The pressure exerted by a gas on the walls of a container results from the momentum transfer during the collisions of the gas molecules with the walls. The normal force on the wall by all the molecules, per unit wall area, is the gas pressure on the wall.
Cartesian components of a molecular velocity
Consider a very dilute gas at a steady-state temperature enclosed in a cubical container of side L. The coordinate axes are aligned with the edges of this cube, as shown in above figure. Suppose that there are N molecules in the container, each of mass m. Consider a molecule moving with velocity .
\(\vec{v_1}\) = v1x \(\hat i\) + v1y\(\hat j\) + v1z\(\hat k\)......(1)
where v, v1x and v1z are the velocity components along the x-, y- and z-axes, respectively, so that
v12 = v1x2 + v1y2 + v1z2
The change in momentum of the molecule on collision with the right wall
= final momentum – initial momentum
= (-m|v1x|) – m|v1x| = -2m|v1x|
Assuming the collision to be elastic, the change in momentum of the right wall due to this collision is 2m|v1x|.
Two dimensional elastic collision of a gas molecule in which x-component of the velocity of the molecule is reversed on the collision \(\vec{v_1}\) is the xy plane
Since the distance between the right and left walls is L, the time between successive collisions with the right wall is,
∆t = \(\frac{2L}{|v_{1x}|}\)
Therefore, the frequency with which the molecule collides with the right wall is \(\frac{|v_{1x}|}{2L}\).
∴ Rate of change of momentum of the right wall = (change in momentum in one collision) (frequency of collision)
= (2m|v1x|). \(\frac{|v_{1x}|}{2L}\) = \(\frac{{mv}_{1x}^2}{L}\)…. (2)
By Newton’s second law of motion, this is the force f1x exerted by the molecule on the right wall.
Thus, the net force on the right wall by all the N molecules is
The pressure Px exerted by all the molecules on a wall perpendicular to the x-axis is
where V = L3 is the volume of the container.
Similarly, Py = \(\frac{m}V\) (v1y2 + v2y2 +....+ vN2)..........(5)
and Pz = \(\frac{m}V\)(v1z2 + v2z2 +....+ vNz2)..............(6)
As pressure is the same in all directions,
By definition, the mean square speed of the gas molecules is
Here, Nm = (number of molecules).
(molecular mass)
= mass of the gas, M
\(\therefore\) \(\frac MV\) = density, \(\rho\)
\(\therefore\) = \(\frac 13\)\(\rho\)\(\overline{v^2}\).............(11)
Writing \(\overline{v^2}\) = v2rms,
P = \(\frac 13\)\(\rho\)\(\overline{v^2}\)rms ........(12)
where vrms is the root-mean-square speed of molecules of the gas at a given temperature. Equation (12) gives the pressure exerted by an ideal gas in terms of its density and the rootmean-square speed of its molecules.
