Calculating Pressure Of An Ideal Gas

Imagine an ideal gas contained in a cubical container. Let one corner of the cube be the origin O, and let the x, y, and z-axes along the edges. Let A1and A2 be the parallel faces perpendicular to the x-axis. Consider a molecule moving with velocity v in the container. The components of velocity along the axes are vx, vy, and vz. Now, when the molecule is colliding with face A1, the x component of velocity is reversed. In contrast, the y and z component of velocity remains unchanged (as per our assumption that the collisions are elastic).

Ideal Gas

The change in momentum of the molecule is:

\(\begin{array}{l}ΔP = (-m v_x) – (m v_x) = -2m v_x ………(1)\end{array} \)

Since the momentum remains conserved, the change in momentum of the wall is 2mvx

After the collision, the molecule travels towards the face A2 with x component of the velocity equal to – vx.

Distance travelled from A1 to A2 = L

Therefore, time = L/vx

After collision with A2, the molecule again travels to A1. Thus, the time between two collisions is 2L/vx.

Therefore, the number of collisions of the molecule per unit of time:

\(\begin{array}{l}n = \frac{v_x}{2L} ………………… (2)\end{array} \)

Using equations 1 and 2,

The momentum imparted per unit time to the wall by the molecule:

\(\begin{array}{l}ΔF = nΔp = \frac{m}{L} v_x^2\end{array} \)

Therefore, the total force on wall A1 due to all the molecules is

\(\begin{array}{l}F = Ʃ \frac{m}{L} v_x^2\end{array} \)

\(\begin{array}{l}=\frac{m}{L} Ʃ{v_x}^2\end{array} \)

\(\begin{array}{l}Ʃ{v_x}^2 = Ʃ{v_y}^2 = Ʃ{v_z}^2… (symmetry)\end{array} \)

\(\begin{array}{l}=\frac{1}{3} Ʃv^2\end{array} \)

Therefore,

\(\begin{array}{l}F = \frac{1}{3} \frac{m}{L} Ʃv^2\end{array} \)

Pressure is force per unit area so that

\(\begin{array}{l}P = \frac{F}{L^2}\end{array} \)
\(\begin{array}{l}= \frac{1}{3} \frac{M}{L^{3}} \frac{\sum v^{2}}{N}\end{array} \)
\(\begin{array}{l}=\frac{1}{3} \rho \frac{\sum v^{2}}{N}\end{array} \)

Where,

M = total mass of gas

ρ = density of the gas

\(\begin{array}{l}Now\, \frac{\sum v^{2}}{N} \,is\,written\,as\,v^2\, and\,is\,called\,the mean\,square\,speed.\end{array} \)
\(\begin{array}{l}P=\frac{1}{3}\rho v^{2}\end{array} \)

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