The energy given by irradiation is absorbed by nuclei or by electrons? 
 A: If we stick to the kinetic theory of gases, the atoms are indivisible entities with their energy being purely kinetic. So the energy of radiation is transferred into the kinetic energy of the atoms.
How exactly this happens is an interesting question. One widely used technique of changing the energy of atoms via irradiation is laser cooling used in the experiments with ultracold atoms. The atoms absorb incident laser light and then re-emit it at slightly different frequency (and momentum), depending on whether the laser is tuned slightly above or below the peak of the atomic absorption line:
$$\hbar\omega_{incident} = \hbar\omega_{re-emitted} + \Delta E,\\
\hbar\mathbf{k}_{incident} = \hbar\mathbf{k}_{re-emitted} + \Delta \mathbf{p}.$$
Thus, only a part of the radiation is converted into the kinetic energy of the atoms (or removed from the atoms in the case of cooling). This energy is transferred to atoms as a whole, i.e. it can be though of as the energy of the center-of-mass of nucleus + electrons, whereas the electronic states in an atom or the nucleon stats in the nucleus remain unchanged.
A: A free electron (not bound to a nuclei) has a continuous energy spectrum, because it is allowed to travel with any velocity $v$. However, a free electron is unable to absorb the photon: Electrons are fundamental (unstructured) entities -- they are not composed of smaller parts. However, in order to absorb the photon the electron would have to satisfy the energy and momentum conservation simultaneously. This is impossible for an unstructured entity. Hence, a free electron is unable to absorb a photon.
In contrast, an atom is a composite (=structured) entity, which possesses certain stable energy levels. Therefore, the atom is able to absorb photon:


*

*If the wavelength of the photon is around $500nm$ (energy range of several $eV$) we say that the electron has absorbed the photon.

*In contrast, if the energy of the photon is in the range several $keV$ the photon was either absorbed by the nucleus or (more probably) an electron was removed from the atom and we are left with an ion and a free electron.


The statement that "an electron absorbs the photon and jumps to an excited state" (or a free electron state) is helpful, because it simplifies the picture. Nevertheless, as should be clear from the "free electron" paragraph, the existence of the nucleus is important for the absorption of the photon. Without the nucleus the electron is unable to satisfy the energy and momentum conservation simultaneously. 
If we try to connect this description of an atom with your question regarding the kinematic gas theory, I believe we have to clarify one thing first:
In the kinematic gas theory the temperature is not related to the "absolute velocity", but only to the random part of the velocity. Hence, if you take a container with gas at temperature $T$ and you consider it in two different coordinate frames -- (1) the rest frame of the container  and (2) in a frame where the container moves with velocity $v>0$ -- the temperature of the gas is the same in both frames. 
Now, coming back to the illumination of the gas at temperature $T$, the answer depends on the context:


*

*A very efficient way to cool down atoms is to use either the doppler effect  to obtain a directional dependency for the absorption of photons: If only atoms with a velocity ant-parallel to a photon absorb the photon, the atoms are slowed down due to the momentum of the photon, $p=E/c = \hbar \omega/c$. The cold atom community usually uses so called Zeeman slowers and optical molasses: Although these two methods differ in their details the net effect is that the atoms absorb photons, which posses a velocity opposite to their own velocities. The photons are said to be red shifted w.r.t. the transition frequency of the atom.

*If we instead consider a gas at room temperature which absorbs photons from every direction with the same probability,  the average temperature of the gas doe not change: The absorption of the photon decreases the atomic velocity is with the same probability as it increases it. Hence, the random part of the velocity does not change. Thus, the temperature remains constant.

*Finally, we could consider blue shifted photons. Now the atom absorbs photons, which travel parallel to the atomic velocity, with a higher probability. In this case, the atom gets kicks in the direction which increases their original velocity. The temperature of the gas increases. 

