A Level H2 Physics Thermal Physics Quiz

<!-- TuitionGoWhere generation metadata: stage=5-1; model=tencent/hy3:free; model_label=Tencent HY3 Free; generated=2026-07-09; Sources: Stage 4-0 LLM templates, syllabus context, and Stage 2 evidence where available. -->

A-Level Physics H2 Quiz - Thermal Physics

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: _______ / 40

Duration: 60 minutes
Total Marks: 40
Instructions: Answer all 20 questions. Show your working clearly where calculations are required. Use the data booklet values where needed: $k = 1.38 \times 10^{-23}\ \text{J K}^{-1}$, $N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$, $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$.

---

Section A: Temperature and Ideal Gases (Questions 1–7)

1. The thermodynamic temperature of a sample is $300\ \text{K}$. Convert this temperature to the Celsius scale. [1]

---

2. State the equation of state for an ideal gas in terms of the number of molecules $N$, using the Boltzmann constant $k$. [1]

---

3. A sealed container holds $2.0 \times 10^{23}$ molecules of an ideal gas at a pressure of $1.5 \times 10^{5}\ \text{Pa}$ and temperature $320\ \text{K}$. Calculate the volume of the container. [3]

---

4. State two basic assumptions of the kinetic theory of gases. [2]

---

---

5. Derive the expression for the pressure $p$ exerted by an ideal gas in terms of the mean square speed $\langle c^2 \rangle$, number density $n$, and mass of one molecule $m$. [3]

---

---

---

6. The mean translational kinetic energy of a gas molecule is given by $\frac{1}{2}m\langle c^2\rangle = \frac{3}{2}kT$. Calculate the mean translational kinetic energy of a molecule at $400\ \text{K}$. [2]

---

7. A gas cylinder contains $0.50\ \text{mol}$ of helium at $290\ \text{K}$. Using $pV = nRT$, calculate the volume if the pressure is $2.0 \times 10^{5}\ \text{Pa}$. [2]

---

---

Section B: Thermodynamic Systems (Questions 8–14)

8. State the zeroth law of thermodynamics. [1]

---

9. Define internal energy of a thermodynamic system. [2]

---

10. State the first law of thermodynamics using the sign convention where $W$ is work done on the gas. [1]

---

11. A gas is compressed so that $120\ \text{J}$ of work is done on it, and it loses $40\ \text{J}$ of heat to the surroundings. Calculate the change in internal energy $\Delta U$. [2]

---

12. For an ideal gas expanding at constant pressure $p$, the work done on the gas is $W = -p\Delta V$. A gas expands from $0.020\ \text{m}^3$ to $0.050\ \text{m}^3$ at $1.0 \times 10^{5}\ \text{Pa}$. Calculate the work done by the gas. [2]

---

13. Define specific heat capacity. [2]

---

14. A $1.2\ \text{kg}$ block of copper (specific heat capacity $390\ \text{J kg}^{-1}\text{K}^{-1}$) is heated from $20^\circ\text{C}$ to $45^\circ\text{C}$. Calculate the heat energy required. [2]

---

---

Section C: Data Interpretation and Synthesis (Questions 15–20)

15. The specific latent heat of vaporisation of water is $2.26 \times 10^{6}\ \text{J kg}^{-1}$. Calculate the energy needed to vaporise $0.020\ \text{kg}$ of water at $100^\circ\text{C}$. [2]

---

16. A student measures the pressure and volume of a fixed mass of gas at constant temperature. The data are shown below.

$p$ / $10^{5}\ \text{Pa}$	$V$ / $10^{-3}\ \text{m}^3$
1.0	4.0
2.0	2.0
4.0	1.0

State the relationship between $p$ and $V$ demonstrated, and calculate $pV$ in $\text{Pa m}^3$. [3]

---

---

17.

Image pending generation: graph for Q17.

Using the diagram, calculate the work done on the gas during process B → C. [2]

---

18. Explain why the internal energy of an ideal gas depends only on its temperature. [2]

---

19. A refrigerator transfers $80\ \text{J}$ of heat from its cold interior to the warmer room while $30\ \text{J}$ of work is done on the refrigerant. Using $\Delta U = Q + W$, calculate $\Delta U$ of the refrigerant. [2]

---

20. A steel component (mass $0.80\ \text{kg}$, specific heat capacity $460\ \text{J kg}^{-1}\text{K}^{-1}$) cools from $150^\circ\text{C}$ to $30^\circ\text{C}$. Calculate the heat lost. [2]

---

<!-- TuitionGoWhere generation metadata: stage=5-1; model=tencent/hy3:free; model_label=Tencent HY3 Free; generated=2026-07-09; Sources: Stage 4-0 LLM templates, syllabus context, and Stage 2 evidence where available. -->

A-Level Physics H2 Quiz - Thermal Physics (Answer Key)

Total Marks: 40
Topic: Thermal Physics (Syllabus 9478, Topics 12–13)

---

Section A: Temperature and Ideal Gases

1. [1 mark]
$T_C = T_K - 273 = 300 - 273 = 27^\circ\text{C}$.
Teaching note: Thermodynamic (Kelvin) and Celsius scales differ by 273.15; at this level 273 is acceptable.
Common mistake: Adding 273 instead of subtracting.

