Physics 9702 · AS & A Level · Kinetic theory of gases

Kinetic theory of gases — practice question

A basic assumption in the kinetic theory of gases is that every collision between gas molecules is elastic.
(a(i))[1]

What is meant by an elastic collision?

(a(ii))[2]

State two further assumptions of the kinetic theory of gases.

(b(i))[1]

An ideal gas molecule has mass $m$ and is inside a cube of side length $L$. The molecule moves with velocity $u$ towards the face of the box shaded in Fig. 3.1. It collides elastically with the shaded face and then the opposite face alternately. Deduce an expression, in terms of $m$, $u$ and $L$, for the magnitude of the change in momentum of the molecule on colliding with a face.

(b(ii))[1]

Deduce an expression, in terms of $m$, $u$ and $L$, for the time between consecutive collisions of the molecule with the shaded face.

(b(iii))[1]

Deduce an expression, in terms of $m$, $u$ and $L$, for the average force exerted by the molecule on the shaded face.

(b(iv))[1]

Deduce an expression, in terms of $m$, $u$ and $L$, for the pressure on the shaded face if the force in (iii) is exerted over the whole area of the face.

(c)[2]

For the model in (b) extended to three dimensions and to a gas containing $N$ molecules, each of mass $m$, moving with mean-square speed $\langle c^2 \rangle$, it can be shown that $pV = \frac{1}{3}Nm\langle c^2 \rangle$, where $p$ is the pressure exerted by the gas and $V$ is the volume of the gas. Use this expression, together with the equation of state of an ideal gas, to show that the average translational kinetic energy $E_k$ of a molecule of an ideal gas is given by $E_k = \frac{3}{2}kT$, where $T$ is the thermodynamic temperature of the gas and $k$ is the Boltzmann constant.

(d)[2]

The mass of a hydrogen molecule is $3.34 \times 10^{-27}\,\text{kg}$. Use the expression for $E_k$ in (c) to determine the root-mean-square (r.m.s.) speed of a molecule of hydrogen gas at $25\,^{\circ}\text{C}$.

Worked solution & mark scheme

This 11-mark question has a full step-by-step worked solution and mark scheme. One marking point: No kinetic energy is lost

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