Thermal physics (ideal gases, kinetic theory, specific heat & latent heat) – A-level Physics exam questions with mark scheme

1Core3 marks

An electric kettle has a power of 2.0 kW. It contains 0.50 kg of water at 18 °C.

Calculate the minimum time taken to heat the water to 100 °C. Specific heat capacity of water = 4200 J kg$^{-1}$ K$^{-1}$. [3]

$Q = mc\Delta\theta = 0.50 \times 4200 \times 82$ = 172 200 J (Δθ = 82 K seen)(1)
$t = Q/P = 172\,200 / 2000$(1)
$t$ = 86 s(1)
Answer: 86 s
Examiner tip: Use the temperature change (82 K), not the final temperature, and convert kW to W.

2Standard4 marks

A 0.040 kg ice cube at 0 °C is dropped into 0.30 kg of water at 25 °C in an insulated cup. All of the ice melts.

Calculate the final temperature of the water. Assume the cup has negligible heat capacity.

Specific heat capacity of water = 4200 J kg$^{-1}$ K$^{-1}$; specific latent heat of fusion of ice = 3.34 × 10$^5$ J kg$^{-1}$. [4]

energy to melt ice $= mL = 0.040 \times 3.34\times10^5$ = 13 400 J (13 360)(1)
energy lost by warm water = energy to melt ice + energy to warm melted ice: $0.30 \times 4200 \times (25 - T) = 13\,360 + 0.040 \times 4200 \times T$(1)
rearranges correctly, e.g. $1428\,T = 18\,140$(1)
$T$ = 13 °C (12.7)(1)
Answer: 13 °C (12.7 °C)
Examiner tip: Don't forget that the melted ice (now water at 0 °C) also has to warm up to the final temperature.

3Core3 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).

4Standard4 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.

5Hard4 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.

6Hard6 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.

7Stretch5 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.

8Standard6 marks

A 50 W heater is used to heat a 0.20 kg sample of a solid in an insulated container. Assume that all the energy supplied by the heater is transferred to the sample. The graph shows how the temperature of the sample varies with time.

(a) Explain why the temperature stays constant between 120 s and 520 s. [2]

(b) Calculate the specific latent heat of fusion of the substance. [2]

(a) the substance is melting / changing state(1)
(a) energy supplied increases the potential energy of the particles (breaking bonds / increasing separation), not their kinetic energy, so temperature does not change allow 'energy is used to break bonds'; reject 'no energy is supplied'(1)
(b) energy supplied = $Pt = 50 \times 400$ = 20 000 J (time 400 s read from graph)(1)
(b) $L = \frac{20\,000}{0.20}$ = 1.0 × 10$^5$ J kg$^{-1}$(1)
(c) $c = \frac{Pt}{m\Delta\theta} = \frac{50 \times 120}{0.20 \times 60}$ (Δθ = 60 K)(1)
(c) $c$ = 500 J kg$^{-1}$ K$^{-1}$(1)
Answer: (b) 1.0 × 10$^5$ J kg$^{-1}$ (c) 500 J kg$^{-1}$ K$^{-1}$
Examiner tip: Read the duration of the flat section (520 − 120 = 400 s), not the time at which it ends.

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