In this problem I aim to show that the expression for the average velocity of particles in a gas can be derived from the Maxwell velocity distribution (problem 1C.1(a) Transport Phenomena).

In the kinetic molecular theory of gases, velocities are randomly distributed and have an average magnitude given by: $$ \bar{u} = \left( {8 \kappa T} \over { \pi m} \right)^\frac{1}{2} $$

This can be derived from the Maxwell distribution of molecular velocities, *f(u)* in the following way:

$$ \bar{u}= {{ \int \limits_0^\infty u \cdot f(u) du } \over { \int \limits_0^\infty f(u) du }} $$

Why? Well short hand-wavy answer is that when you take the Maxwell distribution and normalize it the result is the probability density function for the velocity. The equation above is simply the expected value of the velocity.

Anyways, the Maxwell distribution of velocities with magnitudes between *u* and *u + du* is:
$$ f(u) = 4 \pi n \left( m \over {2 \pi \kappa T} \right)^\frac{3}{2} \cdot u^2 \cdot \exp \left( { -m \over {2 \kappa T}} u^2 \right) $$

To solve this problem I need to find two integrals:

- $ \int \limits_0^\infty u \cdot f(u) du $
- $ \int \limits_0^\infty f(u) du $

First the integral $ \int \limits_0^\infty u \cdot f(u) du $

Now the second integral $ \int \limits_0^\infty f(u) du $

Finally combining them all together