# The mean kinetic energy of a gas particle

I'm in undergraduate stat mech/thermo. In the context of the Maxwell-Boltzmann distribution, the mean kinetic energy of a gas particle is $$\langle KE \rangle = \frac{1}{2}m \langle v^2 \rangle$$.

I do not see why we use $$\langle v^2 \rangle$$, and not $$\langle v \rangle ^2$$. I understand that they are different terms mathematically, just can't figure out what necessitates the use of one over the other. Thank you.

• By definition, kinetic energy is $K = \frac{mv^2}{2}$, so the average, $\langle K \rangle$ is $\langle mv^{2}/2\rangle = \frac{m}{2}\langle v^2\rangle$, because the ideal gas is thought to be of one component so the mass is just a constant. Just that. Commented Feb 21 at 5:38
• Assuming we are in the rest frame of the system $\langle v \rangle$ is zero so it wouldn't be very useful for calculating the internal energy. Commented Feb 21 at 7:38
• @JohnRennie These comments together are the answer. Commented Feb 22 at 11:35

In the kinetic theory of gases, the average velocity of a particle is derived by considering the molecules of gas to be point particles with velocities $$v_{i}$$, where $$i\, =\, 1,\,2, \, \ldots,\,N$$ is the particle index and $$N$$ is the total number of particles. In this context, $$\langle v \rangle$$ is the average velocity of the collection:

$$\langle v \rangle^2 = \left( \frac{1}{N} \sum_{i=1}^{N} v_i \right)^2$$

and

$$\langle v^{2} \rangle$$ is:

$$\langle v^2 \rangle = \frac{1}{N} \sum_{i=1}^{N} v_i^2.$$

Expanding a few terms reveals the problem:

$$\langle v \rangle^2 = \frac{1}{N^2} (v_1 + v_2 + v_3 + \ldots + v_N)^2,$$

whereas:

$$\langle v^2 \rangle = \frac{1}{N} \sum_{i=1}^{N} v_i^2 = \frac{1}{N} (v_1^2 + v_2^2 + v_3^2 + \ldots + v_N^2).$$

I hope this helps!