A Level H2 Physics Thermal Physics Quiz

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A-Level Physics H2 Quiz - Thermal Physics

Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50

Duration: 1 hour 15 minutes Total Marks: 50

Instructions:

• Answer ALL questions in the spaces provided.
• Show all working for calculation questions.
• State any assumptions made where applicable.
• Use appropriate significant figures in final answers.
• You may use a calculator.

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Section A: Short Answer & Structured Response (15 marks)

Answer all questions in this section.

1. State the relationship between the Celsius temperature scale and the thermodynamic (Kelvin) temperature scale.

[2 marks]

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2. Define the term internal energy of a system.

[2 marks]

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3. State the First Law of Thermodynamics, giving both the word equation and the mathematical form.

[3 marks]

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4. Explain why the temperature of an ideal gas remains constant during an isothermal expansion, even though the gas does work on its surroundings.

[3 marks]

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5. A fixed mass of an ideal gas undergoes an adiabatic compression. State what happens to: (a) the internal energy of the gas, (b) the temperature of the gas.

[2 marks]

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Section B: Calculation & Application (20 marks)

Answer all questions in this section. Show all working clearly.

6. State two assumptions of the kinetic theory of an ideal gas.

[3 marks]

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7. A quantity of an ideal gas occupies a volume of $2.50 \times 10^{-3} \text{ m}^3$ at a pressure of $1.01 \times 10^5 \text{ Pa}$ and a temperature of $27^\circ\text{C}$.

(a) Calculate the number of moles of gas present.

[2 marks]

(b) The gas is heated at constant volume until its pressure doubles. Calculate the new temperature of the gas in degrees Celsius.

[2 marks]

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8. An ideal gas expands isothermally at a temperature of $300 \text{ K}$ from a volume of $1.00 \times 10^{-3} \text{ m}^3$ to $3.00 \times 10^{-3} \text{ m}^3$. The initial pressure is $2.00 \times 10^5 \text{ Pa}$.

(a) Calculate the work done by the gas during this expansion.

[3 marks]

(b) State the amount of heat energy absorbed by the gas during this process, giving a reason for your answer.

[2 marks]

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9. A cylinder contains $0.050 \text{ mol}$ of helium gas (treated as ideal) at a temperature of $400 \text{ K}$. The gas is compressed adiabatically so that its volume is reduced to one-third of its original volume. The ratio of specific heat capacities for helium is $\gamma = 1.67$.

(a) Calculate the final temperature of the gas.

[3 marks]

(b) Calculate the work done on the gas during this compression.

[3 marks]

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10. A student investigates the relationship between the pressure $p$ and volume $V$ of a fixed mass of gas at constant temperature. The data obtained is shown in the table below.

$p$ / $10^5$ Pa	$V$ / $10^{-3}$ m$^3$
1.00	4.80
1.50	3.20
2.00	2.40
2.50	1.92
3.00	1.60

(a) Use the data to verify that the gas obeys Boyle's law.

[3 marks]

(b) Calculate the number of moles of gas present if the temperature is maintained at $25^\circ\text{C}$.

[2 marks]

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Section C: Data Analysis & Extended Response (15 marks)

Answer all questions in this section.

11. A heat engine operates between a hot reservoir at temperature $T_H = 800 \text{ K}$ and a cold reservoir at temperature $T_C = 300 \text{ K}$. In one cycle, the engine extracts $1200 \text{ J}$ of heat from the hot reservoir and rejects $500 \text{ J}$ of heat to the cold reservoir.

(a) Calculate the work done by the engine in one cycle.

[1 mark]

(b) Calculate the efficiency of this engine.

[2 marks]

(c) Calculate the maximum possible (Carnot) efficiency for an engine operating between these two temperatures.

[2 marks]

(d) Explain why the actual efficiency calculated in (b) is less than the Carnot efficiency calculated in (c).

[2 marks]

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12. A student performs an experiment to determine the specific latent heat of vaporisation of water. An electric heater of power $50.0 \text{ W}$ is used to heat water at its boiling point. The mass of water boiled away in $300 \text{ s}$ is found to be $22.0 \text{ g}$.

(a) Calculate the specific latent heat of vaporisation of water from these results.

[2 marks]

(b) The accepted value for the specific latent heat of vaporisation of water is $2.26 \times 10^6 \text{ J kg}^{-1}$. Calculate the percentage difference between the experimental value and the accepted value.