2. [1 mark]
$pV = NkT$
Teaching note: $N$ = number of molecules, $k$ = Boltzmann constant. This is the microscopic form of the ideal gas law.

3. [3 marks]
$pV = NkT \Rightarrow V = \frac{NkT}{p}$
$V = \frac{(2.0\times10^{23})(1.38\times10^{-23})(320)}{1.5\times10^{5}}$
$= \frac{883.2}{1.5\times10^{5}} = 5.89\times10^{-3}\ \text{m}^3$
Mark breakdown: 1 for rearranging, 1 for substitution, 1 for answer with unit.
Common mistake: Wrong power of ten from $k$.

4. [2 marks]
Any two of:

• Molecules are in continuous random motion.
• Volume of molecules negligible compared to container.
• No intermolecular forces except during elastic collisions.
• Collisions with walls and each other are perfectly elastic.
  (1 mark each)

5. [3 marks]
Pressure from molecular motion: consider $N$ molecules in volume $V$, number density $n = N/V$.
For one molecule mass $m$, momentum change per collision with wall = $2mc_x$.
Rate of collisions $\propto c_x / (2L)$; force from all molecules gives $p = \frac{1}{3} n m \langle c^2 \rangle$.
Mark breakdown: 1 for momentum change, 1 for averaging/rate, 1 for final expression.

6. [2 marks]
$\frac{3}{2}kT = \frac{3}{2}(1.38\times10^{-23})(400) = 8.28\times10^{-21}\ \text{J}$.
Teaching note: Mean translational KE depends only on $T$.

7. [2 marks]
$V = \frac{nRT}{p} = \frac{(0.50)(8.31)(290)}{2.0\times10^{5}} = 6.02\times10^{-3}\ \text{m}^3$.
Common mistake: Using $k$ instead of $R$ with moles.

---

Section B: Thermodynamic Systems

8. [1 mark]
If two systems are each in thermal equilibrium with a third, they are in thermal equilibrium with each other.

9. [2 marks]
Internal energy is the sum of the microscopic kinetic and potential energies of all particles in the system. (1 for kinetic, 1 for potential)

10. [1 mark]
$\Delta U = Q + W$ (where $W$ = work done on gas).

11. [2 marks]
$\Delta U = Q + W = (-40) + (120) = +80\ \text{J}$.
Negative $Q$ because heat lost.

12. [2 marks]
Work done by gas = $p\Delta V = (1.0\times10^{5})(0.050 - 0.020) = 3.0\times10^{3}\ \text{J}$.
(Work done on gas would be $-3.0\times10^{3}\ \text{J}$.)

13. [2 marks]
Specific heat capacity is the heat energy required to raise the temperature of $1\ \text{kg}$ of a substance by $1\ \text{K}$ (or $1^\circ\text{C}$). 1 for mass ref, 1 for temp ref.

14. [2 marks]
$Q = mc\Delta T = (1.2)(390)(45 - 20) = 1.2 \times 390 \times 25 = 11700\ \text{J}$.

---

Section C: Data Interpretation and Synthesis

15. [2 marks]
$Q = ml = (0.020)(2.26\times10^{6}) = 4.52\times10^{4}\ \text{J}$.

16. [3 marks]
Relationship: $p$ is inversely proportional to $V$ at constant $T$ (Boyle’s law).
$pV$ values: $1.0\times10^{5} \times 4.0\times10^{-3} = 400\ \text{Pa m}^3$; similarly all give $400\ \text{Pa m}^3$.
Mark breakdown: 1 relation, 2 for calculation/constant.

17. [2 marks]
B→C is isobaric at $p = 3.0\times10^{5}\ \text{Pa}$, $\Delta V = 3.0\times10^{-3} - 1.0\times10^{-3} = 2.0\times10^{-3}\ \text{m}^3$.
Work done on gas $W = -p\Delta V = -(3.0\times10^{5})(2.0\times10^{-3}) = -600\ \text{J}$.
(Magnitude 600 J done by gas.)

18. [2 marks]
For ideal gas, intermolecular forces are negligible so potential energy is zero; internal energy is only random kinetic energy, which depends on $T$ via $\frac{3}{2}NkT$.

19. [2 marks]
$Q = -80\ \text{J}$ (heat removed from refrigerant), $W = +30\ \text{J}$.
$\Delta U = -80 + 30 = -50\ \text{J}$.

20. [2 marks]
$Q = mc\Delta T = (0.80)(460)(150 - 30) = 0.80 \times 460 \times 120 = 44160\ \text{J}$ lost.