Kinetic theory of gases

Derivation

Consider a mono-atomic gas: N atoms of mass m enclosed in a cubic container of side a. The volume , the area of side . Let the axes of the coordinate system be parallel to the cube sides. When an atom bounces of the container wall perpendicular to the x axis, the atom's momentum change will be

$\begin{displaymath}\Delta q = q_f - q_i = -mv_x - mv_x = -2mv_x \end{displaymath}$

since the momentum of the particle along the x axis changes from the initial value of to the final value of in the process (see introductory physics - particle colliding with a wall).

During a time $\Delta t$ , all particles moving left-to-right with velocity contained within a box of base A and length $v_x \Delta t$ will strike the wall. The volume of the box u is

$\begin{displaymath}u = A v_x \Delta t \end{displaymath}$

Since half of all atoms move left-to-right, and the fraction of all atoms which fall in to the box is u/V, the number of collisions with our selected wall during time $\Delta t$ , if all atoms had same velocity , would be

$\begin{displaymath}n = \frac{1}{2} N\frac{u}{V} = \frac{1}{2} N v_x \Delta t A/V \end{displaymath}$

Newton's second law tells us that the momentum change per unit time is equal to the force. Newton's third law tells us that the force acting on the wall is equal in magnitude but opposite in sign to the force acting on the atoms. The force acting on the wall due to collisions with atoms of velocity is

$\begin{displaymath}f = -\frac{n \Delta q}{\Delta t} = \frac{N m v_x^2 A}{V} \end{displaymath}$

In a gas the atoms have a distribution, or range of velocities. To obtain the average force F acting on the wall from collisions with atoms having all possible velocities we write

$\begin{displaymath}F = \langle f \rangle = \frac{N m A}{V} \langle v_x^2 \rangle \end{displaymath}$

The angled brackets denote averaging over all atoms/molecules, i.e.

$\begin{displaymath}\langle v_x^2 \rangle = \frac{1}{N}\sum_{i=1}^N v_{x,i}^2 \end{displaymath}$

with $v_{x,i}$ denoting the x-component of the velocity of atom i.

We can now calculate the pressure exerted by the gas on the container wall

$\begin{displaymath}p = \frac{F}{A} = \frac{N m}{V} \langle v_x^2 \rangle \end{displaymath}$

Of course each atom moves in three directions x, y, z. We expect the motions in each direction to be the same, i.e.

$\begin{displaymath}\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle \end{displaymath}$

Define the root mean square velocity c :

$\begin{displaymath}c = \langle v^2 \rangle^{1/2} \end{displaymath}$

$\begin{displaymath}c^2 = \langle v^2 \rangle = \langle v_x^2 \rangle + \... ...\rangle + \langle v_z^2 \rangle = 3 \langle v_x^2 \rangle \end{displaymath}$

Then

$\begin{displaymath}p = \frac{N m c^2}{3V} = \frac{n M c^2 }{3V} \end{displaymath}$

We have used the definition of M, the molecular mass

$\begin{displaymath}M = m N_A \end{displaymath}$

with - Avogadro's constant, n the amount of substance and the property that

$\begin{displaymath}N m = n N_A m = n M n = \frac{N}{N_A} \end{displaymath}$

Multiplying both sides of the equation for pressure by V we obtain our final result of the kinetic theory of gases:

$\begin{displaymath}pV = \frac{1}{3} n M c^2 \end{displaymath}$

This is a very important result, because it relates macroscopic quantities p and V, easily measured in a laboratory, to a microscopic quantity c, the RMS velocity of the individual atoms, which is quite difficult to observe directly.

Conclusions

Conclusion #1: root-mean-square velocity c

Using the ideal gas equation of state

$\begin{displaymath}pV = nRT \end{displaymath}$

we obtain the following equation for c

$\begin{displaymath}c = \sqrt{\frac{3RT}{M} } \end{displaymath}$

c is proportional to $\sqrt{T}$ and $\sqrt{1/M}$ .

Thus, quadrupling the temperature will double c for a given gas sample. For two gases, one of which has four times larger M than the other, molecules of the heavier gas will move half as fast as those of the lighter one.

Conclusion #2: temperature T

Transforming the above equation for c gives

$\begin{displaymath}RT = \frac{1}{3} M c^2 = \frac{2}{3} E_{k,m} = \frac{2}{3} N_A e_k \end{displaymath}$

where $E_{k,m}$ is the average kinetic energy per mole

$\begin{displaymath}E_{k,m} = \frac{1}{2} M c^2 = \frac{1}{2} N_A m c^2 = N_A e_k \end{displaymath}$

and

$\begin{displaymath}e_k = \frac{1}{2} m c^2 \end{displaymath}$

is the average kinetic energy per molecule. This leads us to the conclusion that the temperature of a mono-atomic ideal gas is proportional to average kinetic energy per mole (or the average kinetic energy per molecule).

