Derivation of Ideal Gas Equation from Kinetic Theory of Gases

26th May 2019 @ 9 min read

Ideal gas equation is PV = nRT. This equation can easily be derived from the combination of Boyle’s law, Charles’s law, and Avogadro’s law. But here, we will derive the equation from the kinetic theory of gases. The kinetic theory of gases is a very important theory which relates macroscopic quantities like pressure to microscopic quantities like the velocity of gas molecules. This equation is applicable only for ideal gases, but be approximated for real gas under some conditions.

In 1856 German chemist August Kronig had developed the simple gas model. He had only considered the translational motion of gas particles in his model. Later, in 1857 Rudolf Clausius independent of Kronig developed a sophisticated version of the kinetic theory of gases. Clausius had not only considered translational motion but also the rotational and vibrational motion of gas molecules.

Assumptions in Ideal Gas Equation

The behaviour of gases is simplified significantly by making many assumptions. This is necessary to avoid complexity. The assumptions in the theory are as follows:

The gas comprises of very small particle known as molecules.
The molecules of the gas are solid rigid identical spheres.
All the molecules of the gas have the same mass.
The volume of each molecule is negligible in comparison to the total volume of the gas. Or in other words, the molecules do not occupy space relative to the size of a container in which they are stored.
The intermolecular forces among the molecules are zero. The molecules do not exert any forces among them.
The molecules are in constant random motion. They travel in only straight lines.
All collisions among them and between the molecules and wall of the container are perfectly elastic in nature. Thus, there is no loss of kinetic energy in collisions.
The pressure is the net result of collisions between molecules and the wall of the container.
The average kinetic energy is the function of temperature alone.

Derivation of Ideal Gas Equation

Visualise a cube in space as shown in the figure below. L be the length of the cube and Area, A. V be the volume of the cube.

The cube is filled with an ideal gas of pressure P, at temperature T. Let n and N be the moles and the number of molecules of the gas in the cube. m and M are the mass of a molecule and the gas.

A cube with the length L and the surface area A — Figure 1: A cube in space with length L and area A

Consider a single molecule moving with velocity v⃗. Let v⃗_x, v⃗_y, and v⃗_z are the velocity components in the x, y, and z directions respectively.

Step 1: To Determine the Frequency of Collisions.

For simplicity, we will start with the x-direction as depicted in the below figure. Initially, at x = 0 a molecule with mass m moves with velocity v⃗_x. It strikes the wall at x = L. This is the first collision. The total distance travel in the first collision is L₀ and the time taken is t₁. The second collision takes place at t₁ + t, and the distance travel between the 1^st collision and 2^nd collision is 2L. The time interval between the 1^st and 2^nd collision is ∆t = (t₁ + t) −t₁ = t. This is also shown in the figure below.

The figure depicts various instances of molecular collision with time. — Figure 2: Location of a molecule at different instant of time

For the 3^rd collision, the time taken is t₁ + 2t, but the time interval between preceding collision is the same ∆t = (t₁ + 2t) − (t₁ + t) = t. Also, the distance travel by the molecule between the 2^nd and 3^rd collision is the same 2L. This two quantities, time interval and distance travel, will remain the same for all successive collisions. So, we can calculate the speed v_x which is distance travel divided time interval for all collisions as follows:

$v_1=\frac{L_0}{t_1},\, v_2=\frac{2L}{t},\, v_3=\frac{2L}{t},\, v_4=\frac{2L}{t} \ldots$

Here v₁, v₂, v₃, v₄… are the velocities for 1^st, 2^nd, 3^rd, 4^th,… collisions, and they all are equal to v_x.

$v_1=v_2=v_3=v_4\ldots=v_x$

Therefore, the time intervals for respective collisions are

$t_1=\frac{L_0}{v_1}=\frac{L_0}{v_x}\\ t_2=\frac{2L}{v_2}=\frac{L_0}{v_x}\\ t_3=\frac{2L}{v_3}=\frac{2L}{v_x}\\ t_4=\frac{2L}{v_4}=\frac{2L}{v_x}\\ \ldots$

The average time taken t in k collisions is

$t &=\frac{t_1+t_2+t_3+t_4+ \ldots t_k}{k}\\ &=\frac{L_0\big/ v_x +2L\big/ v_x +2L\big/ v_x +2L\big/ v_x + \ldots +2L\big/ v_x}{k}\\ &=\frac{L_0\big/ v_x +2L\big/ v_x(1+1+1+1+ \ldots +1)}{k}\\ &=\frac{L_0\big/ v_x +2L\big/ v_x \times k}{k}\\ &=\frac{L_0 +2kL}{kv_x}$

When k tends to infinity, we get

$t =\lim_{k \rightarrow \infty}\frac{L_0 +2kL}{kv_x} =\frac{2L}{v_x}$

This is the time period for a collision, and the reciprocal of it is frequency (f) of collisions.

