Question

Derive an expression for a pressure exerted by a gas on the basis of kinetic theory of gases.

Hint:

Kinetic theory links macroscopic pressure to microscopic molecular motion; use \( P = \frac{1}{3} \rho v_{\text{rms}}^2 \).

Answer 1

The kinetic theory assumes gas molecules are in random motion, colliding elastically with container walls.
Step 1: Consider a cubic container of side length \( L \) with \( N \) molecules, each of mass \( m \). Assume 1/3 molecules move along each axis.
Step 2: For a molecule moving with velocity \( v_x \) along x-axis, change in momentum upon collision with wall: \( 2 m v_x \). Time between collisions: \( \frac{2L}{v_x} \).
Force per molecule: \( \frac{2 m v_x}{\frac{2L}{v_x}} = \frac{m v_x^2}{L} \).
Total force on wall: \( \frac{N}{3} \cdot \frac{m \langle v_x^2 \rangle}{L} \).
Pressure \( P = \frac{\text{force}}{\text{area}} = \frac{N m \langle v_x^2 \rangle}{3 L^3} = \frac{N m \langle v_x^2 \rangle}{3 V} \).
Step 3: Since \( v_{\text{rms}}^2 = \langle v^2 \rangle = 3 \langle v_x^2 \rangle \), \( \langle v_x^2 \rangle = \frac{v_{\text{rms}}^2}{3} \).
\[ P = \frac{1}{3} \frac{N m v_{\text{rms}}^2}{V}. \] Answer: \( P = \frac{1}{3} \rho v_{\text{rms}}^2 \), where \( \rho = \frac{N m}{V} \).