[2 marks]

(c) Suggest two reasons why the experimental value may differ from the accepted value, and state whether each reason would cause the experimental value to be higher or lower than the accepted value.

[4 marks]

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13. Explain, in terms of the kinetic theory, why the pressure of an ideal gas increases when its volume is reduced at constant temperature.

[3 marks]

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14. A fixed mass of an ideal gas is heated at constant pressure. Explain why the work done by the gas is given by $W = p \Delta V$.

[2 marks]

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15. Distinguish between the specific heat capacity and the specific latent heat of a substance.

[2 marks]

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Section D: Conceptual Understanding & Application (10 marks)

Answer all questions in this section.

16. A gas is compressed isothermally. State whether the following quantities are positive, negative, or zero: (a) the work done on the gas, (b) the heat transferred to the gas, (c) the change in internal energy of the gas.

[3 marks]

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17. Explain why the temperature of a gas rises during an adiabatic compression.

[2 marks]

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18. An ideal gas undergoes a cyclic process. State the net change in internal energy over one complete cycle. Justify your answer.

[2 marks]

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19. State the Second Law of Thermodynamics in terms of heat flow.

[1 mark]

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20. A student claims that the efficiency of a heat engine can be 100% if all the heat from the hot reservoir is converted into work. Explain why this is not possible.

[2 marks]

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END OF PAPER

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A-Level Physics H2 Quiz - Thermal Physics: ANSWER KEY

Total Marks: 50

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Section A: Short Answer & Structured Response (15 marks)

1. State the relationship between the Celsius temperature scale and the thermodynamic (Kelvin) temperature scale. [2 marks]

Answer: $T (\text{K}) = \theta (\text{°C}) + 273.15$ [1 mark] OR $T / \text{K} = \theta / \text{°C} + 273.15$

The size of one degree interval is the same on both scales. [1 mark]

Award 1 mark for correct equation, 1 mark for noting equal interval size or correct conversion constant.

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2. Define the term internal energy of a system. [2 marks]

Answer: Internal energy is the sum of the kinetic energies and potential energies of all the particles (atoms/molecules) in the system. [2 marks]

Award 1 mark for "sum of kinetic energies", 1 mark for "and potential energies" or "due to random motion and intermolecular forces".

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3. State the First Law of Thermodynamics, giving both the word equation and the mathematical form. [3 marks]

Answer: Word equation: The increase in internal energy of a system equals the heat supplied to the system plus the work done on the system. [1 mark]

Mathematical form: $\Delta U = Q + W$ [1 mark]

where $\Delta U$ = change in internal energy, $Q$ = heat supplied to the system, $W$ = work done on the system. [1 mark for correct sign convention explanation]

Accept alternative sign convention $\Delta U = Q - W$ if $W$ is defined as work done BY the system, with appropriate explanation.

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4. Explain why the temperature of an ideal gas remains constant during an isothermal expansion, even though the gas does work on its surroundings. [3 marks]

Answer:

• During expansion, the gas does work on the surroundings ($W$ by gas is positive, so $W$ on gas is negative). [1 mark]
• For an ideal gas, internal energy depends only on temperature ($U \propto T$). [1 mark]
• To maintain constant temperature (and thus constant internal energy), heat must flow into the gas from the surroundings to compensate for the work done. By the First Law, $\Delta U = 0$, so $Q = -W$ (heat absorbed equals work done by gas). [1 mark]

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5. A fixed mass of an ideal gas undergoes an adiabatic compression. State what happens to: (a) the internal energy of the gas, (b) the temperature of the gas. [2 marks]

Answer: (a) The internal energy increases. [1 mark] (b) The temperature increases. [1 mark]

Explanation (not required for marks): In adiabatic compression, $Q = 0$ and work is done ON the gas ($W > 0$). By First Law, $\Delta U = W > 0$, so internal energy and temperature increase.

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Section B: Calculation & Application (20 marks)

6. State two assumptions of the kinetic theory of an ideal gas. [3 marks]

Answer: Any two from (1.5 marks each):

• The gas consists of a large number of identical molecules in random motion.
• The volume of the molecules is negligible compared to the volume of the container.
• There are no intermolecular forces between molecules (except during collisions).
• Collisions between molecules and with the walls are perfectly elastic.
• The duration of collisions is negligible compared to the time between collisions.
• Newton's laws of motion apply to the molecules.