It turns out that that this same relation between the average kinetic energy and temperature can be obtained for all bodies - gases, liquids or solids using the theory of classical mechanics. The application of quantum mechanics to this problem adds a little complication. Things stay the same for the mono-atomic ideal gas; for more complicated systems (e.g. diatomic gas, liquids or solids) the average kinetic energy and T are still related, but proportionality holds only at high temperatures. That is why our Microscopic Definition of T is worded more carefully:

The temperature of a body is a measure of the average kinetic energy per molecule.

Conclusion #3: Internal energy U of mono-atomic ideal gas

The only form of energy in an ideal gas is the kinetic energy. Using the kinetic theory of gases equation, we find for the internal energy U of a mono-atomic gas

$\begin{displaymath}U(T) = U(0) + E_k = U(0) + \frac{3}{2}nRT \end{displaymath}$

where U(0) is the energy of the gas at the temperature of 0 K. Thus, the energy of a mono-atomic ideal gas depends on the temperature of the gas only (p or V do not appear in the formula) and U varies linearly with T.

We can use the above formula to calculate the heat capacity of the ideal mono-atomic gas

$\begin{displaymath}C_V = \left(\frac{\partial U}{\partial T}\right)_V = \frac{3}{2}nR \end{displaymath}$

and it's molar heat capacity

$\begin{displaymath}C_{V,m} = \frac{C_V}{n} = \frac{3}{2}R \end{displaymath}$

Mean velocity and effusion

The mean velocity in a system of N atoms is defined as

$\begin{displaymath}\overline{c} = \frac{1}{N}\sum_{i=1}^N v_i \end{displaymath}$

where is the velocity of atom/molecule i. This quantity is similar to our RMS velocity c. Using the Maxwell distribution of molecular velocities, it can be shown that (more details below)

$\begin{displaymath}\overline{c} = \sqrt{\frac{8RT}{\pi M}} = c\sqrt{\frac{8}{3\pi}} \approx 0.92 c \end{displaymath}$

Effusion is a process where gas escapes from a container into vacuum through a small hole. Experimental observations led to Graham's Law of Effusion: the effusion rate is proportional to $M^{-1/2}$ . We can justify this formula by observing that we would expect the effusion rate to be proportional to the mean velocity, which in turn is proportional to the RMS velocity c.

Interesting fact. Our atmosphere contains very little of the lightest gases, hydrogen and helium. These gases have the largest atomic/molecular speeds and are able to escape the gravitational pull of the Earth into outer space.

Atomic motions in gases

Motions of atoms in gases may be described as a series of free motions at constant velocity in a straight line interspersed with collisions with neighbors which suddenly change the direction of the moving particle. On the straight line paths the atoms move with an average velocity $\overline{c}$ . This motion is quite fast, at hundreds or thousands m/s, which corresponds to the velocity of a bullet fired from a gun. The average distance traveled by an atom between collisions is called the mean free path. The mean free path of oxygen at 1 atm pressure is 70 nm, which is about 500 times the molecular diameter. Thus, the atoms spend most of their time flying in straight lines. Still, each atom undergoes about a billion collisions per second. The average time between collisions defines a unit of time called "collision time". The average distance travelled by a particle over a time t shorter than the collision time is

$\begin{displaymath}d = \overline{c} t \end{displaymath}$

Although the atoms move very fast, they constantly change the direction of their motion. Thus it takes them a long time to move significantly away from their starting point. The process of atomic motion over a time scale much longer than the collision time is called diffusion. Atoms observed on this time scale follow the following law:

$\begin{displaymath}d^2 = 6 D t or d = \sqrt{6Dt} \end{displaymath}$

where is the square of the average distance traveled during time t and D is the diffusion coefficient. In a gas D can be calculated in terms of the mean free path $\lambda$ and the mean velocity:

$\begin{displaymath}D = \frac{1}{3}\lambda \overline{c} \end{displaymath}$

Diffusive motion, also called Brownian motion, occurs in both gases and liquids, in which particle collisions are a fact of life. In liquids the collisions happen more often, and the mean free path is comparable to the molecular size. Diffusion is quite slow over large distances - we need to quadruple the observation time to observe a doubling of the distance traveled. This is quite different than the straight-line motions at constant velocity observed between collisions, for which the distance traveled is proportional to time.

