Pressure of Ideal Gas

Set-up and assumptions

Let there be $n$ moles of an ideal gas enclosed in a cubical box of volume $V = L^{3}$, with the sides of the box parallel to the coordinate axes. The walls of the box are kept at a constant temperature $T$. We use the word molecular speed rather than molecular velocity, since the kinetic energy of a molecule depends on the magnitude of the velocity irrespective of its direction.

• The gas molecules are in continuous random motion in various directions, colliding with one another and hitting the walls of the box and bouncing back.

• As a first approximation we neglect intermolecular collisions and consider only elastic collisions with the walls. This is physically reasonable, because the mean free path increases as the pressure is reduced; the pressure is imagined so adjusted that the molecules do not collide with each other but do collide with the walls.

Diagram in words

A cube of side $L$ with its faces parallel to the coordinate planes. One face — the shaded wall — is parallel to the $yz$-plane. A single molecule of mass $m$ is shown inside the cube, moving with velocity $\vec{v}$ towards this shaded wall, with the components $v_{x}$, $v_{y}$, $v_{z}$ marked. After the elastic collision the molecule leaves with $v_{x}$ reversed and $v_{y}$, $v_{z}$ unchanged.

Derivation

i. A typical molecule moves with velocity $\vec{v}$ and is about to collide elastically with the shaded wall of the cube, which is parallel to the $yz$-plane.

ii. During the elastic collision the component $v_{x}$ of the velocity gets reversed, while $v_{y}$ and $v_{z}$ remain unaltered. Considering all the molecules, their average $y$ and $z$ components of velocity are unchanged by collisions with this wall, because the gas molecules remain evenly distributed throughout the volume and gain no additional motion along $\pm y$ or $\pm z$.

iii. Hence the change in momentum of the particle is only in the $x$ component:

$$\Delta p_{x} = \text{final momentum} - \text{initial momentum} = \left(-mv_{x}\right) - \left(mv_{x}\right) = -2mv_{x} \qquad ....(1)$$

iv. Thus the momentum transferred to the wall during the collision is $+2mv_{x}$. The rebounded molecule then travels to the opposite wall and collides with it.

v. After colliding with the shaded wall, the molecule travels to the opposite wall, is reflected back, and travels towards the shaded wall again. It therefore covers a distance $2L$ between two successive collisions with the shaded wall. As $L$ is the length of the cubical box, the time for the molecule to travel back and forth to the shaded wall is

$$\Delta t = \dfrac{2L}{v_{x}}$$

vi. The average force exerted on the shaded wall by molecule 1 equals the average rate of change of momentum:

$$F_{avg} = \dfrac{2mv_{x_{1}}}{2L/v_{x_{1}}} = \dfrac{mv_{x_{1}}^{2}}{L} \qquad ....(2)$$

where $v_{x_{1}}$ is the $x$ component of the velocity of molecule 1.

vii. Considering the other molecules 2, 3, 4, ... with respective $x$ components $v_{x_{2}}, v_{x_{3}}, v_{x_{4}}, \ldots$, the total average force on the wall is

$$F_{avg} = \dfrac{m}{L}\left(v_{x_{1}}^{2} + v_{x_{2}}^{2} + v_{x_{3}}^{2} + \cdots\right)$$

viii. The average pressure is the average force divided by the area $L^{2}$ of the shaded wall:

$$P = \dfrac{\text{Average force}}{\text{Area of shaded wall}} = \dfrac{m\left(v_{x_{1}}^{2} + v_{x_{2}}^{2} + \cdots\right)}{L \times L^{2}}$$

ix. The average of the square of the $x$ component of the velocities is

$$\overline{v_{x}^{2}} = \dfrac{v_{x_{1}}^{2} + v_{x_{2}}^{2} + v_{x_{3}}^{2} + \cdots + v_{x_{N}}^{2}}{N}$$

so that, with $V = L^{3}$,

$$P = \dfrac{mN\overline{v_{x}^{2}}}{V}$$

x. Now $\overline{v^{2}} = \overline{v_{x}^{2}} + \overline{v_{y}^{2}} + \overline{v_{z}^{2}}$. By symmetry, since the molecules have no preferred direction to move,

$$\overline{v_{x}^{2}} = \overline{v_{y}^{2}} = \overline{v_{z}^{2}} = \dfrac{1}{3}\overline{v^{2}}$$

xi. Therefore the average pressure is

$$P = \dfrac{1}{3}\dfrac{N}{V}m\overline{v^{2}} \qquad ....(3)$$

Remarks the board expects

• Equation (3) has been obtained for a cubical container, but it can be shown to be valid for containers of any shape.

• The neglect of intermolecular collisions does not affect the result. The number of molecules is of the order of $10^{23}$, and if a molecule acquires a velocity with components different from $v_{x}$, $v_{y}$, $v_{z}$ after a collision, there will invariably be some other molecule of different initial velocity that now acquires those components. Since the gas is in equilibrium this must be happening, so the expression is unchanged.

Standard results

$$P = \dfrac{1}{3}\dfrac{N}{V}m\overline{v^{2}} = \dfrac{1}{3}\rho\overline{v^{2}}, \qquad \rho = \dfrac{Nm}{V}$$

$$P = \dfrac{2}{3}\dfrac{N}{V}\left(\dfrac{1}{2}m\overline{v^{2}}\right)$$

• $\Delta p_{x} = -2mv_{x}$; momentum given to the wall $= +2mv_{x}$; $\Delta t = \dfrac{2L}{v_{x}}$; $\overline{v_{x}^{2}} = \dfrac{1}{3}\overline{v^{2}}$.