d
For $1$ mole of an ideal gas
$pV = RT \text {...(i) }$
at constant pressure:
$PdV = RdT \text {....(ii) }$
From Eqs. $(i)$ and $(ii)$, we get
$\frac{d V}{V}=\frac{d T}{T}$
The coefficient of volume expansion at constant pressure is given by $\frac{ dV }{ VdT }=\frac{1}{ T }$
same for all gases at same temperature.
The average translational kinetic energy per molecule is $(3 / 2) kT$ and not $3 kT$. With decrease in pressure, volume of the gas increase so its mean free path increases.
The average translational kinetic energy of the molecules is independent of their nature, so each component of the gaseous mixture will have the same value of average translational kinetic energy.