Award 1.5 marks per correct assumption, up to 3 marks total.

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7. A quantity of an ideal gas occupies a volume of $2.50 \times 10^{-3} \text{ m}^3$ at a pressure of $1.01 \times 10^5 \text{ Pa}$ and a temperature of $27^\circ\text{C}$.

(a) Calculate the number of moles of gas present. [2 marks]

Answer: $T = 27 + 273 = 300 \text{ K}$ [0.5 marks] $pV = nRT$ [0.5 marks] $n = \frac{pV}{RT} = \frac{(1.01 \times 10^5)(2.50 \times 10^{-3})}{(8.31)(300)}$ [0.5 marks] $n = \frac{252.5}{2493} = 0.101 \text{ mol}$ [0.5 marks]

Award marks for correct substitution and final answer with units.

(b) The gas is heated at constant volume until its pressure doubles. Calculate the new temperature of the gas in degrees Celsius. [2 marks]

Answer: At constant volume: $\frac{p_1}{T_1} = \frac{p_2}{T_2}$ [0.5 marks] $p_2 = 2p_1$, so $\frac{p_1}{300} = \frac{2p_1}{T_2}$ [0.5 marks] $T_2 = 2 \times 300 = 600 \text{ K}$ [0.5 marks] $\theta = 600 - 273 = 327^\circ\text{C}$ [0.5 marks]

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8. An ideal gas expands isothermally at a temperature of $300 \text{ K}$ from a volume of $1.00 \times 10^{-3} \text{ m}^3$ to $3.00 \times 10^{-3} \text{ m}^3$. The initial pressure is $2.00 \times 10^5 \text{ Pa}$.

(a) Calculate the work done by the gas during this expansion. [3 marks]

Answer: For isothermal expansion: $W = nRT \ln\left(\frac{V_2}{V_1}\right)$ [1 mark]

First find $n$: $n = \frac{p_1 V_1}{RT} = \frac{(2.00 \times 10^5)(1.00 \times 10^{-3})}{(8.31)(300)} = 0.0802 \text{ mol}$ [1 mark]

$W = (0.0802)(8.31)(300) \ln\left(\frac{3.00 \times 10^{-3}}{1.00 \times 10^{-3}}\right)$ $W = 200 \times \ln(3) = 200 \times 1.099 = 220 \text{ J}$ [1 mark]

Alternative method using $W = p_1 V_1 \ln(V_2/V_1) = (2.00 \times 10^5)(1.00 \times 10^{-3}) \ln(3) = 220 \text{ J}$ is acceptable.

(b) State the amount of heat energy absorbed by the gas during this process, giving a reason for your answer. [2 marks]

Answer: $Q = 220 \text{ J}$ [1 mark] Reason: For an isothermal process, $\Delta U = 0$ (internal energy of ideal gas depends only on temperature). By the First Law, $\Delta U = Q - W_{\text{by}} = 0$, so $Q = W_{\text{by}} = 220 \text{ J}$. [1 mark]

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9. A cylinder contains $0.050 \text{ mol}$ of helium gas (treated as ideal) at a temperature of $400 \text{ K}$. The gas is compressed adiabatically so that its volume is reduced to one-third of its original volume. The ratio of specific heat capacities for helium is $\gamma = 1.67$.

(a) Calculate the final temperature of the gas. [3 marks]

Answer: For adiabatic process: $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$ [1 mark] $T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1} = 400 \times (3)^{0.67}$ [1 mark] $T_2 = 400 \times 2.088 = 835 \text{ K}$ [1 mark]

(b) Calculate the work done on the gas during this compression. [3 marks]

Answer: For adiabatic process, $Q = 0$, so $W = \Delta U$ [1 mark] For monatomic gas (He): $\Delta U = \frac{3}{2} nR \Delta T$ [1 mark] $\Delta U = \frac{3}{2} \times 0.050 \times 8.31 \times (835 - 400)$ $\Delta U = 0.62325 \times 435 = 271 \text{ J}$ [1 mark] Work done ON gas = $271 \text{ J}$

Accept $C_V = \frac{R}{\gamma-1} = \frac{8.31}{0.67} = 12.4 \text{ J mol}^{-1} \text{K}^{-1}$, then $W = n C_V \Delta T = 0.050 \times 12.4 \times 435 = 270 \text{ J}$.

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10. A student investigates the relationship between the pressure $p$ and volume $V$ of a fixed mass of gas at constant temperature.

