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, 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 while 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:
\(ΔP\) = \((-m v_x) – (m v_x)\) = \(-2m v_x\) ……….. (1)

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

After the collision, the molecule travels towards the face \(A_2\) with x component of the velocity equal to \(- v_x\).

Distance travelled from \(A_1\) to \(A_2\) = \(L\)
Therefore, time = \(\frac{L}{v_x}\)

After collision with \(A_2\), the molecule again travels to \(A_1\). Thus, the time between two collisions is \(\frac{2L}{v_x}\)

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

\(n\) = \(\frac{v_x}{2L}\) ………………… (2)

Using equations 1 and 2,

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

\(ΔF\) = \(nΔp\)

= \(\frac{m}{L} v_x^2\)

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

\(F\) = \(Ʃ \frac{m}{L} vx^2\)

= \(\frac{m}{L} Ʃ{v_x}^2\)

Now, \(Ʃ{v_x}^2\) = \(Ʃ{v_y}^2\) = \(Ʃ{v_z}^2\) (symmetry)

= \(\frac{1}{3} Ʃv^2\)

Therefore, \(F\) = \(\frac{1}{3} \frac{m}{L} Ʃv^2\)

Pressure is force per unit area so that

\(P\) = \(\frac{F}{L^2}\)

= \(\frac{1}{3} \frac{M}{L^{3}} \frac{\sum v^{2}}{N}\)

= \(\frac{1}{3} \rho \frac{\sum v^{2}}{N}\)


\( M \)= total mass of gas

\(\rho\) = density of the gas

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

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