Explanation
The average translational kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to $$\dfrac{3}{2}k_BT $$. It depends only on temperature and is independent of the nature of the gas.
Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is $$1:1$$.
There are always $$3N$$ total independent degrees of freedom for a molecule, where $$N$$ is the number of atoms. These come about because when each atom moves, it has three independent degrees of freedom: its position in each of the $$x, y, z$$ directions.
Now, having independent degrees of freedom for each atom isn't all that useful. Instead, we can make combinations of different degrees of freedom. The important thing when doing so is that the number of independent degrees of freedom are preserved: it's just that what a particular degree of freedom does to the atoms changes.
The standard breakdown of degrees of freedom subtracts out global movement in each of the three directions. So you have $$3N$$ total degrees of freedom, but you can set aside $$3$$ of them as translation of the whole molecule in each of the $$x, y, z$$ directions, leaving $$(3N-3)$$ degrees of freedom.
Likewise, it's standard to subtract out the whole molecule rotation. For most larger molecules, there's three different degrees of rotational freedom: rotation around each of the $$x, y, z$$ directions. But for linear molecules like $$CO_2$$, one of those rotations (around the axis of the molecule) doesn't actually change the position of the atoms. Therefore it's not a "degree of freedom" which counts against the $$3N$$ total. So while for non-linear molecules there are $$(3N-3-3) = (3N-6)$$ degrees of freedom which are independent from the global rotational and translational ones, for linear molecules there are $$(3N-3-2) = (3N-5)$$ degrees of freedom which are independent from the global rotational and translational ones. -- A quick clarification. The reason why we ignore this rotation is not because the center of mass doesn't move. The center of mass doesn't move for $$any$$ of the global rotations: in the typical assignment of degrees of freedom the axis of rotation goes through the center of mass. Instead, the reason the rotation is ignored is that $$none$$ of the atoms move due to the "rotation".
So since $$CO_2$$ has three atoms and is linear, it has ($$3\times3 - 5 = 4 $$)degrees of freedom which are independent of the global rotation and translation. We call these the vibrational modes.
(I) Translation kinetic energy 7/5
(II) Rotational kinetic energy of $$CO_2$$ Function of T only 2. 15/13
(III) Translation kinetic energy per unit volume RT
(IV) $$\lambda$$ for $$CO_2$$ at very high temperature $$\frac{3}{2}P$$
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