Match each kinetic theory concept or microscopic property of an ideal gas on the left with… — JAMB UTME

Match each kinetic theory concept or microscopic property of an ideal gas on the left with its corresponding mathematical expression or derivation result on the right.

Magnitude of momentum change $(\Delta p_x)$ for a gas molecule of mass $m$ colliding elastically with a container wall perpendicular to the x-axis at speed $v_x$ $2 m v_x$
Average force $(F_x)$ exerted by a single gas molecule moving back and forth between two parallel walls separated by length $L$ $\frac{m v_x^2}{L}$
Translational kinetic energy per unit volume $\left(\frac{E_k}{V}\right)$ of an ideal gas operating at pressure $P$ $\frac{3}{2} P$
Root-mean-square speed $(v_{\text{rms}})$ of an ideal gas molecule in terms of molar mass $M$ , universal gas constant $R$ , and absolute temperature $T$ $\sqrt{\frac{3 R T}{M}}$

Answer

The correct matches pair the momentum change per collision with

2 m v_x

, the single-molecule average wall force with

\frac{m v_x^2}{L}

, the kinetic energy density with

\frac{3}{2} P

, and the root-mean-square speed with

\sqrt{\frac{3 R T}{M}}

Each kinetic theory quantity is derived directly from fundamental principles of mechanics applied to gas particles. Elastic collision with a wall yields a momentum reversal of magnitude

2 m v_x

. Taking the round-trip collision frequency over length

L

yields an average force of

\frac{m v_x^2}{L}

. Linking microscopic kinetic energy density to pressure gives

\frac{E_k}{V} = \frac{3}{2} P

, and linking pressure to the ideal gas law for one mole yields

v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}

Step-by-Step Solution

Analyze momentum transfer during elastic collision of a molecule with a wall.

Initial momentum along the x-axis is

p_i = m v_x

and final momentum after elastic reflection is

p_f = -m v_x

. The change in momentum is

\Delta p_x = p_f - p_i = -2 m v_x

, which has a magnitude of

2 m v_x

Elastic collision conserves kinetic energy and reverses velocity direction perpendicular to the wall.

Calculate the time rate of momentum transfer to determine average force.

The round-trip distance between opposite walls separated by length

L

2L

, so the time between collisions with the same wall is

\Delta t = \frac{2L}{v_x}

. The average force is

F_x = \frac{\Delta p}{\Delta t} = \frac{2 m v_x}{2L / v_x} = \frac{m v_x^2}{L}

Newton's second law expresses force as the average rate of change of momentum.

Relate total translational kinetic energy density to gas pressure.

From kinetic theory, gas pressure is given by

P = \frac{1}{3} \frac{N m}{V} v_{\text{rms}}^2

. Since total kinetic energy

E_k = \frac{1}{2} N m v_{\text{rms}}^2

, we can express pressure as

P = \frac{2}{3} \left(\frac{E_k}{V}\right)

. Rearranging gives energy density

\frac{E_k}{V} = \frac{3}{2} P

Translational kinetic energy density is directly proportional to pressure with a factor of 3/2.

Derive the formula for root-mean-square velocity from macroscopic and microscopic gas equations.

Substitute density

\rho = \frac{M}{V}

(where

M

is molar mass) into

P = \frac{1}{3} \rho v_{\text{rms}}^2

, obtaining

P = \frac{M v_{\text{rms}}^2}{3 V}

. Since

P V = R T

for one mole of ideal gas,

R T = \frac{1}{3} M v_{\text{rms}}^2 \implies v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}

Connects microscopic speed distribution parameter with thermodynamic temperature and molar mass.

Key Concept

Kinetic Theory of Matter and Pressure of Gases

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