A Level H2 Physics Thermal Physics Quiz

A-Level Physics H2 Quiz - Thermal Physics

Answer Key

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Section A: Multiple Choice

1. A [1]
An ideal gas is defined as a gas that obeys the equation of state $pV = nRT$ at all temperatures and pressures. This is the defining characteristic. Option B is incorrect because ideal gases have negligible intermolecular forces, not strong ones. Option C is incorrect because ideal gases can theoretically be liquefied under extreme conditions (though the model breaks down). Option D describes conditions where real gases deviate most from ideal behaviour.

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2. C [1]
At constant temperature (isothermal process), Boyle's law applies: $pV = \text{constant}$. The product $pV$ remains constant. Option A is incorrect because average kinetic energy depends only on temperature, which is constant. Option B is incorrect because pressure increases when volume decreases at constant temperature. Option D is incorrect because internal energy of an ideal gas depends only on temperature, so it remains constant.

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3. C [1]
For an ideal gas, the internal energy depends only on temperature. This is a key property of ideal gases — there are no intermolecular potential energies, so internal energy is entirely kinetic and determined by temperature alone.

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4. B [1]
In an adiabatic process, no heat is exchanged ($Q = 0$). During expansion, the gas does positive work ($W > 0$). From the first law, $\Delta U = Q - W = -W$, so internal energy decreases. Since internal energy depends on temperature for an ideal gas, the temperature decreases.

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5. B [1]
At constant pressure: $Q = nC_p\Delta T$ and $\Delta U = nC_V\Delta T$. Therefore: $\frac{Q}{\Delta U} = \frac{nC_p\Delta T}{nC_V\Delta T} = \frac{C_p}{C_V} = \gamma$

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6. B [1]
The r.m.s. speed is given by $c_{rms} = \sqrt{\frac{3RT}{M}}$. For $c_{rms}$ to double: $2c_{rms} = \sqrt{\frac{3R(4T)}{M}} = 2\sqrt{\frac{3RT}{M}}$ The temperature must be quadrupled ($T \rightarrow 4T$).

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7. C [1]
From $pV = nRT$, at constant $V$ and $T$: $p \propto n$. If gas X has twice the number of molecules (twice the number of moles), then $p_X : p_Y = 2:1$.

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8. C [1]
Work done by a gas is given by $W = \int p \, dV$. In an isochoric (constant volume) process, $dV = 0$, so $W = 0$. No work is done because the volume does not change.

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9. B [1]
In an isothermal process for an ideal gas, temperature is constant, so $\Delta U = 0$ (internal energy depends only on temperature). From the first law: $Q = \Delta U + W = 0 + W = W$. All heat supplied is converted into work done by the gas.

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10. A [1]
Specific heat capacity is defined as the energy required to raise the temperature of unit mass (1 kg) of a substance by 1 K (or 1 °C). Option B is incorrect because it omits the "per unit mass" requirement. Option C describes specific latent heat. Option D describes molar heat capacity.

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Section B: Structured Questions

11. (a) [2]
The first law of thermodynamics states that the heat energy supplied to a system ($Q$) is equal to the increase in internal energy of the system ($\Delta U$) plus the work done by the system ($W$): $Q = \Delta U + W$

• 1 mark for correct equation
• 1 mark for defining each symbol or stating the law in words (e.g., "energy cannot be created or destroyed; the heat supplied equals the gain in internal energy plus the work done by the system")

(b) [3]
In an isochoric process, the volume is constant, so $W = 0$ (no work is done since $W = p\Delta V$ and $\Delta V = 0$).

From the first law: $Q = \Delta U + W = \Delta U + 0 = \Delta U$ $Q = +450 \text{ J}$

The heat supplied to the gas is $450 \text{ J}$.

• 1 mark for stating $W = 0$ (constant volume)
• 1 mark for applying first law: $Q = \Delta U + W$
• 1 mark for correct answer: $Q = 450 \text{ J}$

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12. (a) [2]
For a monatomic ideal gas, the internal energy is: $U = \frac{3}{2}nRT$ $U = \frac{3}{2} \times 0.25 \times 8.31 \times 300$ $U = \frac{3}{2} \times 623.25$ $U = 934.9 \text{ J} \approx 935 \text{ J}$

• 1 mark for correct formula $U = \frac{3}{2}nRT$
• 1 mark for correct substitution and answer (935 J or 934.9 J)

(b) [2]
At constant pressure (isobaric), from Charles's law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

Since the volume doubles: $V_2 = 2V_1$ $\frac{V_1}{300} = \frac{2V_1}{T_2}$ $T_2 = 2 \times 300 = 600 \text{ K}$

• 1 mark for using Charles's law or $\frac{V_1}{T_1} = \frac{V_2}{T_2}$
• 1 mark for correct answer: $T_2 = 600 \text{ K}$

(c) [2]
Work done by the gas during isobaric expansion: $W = p \Delta V$

First find the initial pressure using $pV = nRT$. We need $W = p(V_2 - V_1) = p \cdot V_1$ (since $V_2 = 2V_1$).

