Ideal gases and molecular kinetic theory (RP8) – A-level Physics exam questions with mark scheme

1Core3 marks

A gas cylinder of volume 0.020 m$^3$ contains an ideal gas at a pressure of 1.5 × 10$^5$ Pa and a temperature of 20 °C.

(a) Calculate the number of moles of gas in the cylinder. [2]

(b) Calculate the number of gas molecules in the cylinder. [1]

(a) temperature in kelvin: $T = 293$ K and $n = \frac{pV}{RT}$(1)
(a) $n = \frac{1.5\times10^5 \times 0.020}{8.31 \times 293}$ = 1.2 mol (1.23)(1)
(b) $N = nN_A = 1.23 \times 6.02\times10^{23}$ = 7.4 × 10$^{23}$ (ecf from (a); allow $N = pV/kT$)(1)
Answer: (a) 1.23 mol (b) 7.4 × 10$^{23}$
Examiner tip: Convert to kelvin first: using 20 instead of 293 in pV = nRT gives an answer about 15 times too big and loses both marks in (a).

2Standard4 marks

A container holds helium gas at a temperature of 300 K. The molar mass of helium is 4.0 × 10$^{-3}$ kg mol$^{-1}$.

(a) Calculate the mean kinetic energy of a helium atom. [1]

(b) Calculate the root mean square speed of the helium atoms. [3]

(a) $\frac{3}{2}kT = 1.5 \times 1.38\times10^{-23} \times 300$ = 6.2 × 10$^{-21}$ J(1)
(b) mass of one atom $m = \frac{4.0\times10^{-3}}{6.02\times10^{23}}$ = 6.6 × 10$^{-27}$ kg(1)
(b) $\frac{1}{2}m(c_{rms})^2 = 6.21\times10^{-21}$ (or $c_{rms} = \sqrt{3RT/M}$)(1)
$c_{rms}$ = 1.4 × 10$^3$ m s$^{-1}$ (1370)(1)
Answer: (a) 6.2 × 10$^{-21}$ J (b) 1.37 × 10$^3$ m s$^{-1}$
Examiner tip: Use the mass of one atom (molar mass ÷ N_A) with k, or the molar mass with R — mixing them is a frequent error.

3Hard4 marks

A car tyre contains air at an absolute pressure of 2.4 × 10$^5$ Pa when its temperature is 7 °C. After a long journey the temperature of the air is 35 °C and the volume of the tyre has increased by 3.0%.

Calculate the new pressure of the air in the tyre. Assume no air escapes and the air behaves as an ideal gas. [4]

temperatures converted to kelvin: 280 K and 308 K(1)
uses $\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}$(1)
$p_2 = \frac{2.4\times10^5 \times 308}{1.03 \times 280}$ (V₂ = 1.03 V₁ used)(1)
$p_2$ = 2.6 × 10$^5$ Pa (2.56 × 10$^5$)(1)
Answer: 2.56 × 10$^5$ Pa
Examiner tip: Write V₂ as 1.03 V₁ — the V₁ cancels, so you don't need the actual volume.

Stuck on Ideal gases and molecular kinetic theory (RP8)?

Get it sorted in a free 30-minute session

We work through your weakest questions on this topic on Zoom. I reply the same day.

Free 30-min session on this topic Send me 5 questions a day

4Hard6 marks

Explain, using the kinetic theory model of an ideal gas, how a gas exerts a pressure on the walls of its container, and why the pressure increases when the gas is heated at constant volume. [6]

molecules are in constant random motion(1)
molecules collide with the walls and rebound (elastically)(1)
each collision changes the momentum of a molecule (by $2mu$ for a head-on collision)(1)
by Newton's second law, rate of change of momentum = force on molecule; by Newton's third law an equal and opposite force acts on the wall(1)
many collisions per second produce a (steady) total force; pressure = force ÷ area of wall(1)
heating increases mean kinetic energy, so molecules move faster (mean KE ∝ T)(1)
faster molecules hit the walls more often (more collisions per second)(1)
and each collision has a larger change in momentum, so a greater force and greater pressure(1)

Level	Marks	What it looks like
Level 3	5–6	A coherent, logically ordered explanation covering both the origin of pressure (collisions, momentum change, force over area) and the effect of heating (faster molecules, more frequent and harder collisions), linked correctly to the relevant ideas.
Level 2	3–4	Explains how pressure arises with some detail, or gives a partial account of both parts; some links between ideas are missing or unclear.
Level 1	1–2	Isolated relevant statements, e.g. molecules collide with the walls or move faster when hot, with little linking.
Level 0	0	No relevant content.

Answer: Pressure from rate of change of momentum in wall collisions; heating → faster molecules → more frequent, larger momentum changes → higher pressure.
Examiner tip: Top-band answers link momentum change to force via Newton's laws — just saying 'molecules hit the walls' stays in Level 1.

5Stretch5 marks

(a) Use the equations $pV = \frac{1}{3}Nm(c_{rms})^2$ and $pV = NkT$ to show that the mean kinetic energy of a molecule of an ideal gas is $\frac{3}{2}kT$. [2]

(b) A sample of air contains nitrogen (molar mass 28 g mol$^{-1}$) and oxygen (molar mass 32 g mol$^{-1}$) molecules at the same temperature. Calculate the ratio $\frac{c_{rms}\ \text{of nitrogen}}{c_{rms}\ \text{of oxygen}}$. [2]

(c) The temperature of a gas is increased from 27 °C to 327 °C. Calculate the factor by which the rms speed of its molecules increases. [1]

(a) equates: $\frac{1}{3}Nm(c_{rms})^2 = NkT$, so $m(c_{rms})^2 = 3kT$(1)
(a) mean KE $= \frac{1}{2}m(c_{rms})^2 = \frac{1}{2} \times 3kT = \frac{3}{2}kT$(1)
(b) same $T$ so same mean KE: $m_N c_N^2 = m_O c_O^2$, ratio $= \sqrt{\frac{32}{28}}$(1)
(b) ratio = 1.07(1)
(c) $c_{rms} \propto \sqrt{T}$: $\sqrt{600/300}$ = 1.41 (accept √2)(1)
Answer: (b) 1.07 (c) 1.41
Examiner tip: Temperatures must be in kelvin: 327 °C is not 'about 12 times' 27 °C — it is double in kelvin.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session