Does the translational part of internal energy mean the kinetic energy of the gas?
Thanks to @IvanNeretin for pointing me in the right direction.

Imagine a box full of an ideal gas. Let the box and gas together be of a mass $m$.

Now imagine that the box is moving with a velocity $v$. Thus the system will have a kinetic energy $\frac{1}{2}mv^2$. Now this energy is of the system as a whole and is therefore not counted as internal energy.

Similarly imagine the box placed at height $h$. It will have a potential energy $mgh$. Again this will not be counted as internal energy.

When talking about diatomic or triatomic gases, we will say that that the total kinetic energy is the sum of all the possible ways in which the particle can move in space. This includes motion like translational motion, rotational motion and vibrational motion. The number of dimensions of such motion are called the degrees of freedom($f$) of the molecule and the total kinetic energy of the system is given by 

$$U = \frac{f}{2}nRT + \phi$$ 

where the first term represents the total kinetic energy and the last one represents all the other energies associated with the gas molecules

For example for a diatomic gas, $f=5$ because it can move in $3$ dimensions and rotate in $2$. So $$U = \frac{5}{2}nRT + \phi$$

For gases in high temperature, the degree of freedom would increase by due to the vibrational motion of the gas particles.

Wikipedia