Pressure

Suppose that the $x$-$y$ plane actually corresponds to a wall of the container. Consider, again, molecules whose speeds lie between $v$ and $v+dv$, and whose directions of motion subtend an angle lying between $\theta$ and $\theta+d\theta$ with the $z$-axis. Each such molecule that encounters the wall bounces off it in a specular fashion, and its $z$-momentum consequently changes by $2\,m\,v_z$, where $m$ is the molecular mass. Thus, the normal reaction force per unit area acting on the wall is

$$dp =[2\,m\,v_z]\,[n\,F(v)\,dv] \,[g(\theta)\,d\theta]\,[v_z]=\,[2... ...n\,F(v)\,dv\right]\left[\frac{1}{2}\,\sin\theta\,d\theta\right][v\,\cos\theta].$$ (5.173)

[See Equation (5.167).] Hence, the total pressure exerted on the wall is

$$p = n\,m\,\int_0^{\pi/2} \sin\theta\,\cos^2\theta\,d\theta\,\int_0^\infty F(v)\,v^2\,dv,$$ (5.174)

which reduces to

$$p = \frac{1}{3}\,n\,m\left\langle v^2\right\rangle,$$ (5.175)

where

$$\left\langle v^2\right\rangle = \int_0^\infty F(v)\,v^2\,dv.$$ (5.176)

is the mean square molecular speed. (See Section 5.5.9.)

However, we can write

$$n = \frac{\nu\,N_A}{V},$$ (5.177)

where $\nu$ is the number of moles of molecules held inside the container, $V$ is the volume of the container, and $N_A$ is Avogadro's number. Equations (5.175) and (5.177) yield

$$\frac{p\,V}{\nu} = \frac{2}{3}\,N_A\,\langle {\cal K}_{\rm trans}\rangle,$$ (5.178)

where

$$\langle {\cal K}_{\rm trans}\rangle = \frac{1}{2}\,m\left\langle v^2\right\rangle$$ (5.179)

is the mean translational kinetic energy of a molecule in the gas. Equation (5.178) is consistent with the ideal gas law, (5.97), provided that

$$\langle {\cal K}_{\rm trans}\rangle= \frac{1}{2}\,m\left\langle v^2\right\rangle= \frac{3}{2}\,k_B\,T,$$ (5.180)

where $k_B=R/N_A$ is the Boltzmann constant. [See Equation (5.100).]