Example. The diffusion coefficient for a typical small protein in aqueous solution is $D = 10^{-6} cm^2/s$ at room temperature. The time needed for the protein to move across a cell of size $10^{-4} cm$ is thus

$\begin{displaymath}t = \frac{d^2}{6 D} = \frac{10^{-8} cm^2}{6\cdot 10^{-6} cm^2/s} = 2\cdot 10^{-3} s \end{displaymath}$

while the time needed to move across a human body, or about 1 m, would be $2 \cdot 10^{9} s$ or about 60 years.

Perrin's experiments. J. B. Perrin (1860-1942) performed now famous experiments on the Brownian motion of particles. He used a microscope to observe small particles, such as the bright yellow gum resin gamboage, suspended in water. The resin particles were much larger than water molecules, but small enough so that collisions with the water would make them execute diffusive motion. He confirmed that the suspended particles obeyed the diffusive law of motion, with the square of the displacement from the original position proportional to observation time. This confirmed the theoretical predictions of Einstein and Smoluchowski. More importantly, Perrin's results provided a convincing demonstration of the existence of atoms and molecules. At that time (start of XXth century) not all scientists were convinced that matter was built of atoms, as experimental evidence was scarce.

Maxwell's distribution of velocities

A given atom in a gas can have all possible values of velocity. However, following one atom over a period of time, or looking at a large sample of different atoms at a given point in time, we will find that some velocities occur more and others less often. Theoretical studies predict and experiments confirm that the distribution of velocities is given by the Maxwell distribution, obtained by J. C. Maxwell at the end of the XIXth century. The probability distribution for the velocity along the x axis is

$\begin{displaymath}f(v_x) = \left(\frac{m}{2\pi k_BT}\right)^{1/2}e^{-\frac{mv_x^2}{2k_B T}} \end{displaymath}$

where $k_B = \frac{R}{N_A}$ is the Boltzmann constant and m is the atom mass.

The distribution allows us to calculate the probability of finding the velocity of the particle in the range between and as

$\begin{displaymath}f(v_x)dv_x \end{displaymath}$

The factor in front of the exponent is chosen so that the distribution is normalized, i.e. the sum probabilities adds up to one

$\begin{displaymath}\int_{-\infty}^{+\infty} f(v_x) dv_x = 1 \end{displaymath}$

We can understand the form of the Maxwell distribution as a special case of the Boltzmann distribution. According to Boltzmann the distribution of particle energies E at a temperature T is

$\begin{displaymath}f(E) \propto e^{-\frac{E}{k_B T}} and in our case E = \frac{1}{2}mv_x^2 \end{displaymath}$

Since motions along x, y, and z are independent, the distribution of particular values will be:

$\begin{displaymath}f(v_x, v_y, v_z) = f(v_x)f(v_y)f(v_z) = \left(\frac{m}{2... ..._BT}\right)^{3/2} e^{-\frac{m(v_x^2 + v_y^2 + v_z^2)}{2k_B T}} \end{displaymath}$

$\begin{displaymath}f(v_x, v_y, v_z) = \left(\frac{m}{2\pi k_BT}\right)^{3/2} e^{-\frac{mv^2}{2k_B T}} \end{displaymath}$

To calculate the distribution of the magnitude of the velocity v independent of its direction, we have to consider that all values satisfying the equation

$\begin{displaymath}v_x^2 + v_y^2 + v_z^2 = v^2 \end{displaymath}$

lie on the surface of a sphere of radius v, with area $4\pi v^2$ . Thus the distribution of v is

$\begin{displaymath}F(v) = 4\pi \left(\frac{m}{2\pi k_BT}\right)^{3/2} v^2 e^{-\frac{mv^2}{2k_B T}} \end{displaymath}$

The shape of this distribution is an asymmetric peak. The peak is narrow and falls at lower values of v when the temperature is low, and broad and falling at higher values of v when T is high.

If you remember your integral calculus, we can use F(v) to calculate some averages. The mean speed is

$\begin{displaymath}\overline{c} = \langle v \rangle = \int_{0}^{\infty} F(v)v dv = \sqrt{\frac{8k_B T}{\pi m}} \end{displaymath}$

The mean-square speed is

$\begin{displaymath}c^2 = \langle v^2 \rangle = \int_{0}^{\infty} F(v)v^2 dv = \sqrt{\frac{3k_B T}{m}} \end{displaymath}$

The agreement with the kinetic theory of gases may be seen when we notice that

$\begin{displaymath}\frac{R}{M} = \frac{N_A k_B}{ N_A m} = \frac{k_B}{m} \end{displaymath}$

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