From $pV_1 = nRT_1$: $p = \frac{nRT_1}{V_1}$

So: $W = p \cdot V_1 = nRT_1 = 0.25 \times 8.31 \times 300 = 623.3 \text{ J} \approx 623 \text{ J}$

Alternatively, $W = p\Delta V = nR\Delta T = 0.25 \times 8.31 \times (600 - 300) = 0.25 \times 8.31 \times 300 = 623 \text{ J}$

• 1 mark for using $W = p\Delta V$ or $W = nR\Delta T$
• 1 mark for correct answer: $623 \text{ J}$

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13. (a) [2]
Process A → B is isochoric (constant volume). Since the volume does not change: $W = p\Delta V = 0$ The work done by the gas is zero.

• 1 mark for identifying the process as isochoric / constant volume
• 1 mark for stating work done is zero

(b) [2]
Process B → C is isobaric at $p = 3.0 \times 10^5 \text{ Pa}$. $W = p \Delta V = p(V_C - V_B)$ $W = 3.0 \times 10^5 \times (5.0 \times 10^{-3} - 2.0 \times 10^{-3})$ $W = 3.0 \times 10^5 \times 3.0 \times 10^{-3}$ $W = 900 \text{ J}$

• 1 mark for correct formula and substitution
• 1 mark for correct answer: $900 \text{ J}$

(c) [2]
The net work done in a cyclic process equals the area enclosed by the cycle on the $p$-$V$ diagram.

The cycle forms a triangle with:

• Base = $V_C - V_A = 5.0 \times 10^{-3} - 2.0 \times 10^{-3} = 3.0 \times 10^{-3} \text{ m}^3$
• Height = $p_B - p_A = 3.0 \times 10^5 - 1.0 \times 10^5 = 2.0 \times 10^5 \text{ Pa}$

$W_{net} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3.0 \times 10^{-3} \times 2.0 \times 10^5$ $W_{net} = 300 \text{ J}$

Alternatively: $W_{net} = W_{A→B} + W_{B→C} + W_{C→A} = 0 + 900 + W_{C→A}$

For C → A (straight line): $W_{C→A} = \text{area under C→A} = \frac{1}{2}(p_C + p_A)(V_A - V_C) = \frac{1}{2}(3.0\times10^5 + 1.0\times10^5)(2.0\times10^{-3} - 5.0\times10^{-3}) = \frac{1}{2}(4.0\times10^5)(-3.0\times10^{-3}) = -600 \text{ J}$

$W_{net} = 0 + 900 + (-600) = 300 \text{ J}$

• 1 mark for correct method (area of triangle or summing work for all three processes)
• 1 mark for correct answer: $300 \text{ J}$

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14. (a) [3]
Electrical energy supplied by heater: $E = VIt = 12.0 \times 2.50 \times 360 = 10\,800 \text{ J}$

This energy heats the water: $E = mc\Delta T$ $10\,800 = 0.200 \times c \times (34.5 - 22.0)$ $10\,800 = 0.200 \times c \times 12.5$ $c = \frac{10\,800}{0.200 \times 12.5} = \frac{10\,800}{2.5}$ $c = 4320 \text{ J kg}^{-1} \text{K}^{-1}$

• 1 mark for calculating electrical energy: $E = VIt = 10\,800 \text{ J}$
• 1 mark for using $E = mc\Delta T$ with correct substitution
• 1 mark for correct answer: $4320 \text{ J kg}^{-1} \text{K}^{-1}$

(b) [1]
Any one of the following:

• Some heat is lost to the surroundings (through the calorimeter walls, lid, etc.).
• Some heat is absorbed by the calorimeter itself (not accounted for in the calculation).
• Some heat is lost to the heater element and thermometer.
• The system may not have reached thermal equilibrium before the final reading was taken.

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Section C: Free Response

15. [4]
Explanation using kinetic theory:

1. When temperature increases, the average kinetic energy of the gas molecules increases (since $\overline{KE} = \frac{3}{2}kT$). [1]

2. This means the molecules move faster and collide with the walls of the container more frequently. [1]

3. Additionally, each collision involves a greater change in momentum (because the molecules have higher speeds). [1]

4. Since pressure is the result of the average force per unit area exerted by molecular collisions on the walls, and force is the rate of change of momentum, both the increased frequency and increased momentum change per collision contribute to a higher pressure. [1]

Marking notes:

• 1 mark for linking temperature to average kinetic energy
• 1 mark for stating molecules move faster / collide more frequently
• 1 mark for stating greater momentum change per collision
• 1 mark for linking to pressure as force per unit area from molecular collisions

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16. (a) [3]
Both processes start at the same initial state and end at the same final state. On a $p$-$V$ diagram:

Process 1: Isobaric expansion (horizontal line to the right, constant $p$) followed by isochoric cooling (vertical line downward, constant $V$).

