# Does the translational part of internal energy mean the kinetic energy of the gas?

Internal energy of an ideal gas consists of energy due to translational, rotational , vibrational etc.

This line from WP says:

In thermodynamics, the internal energy of a system is the energy contained within the system, excluding the kinetic energy of motion of the system as a whole and the potential energy of the system as a whole due to external force fields

At the same time it says:

The ideal gas is a gas of particles considered as point objects ... Such systems are approximated by the monatomicgases, helium and the other noble gases. Here the kinetic energy consists only of the translationalenergy of the individual atoms. Monatomic particles do not rotate or vibrate, and are not electronically excited to higher energies except at very high temperatures. Therefore, internal energy changes in an ideal gas may be described solely by changes in its kinetic energy. Kinetic energy is simply the internal energy of the perfect gas and depends entirely on its pressure, volume and thermodynamic temperature.

So my question is:

Does the translational internal energy of the diatomic or polyatomic molecules also refer to their kinetic energies?

• The words "as a whole" are the key. – Ivan Neretin Oct 6 '18 at 9:34

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

• thanks for the quick reply. That's what I want to stress on...while internal energy doesn't involve the KE due to the motion of the system in total...it involves the KE of the gases in case of ideal gases...this is what the kinetic equation of ideal gas also says.But my question is if I were asked to calculate the KE of diatomic or polyatomic gases (which are not deemed ideal) then is it Ok to say that the translational part of internal energy is the Kinetic energy of the gas molecules? – user 33690 Oct 6 '18 at 10:05
• The kinetic energy would also depend on the rotational motion and the vibrations. So only the translational part of the internal energy cannot be called the total kinetic energy. – Harshit Joshi Oct 6 '18 at 10:24