$f=\frac{1}{t}=\frac{v_x}{2L} \qquad \text{(1)}$

Step 2: To Derive Force

The force F is defined as the rate of change of momentum ∆p.

$F=\frac{\Delta p}{\Delta t} = \frac{\Delta p}{f}$

The change in momentum is difference in momentum after a collision (−mv_x) minus before a collision (mv_x).

$\Delta p= -mv_x-mv_x=-2mv_x$

Note: The negative momentum means the molecule has lost the momentum after the collision.

The force exerted by wall F_wm on the molecule is given as:

$F_\text{wm} =\Delta{p}f = -2mv_xf$

The Newton’s 3^rd law says at every action there is an equal and opposite reaction. So, the force exerted by the molecule on the wall F_mw is opposite of the force exerted by the wall on the molecule F_wm.

$F_\text{mw} =-F_\text{wm} =2mv_xf \qquad (2)$

Using equation (1) and (2),

$F_\text{mw}=2mv \times \frac{v_x}{2L}=\frac{m{v^2_x}}{L}$

In the beginning, we have considered only one molecule; So, the above force equation is for only one molecule. The force by N molecules moving in x-direction is,

$F_\text{mw} &=\frac{m v_{x_1}^2}{L} +\frac{m v_{x_2}^2}{L} +\frac{m v_{x_3}^2}{L} \ldots +\frac{m v_{x_N}^2}{L}\\ &=\frac{m}{L}\left(v_{x_1}^2+ v_{x_2}^2+ v_{x_3}^2+\ldots+ v_{x_N}^2\right)\\ &=\frac{m}{L}\left(\sum_{i=1}^{N}v_{x_i}^2\right) \qquad\qquad\qquad \qquad\qquad\qquad\qquad (3)$

Root mean square velocity v̄² is given as

${v_\text{rms}}^2 =\bar{v}^2 =\frac{1}{N}\sum_{i=1}^{N} v_{x_i}^2$

Substituting the above equation in (3),

$F_\text{mw} =\frac{Nm\bar{v}^2}{L}$

Note: This force is a force exerted by a molecule on one wall of the cube.

Step 3: To Determine the Pressure of the Gas

By knowing the force on the wall, we can determine pressure (P), which is force per unit area.

$P =\frac{F_\text{mw}}{A} =\frac{Nm \bar{v}_x^2}{AL} =\frac{Nm \bar{v}_x^2}{V} \qquad (4)$

Here, V is the volume of the cube.

In equation (4), we have only incorporated the x direction. We need to replace v̄²
_x by v̄².

From mathematics, the relation between v̄²
_x, v̄²
_y, and v̄²
_z and v̄² is given as:

$\bar{v}^2 =\bar{v}_x^2 +\bar{v}_y^2 +\bar{v}_z^2 \qquad (5)$

Also, each of these velocities is equal.

$\bar{v}_x^2 =\bar{v}_y^2 =\bar{v}_z^2 \qquad(6)$

From equation (5) and (6),

$\bar{v}^2 =\bar{v}_x^2 +\bar{v}_x^2 +\bar{v}_x^2$

$\bar{v}^2 =3\bar{v}_x^2 \qquad (7)$

Substituting equation (7) in (4),

$P=\frac{Nm \bar{v}^2}{3V} \qquad (8)$

The equation (8) relates macroscopic pressure to microscopic velocity.

Step 4: To Determine Energy associated with Ideal Gas

The problem with equation (8) is that the rms velocity v̄² is not a convenient quantity to measure. We will replace it with a more convenient quantity: temperature which can easily measure by a thermometer.

The kinetic energy equation from classical mechanics is half of the mass times square of velocity.

$E_\text{k}=\frac{1}{2}m \bar{v}^2 \qquad (9)$

Also, from statistical mechanics, the kinetic energy for monatomic gas is given as:

$E_\text{k}=\frac{3}{2} kT \qquad (10)$

Here, k is the Boltzmann constant and its approximate value is 1.380 × 10⁻²³ J K⁻¹.

Using equation (9) and (10)

$E_\text{k}=\frac{3}{2} kT =\frac{1}{2}m \bar{v}^2$

$m \bar{v}^2=3kT \qquad (11)$

Step 5: To Determine Ideal Gas Equation

From equation (8) and (11),

$P=\frac{N 3 kT}{3V}=\frac{NkT}{V}$

The Boltzmann constant (k) is also ratio of the Gas constant (R) to the Avogadro’s constant (N_A). Substituting $k=\frac{R}{N_\text{A}}$ in equation the above equation,