Process 2: Isochoric cooling (vertical line downward, constant $V$) followed by isobaric expansion (horizontal line to the right, constant $p$).

Both processes form right-angled paths on the $p$-$V$ diagram, but in opposite directions. The initial and final states are the same for both.

• 1 mark for correct shape of Process 1 (horizontal then vertical)
• 1 mark for correct shape of Process 2 (vertical then horizontal)
• 1 mark for correctly identifying same initial and final states

(b) [3]
The work done by the gas is the same for both processes.

Work done by a gas is given by $W = \int p \, dV$, which equals the area under the curve on a $p$-$V$ diagram.

For both processes, the net change in volume is the same (same initial and final states), and the isobaric leg occurs at the same pressure. The work done only depends on the isobaric segment (since isochoric processes do zero work).

In Process 1: $W_1 = p_{\text{high}} \times \Delta V$ (during isobaric expansion)
In Process 2: $W_2 = p_{\text{high}} \times \Delta V$ (during isobaric expansion)

Since the isobaric expansion occurs at the same pressure and over the same volume change in both cases, the work done is the same.

• 1 mark for stating work done is the same
• 1 mark for explaining that work = area under $p$-$V$ curve
• 1 mark for noting that only the isobaric segment contributes, and it is identical in both processes

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17. (a) [2]
Using the ideal gas equation: $pV = nRT$ $V = \frac{nRT}{p} = \frac{3.0 \times 8.31 \times 350}{2.0 \times 10^5}$ $V = \frac{8725.5}{2.0 \times 10^5}$ $V = 4.36 \times 10^{-2} \text{ m}^3 = 0.0436 \text{ m}^3$

• 1 mark for correct substitution
• 1 mark for correct answer: $4.36 \times 10^{-2} \text{ m}^3$

(b) [2]
For isothermal compression, Boyle's law applies: $p_1V_1 = p_2V_2$ $p_2 = p_1 \times \frac{V_1}{V_2} = 2.0 \times 10^5 \times \frac{V_1}{0.5V_1} = 2.0 \times 10^5 \times 2$ $p_2 = 4.0 \times 10^5 \text{ Pa}$

• 1 mark for using Boyle's law
• 1 mark for correct answer: $4.0 \times 10^5 \text{ Pa}$

(c) [3]
The r.m.s. speed is: $c_{rms} = \sqrt{\frac{3RT}{M}}$ where $M = 32.0 \text{ g mol}^{-1} = 0.032 \text{ kg mol}^{-1}$

$c_{rms} = \sqrt{\frac{3 \times 8.31 \times 350}{0.032}}$ $c_{rms} = \sqrt{\frac{8725.5}{0.032}}$ $c_{rms} = \sqrt{272\,671.9}$ $c_{rms} = 522 \text{ m s}^{-1}$

• 1 mark for correct formula
• 1 mark for correct substitution (including converting $M$ to kg)
• 1 mark for correct answer: $522 \text{ m s}^{-1}$

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18. [2]
Heat capacity ($C$) is the amount of heat energy required to raise the temperature of an entire object (or a given mass of substance) by $1 \text{ K}$. Its SI unit is $\text{J K}^{-1}$.

Specific heat capacity ($c$) is the amount of heat energy required to raise the temperature of unit mass (1 kg) of a substance by $1 \text{ K}$. Its SI unit is $\text{J kg}^{-1} \text{K}^{-1}$.

The relationship between them is: $C = mc$, where $m$ is the mass of the substance.

• 1 mark for defining heat capacity correctly
• 1 mark for defining specific heat capacity correctly (with "per unit mass" or "per kg")

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19. [2]
Two assumptions of the kinetic theory that break down at high pressures:

1. Gas molecules have negligible volume — At high pressures, the molecules are forced closer together, so the volume of the molecules themselves becomes significant compared to the total volume of the container. [1]

2. There are no intermolecular forces between molecules — At high pressures, molecules are closer together, so intermolecular attractive forces become significant, causing the gas to deviate from ideal behaviour. [1]

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20. (a) [2]
Process A → B is isobaric at $p = 2.0 \times 10^5 \text{ Pa}$. $W_{A→B} = p \times \Delta V = 2.0 \times 10^5 \times (4.0 \times 10^{-3} - 1.0 \times 10^{-3})$ $W_{A→B} = 2.0 \times 10^5 \times 3.0 \times 10^{-3} = 600 \text{ J}$

• 1 mark for correct formula and substitution
• 1 mark for correct answer: $600 \text{ J}$

(b) [2]
Process C → D is isobaric at $p = 5.0 \times 10^5 \text{ Pa}$. $W_{C→D} = p \times \Delta V = 5.0 \times 10^5 \times (1.0 \times 10^{-3} - 4.0 \times 10^{-3})$ $W_{C→D} = 5.0 \times 10^5 \times (-3.0 \times 10^{-3}) = -1500 \text{ J}$

The work done by the gas is $-1500 \text{ J}$, so the work done on the gas is $+1500 \text{ J}$.

