Statistical Thermodynamics and Rate Theories/Principles of the Kinetic Theory of Gases

Kinetic Theory of Gases
The kinetic theory of gases is a simple model used to describe the behavior of molecules in the gas phase. It is based on the following assumptions:
 * 1) The gas is made up of a collection of non-interacting particles separated by a distance larger than its dimensions
 * 2) Particles move in random directions such that direction of motion has a negligible effect on the properties of the gas (it is isotropic).
 * 3) All collisions are perfectly elastic. Particles may collide with the walls of the container or with each other.
 * 4) Particles carry only translational kinetic energies.

Given that translational energy spacing is relatively small compared to KT, particles behave classically and follow Newtons law's of motion. In large systems, in is unrealistic to monitor the velocity of individual particles so instead an average is taken. Of the range of possible velocities some will be more common than others, giving rise to a probability distribution, better known as the Maxwell distribution.

The probability that a particle will have a given velocity is provided by the equation below. Ω is the probability density.

$$\Omega(v_x,v_y,v_z) dv_xdv_ydv_z=f(v_x)dv_xf(v_y)dv_yf(v_z)dv_z$$

Here we have assumed that the probability distributions in each direction are independent of each other, thus it is seperable.

Each particle has translational kinetic energy characterized by

$$\varepsilon_{trans} =\frac{1}{2}m(v_x^2+v_y^2+v_z^2)$$

The energy distribution is expected to have a form similar of that of the canonical ensemble as follows.

$$P(\varepsilon_{trans})\propto e^{-\varepsilon_{trans}/k_BT}$$

By analogy, the velocity distribution is

$$f(v_i)= A e^{-\frac{1}{2}m(v_i^2)/k_BT}$$

A is later found to be $$A=\left(\frac{m}{2\pi k_BT} \right)^{\frac{1}{2}} $$

In order to find the probability of having a particular velocity we must integrate Equation 1 over all space. The function must be normalized such that the probability is unity. The leading coefficient C will take care of this.

$$\textstyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \displaystyle \Omega(v_x,v_y,v_z) dv_xdv_ydv_z$$

Recall that $$e^{a+b}=e^ae^b$$, which leads to the following.

$$\textstyle \int_{-\infty}^{\infty} e^{-\frac{1}{2}mv_x^2/k_BT}dv_x \int_{-\infty}^{\infty} e^{-\frac{1}{2}mv_y^2/k_BT}dv_y\int_{-\infty}^{\infty} \displaystyle e^{-\frac{1}{2}mv_z^2/k_BT} dv_z=\frac{1}{C}=\frac{1}{A^3}$$

The exponentials are of the form of the Gaussian curve and follow

$$\textstyle \int_{-\infty}^{\infty} \displaystyle e^{-ax^2} \textrm{d}x=\sqrt{\frac{\pi}{a}}$$

In this case, $$a= \frac{m}{2k_BT}$$

Evaluating gives

$$C= \left(\frac{m}{2\pi k_BT} \right)^{\frac{3}{2}}$$

Since the gas is isotropic the function Ω must depend only on speed. The magnitude of the velocity vector is given as follows.

$$\mathbf v \cdot \mathbf v= u^2=v_x^2+v_y^2+v_z^2$$

Up to this point, everything has been done using Cartesian coordinates. However due to the spherical symmetry of the system is it much more natural to work in spherical coordinates.

The Maxwell Distribution of Molecular Speeds

$$f(v) dv=4\pi v^2 \left(\frac{m}{2\pi k_BT} \right)^{3/2}e^{-\frac{1}{2}mv^2/k_BT}dv$$

The $$4\pi v^2 $$ term comes from the conversion to spherical coordinates (Integrating over the entire surface).

Example
Find the average value of $$v_x^2 $$ and the kinetic energy associated with motion along the x-direction.

It is best to approach this problem using the Cartesian form since we the integral is simplified considering that the function is independent of y and z.

$$\langle v_x^2 \rangle = \textstyle \int\limits_{-\infty}^{\infty} \displaystyle v_x^2 \Omega(v_x) d x$$

$$\langle v_x^2 \rangle = \textstyle \left(\frac{m}{2\pi k_BT} \right)^{\frac{1}{2}}\int\limits_{-\infty}^{\infty} \displaystyle v_x^2 e^{-\frac{1}{2}m(v_i^2)/k_BT} \textrm{d}x $$

This is an even function, therefore we can integrate over one half of the function then multiply by 2.

$$\langle v_x^2 \rangle = 2 \left(\frac{m}{2\pi k_BT} \right)^{\frac{1}{2}}\int\limits_{0}^{\infty} \displaystyle v_x^2 e^{-\frac{1}{2}m(v_i^2)/k_BT} \textrm{d}x= \frac{k_BT}{m}$$

Multiplying this result by $$\frac{1}{2}m$$ will give a kinetic energy of $$\frac{1}{2}k_BT$$. For 3 independent direction, the total kinetic energy for an ideal gas in equilibrium is $$\frac{3}{2}k_BT$$.