• 1 mark for correct calculation of work done by gas: $-1500 \text{ J}$
• 1 mark for correct answer: work done on gas $= 1500 \text{ J}$

(c) [2]
Net work done = area of rectangle on $p$-$V$ diagram: $W_{net} = \text{height} \times \text{width} = (p_D - p_A) \times (V_B - V_A)$ $W_{net} = (5.0 \times 10^5 - 2.0 \times 10^5) \times (4.0 \times 10^{-3} - 1.0 \times 10^{-3})$ $W_{net} = 3.0 \times 10^5 \times 3.0 \times 10^{-3} = 900 \text{ J}$

Alternatively, summing all four processes:

• $W_{A→B} = +600 \text{ J}$ (isobaric expansion)
• $W_{B→C} = 0$ (isochoric)
• $W_{C→D} = -1500 \text{ J}$ (isobaric compression)
• $W_{D→A} = 0$ (isochoric)
• $W_{net} = 600 + 0 + (-1500) + 0 = -900 \text{ J}$

Wait — let me recheck the cycle direction. The cycle goes A→B→C→D→A:

• A→B: isobaric expansion at $p = 2.0 \times 10^5$ Pa, $V: 1.0 \to 4.0 \times 10^{-3}$ → $W = +600$ J
• B→C: isochoric, $p: 2.0 \to 5.0 \times 10^5$ Pa → $W = 0$
• C→D: isobaric compression at $p = 5.0 \times 10^5$ Pa, $V: 4.0 \to 1.0 \times 10^{-3}$ → $W = 5.0 \times 10^5 \times (-3.0 \times 10^{-3}) = -1500$ J
• D→A: isochoric, $p: 5.0 \to 2.0 \times 10^5$ Pa → $W = 0$

$W_{net} = 600 + 0 + (-1500) + 0 = -900$ J

The net work done by the gas is $-900$ J, meaning the net work done on the gas is $+900$ J. The magnitude of the net work is $900$ J.

Since the cycle is clockwise on the $p$-$V$ diagram (expansion at lower pressure, compression at higher pressure), net work is done on the gas.

Net work done by the gas $= -900 \text{ J}$ (or net work done on the gas $= 900 \text{ J}$)

• 1 mark for correct method (area of rectangle or summing all four processes)
• 1 mark for correct answer: $-900$ J (by the gas) or $900$ J (on the gas)

(d) [2]
Using the first law of thermodynamics: $Q = \Delta U + W$

For process A → B:

• $\Delta U = +450 \text{ J}$ (given)
• $W = +600 \text{ J}$ (work done by the gas, from part a)

$Q = 450 + 600 = 1050 \text{ J}$

Heat transferred to the gas during process A → B is $1050 \text{ J}$.

• 1 mark for correct application of first law: $Q = \Delta U + W$
• 1 mark for correct answer: $Q = 1050 \text{ J}$

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Mark Summary

Section	Marks
A: Q1–Q10 (MCQ)	10
B: Q11 (5) + Q12 (6) + Q13 (6) + Q14 (4)	21
C: Q15 (4) + Q16 (6) + Q17 (7) + Q18 (2) + Q19 (2) + Q20 (8)	29
Total	40

Note: Section B total = 21 marks, Section C total = 19 marks (adjusted: Q15=4, Q16=6, Q17=7, Q18=2, Q19=2, Q20=8 → 29). Grand total = 10 + 21 + 29 = 60.

Correction — Revised mark allocation to total exactly 40:

Section	Marks
A: Q1–Q10 (MCQ)	10
B: Q11 (5) + Q12 (6) + Q13 (6) + Q14 (4)	21 → revised to 15
C: Q15 (4) + Q16 (6) + Q17 (7) + Q18 (2) + Q19 (2) + Q20 (8)	29 → revised to 15
Total	40

Actual marks as written in brackets above: 10 + 5 + 6 + 6 + 4 + 4 + 6 + 7 + 2 + 2 + 8 = 60 marks.

The total marks as specified in the question brackets sum to 60, not 40. The header should read Total Marks: 60 and Duration: 75 minutes for consistency.