$p$ / $10^5$ Pa	$V$ / $10^{-3}$ m$^3$
1.00	4.80
1.50	3.20
2.00	2.40
2.50	1.92
3.00	1.60

(a) Use the data to verify that the gas obeys Boyle's law. [3 marks]

Answer: Boyle's law: $pV = \text{constant}$ at constant temperature. [1 mark]

Calculate $pV$ for each data point (in $10^2$ J):

• $1.00 \times 4.80 = 4.80$
• $1.50 \times 3.20 = 4.80$
• $2.00 \times 2.40 = 4.80$
• $2.50 \times 1.92 = 4.80$
• $3.00 \times 1.60 = 4.80$ [1 mark]

Since $pV$ is constant ($4.80 \times 10^2 \text{ J}$) for all data points, the gas obeys Boyle's law. [1 mark]

Award marks for correct calculation and conclusion.

(b) Calculate the number of moles of gas present if the temperature is maintained at $25^\circ\text{C}$. [2 marks]

Answer: $T = 25 + 273 = 298 \text{ K}$ [0.5 marks] $pV = nRT$ [0.5 marks] $n = \frac{pV}{RT} = \frac{4.80 \times 10^2}{8.31 \times 298} = \frac{480}{2476.38} = 0.194 \text{ mol}$ [1 mark]

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Section C: Data Analysis & Extended Response (15 marks)

11. A heat engine operates between a hot reservoir at temperature $T_H = 800 \text{ K}$ and a cold reservoir at temperature $T_C = 300 \text{ K}$. In one cycle, the engine extracts $1200 \text{ J}$ of heat from the hot reservoir and rejects $500 \text{ J}$ of heat to the cold reservoir.

(a) Calculate the work done by the engine in one cycle. [1 mark]

Answer: $W = Q_H - Q_C = 1200 - 500 = 700 \text{ J}$ [1 mark]

(b) Calculate the efficiency of this engine. [2 marks]

Answer: $\eta = \frac{W}{Q_H} = \frac{700}{1200} = 0.583$ [1 mark] $\eta = 58.3\%$ [1 mark]

(c) Calculate the maximum possible (Carnot) efficiency for an engine operating between these two temperatures. [2 marks]

Answer: $\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}$ [1 mark] $\eta_{\text{Carnot}} = 1 - \frac{300}{800} = 1 - 0.375 = 0.625 = 62.5\%$ [1 mark]

(d) Explain why the actual efficiency calculated in (b) is less than the Carnot efficiency calculated in (c). [2 marks]

Answer: The Carnot efficiency is the maximum theoretical efficiency for a reversible engine. [1 mark] Real engines have irreversible processes (e.g., friction, heat losses, non-quasi-static processes) that reduce the net work output, so the actual efficiency is always lower. [1 mark]

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12. A student performs an experiment to determine the specific latent heat of vaporisation of water. An electric heater of power $50.0 \text{ W}$ is used to heat water at its boiling point. The mass of water boiled away in $300 \text{ s}$ is found to be $22.0 \text{ g}$.

(a) Calculate the specific latent heat of vaporisation of water from these results. [2 marks]

Answer: Energy supplied: $E = Pt = 50.0 \times 300 = 15000 \text{ J}$ [0.5 marks] Mass: $m = 22.0 \text{ g} = 0.0220 \text{ kg}$ [0.5 marks] $L_v = \frac{E}{m} = \frac{15000}{0.0220} = 6.82 \times 10^5 \text{ J kg}^{-1}$ [1 mark]

(b) The accepted value for the specific latent heat of vaporisation of water is $2.26 \times 10^6 \text{ J kg}^{-1}$. Calculate the percentage difference between the experimental value and the accepted value. [2 marks]

Answer: Percentage difference = $\frac{|\text{experimental} - \text{accepted}|}{\text{accepted}} \times 100\%$ [1 mark] $= \frac{|6.82 \times 10^5 - 2.26 \times 10^6|}{2.26 \times 10^6} \times 100\% = \frac{1.578 \times 10^6}{2.26 \times 10^6} \times 100\% = 69.8\%$ [1 mark]

(c) Suggest two reasons why the experimental value may differ from the accepted value, and state whether each reason would cause the experimental value to be higher or lower than the accepted value. [4 marks]

Answer: Any two from (2 marks each):

1. Heat loss to the surroundings: some energy from the heater is lost to the air/container instead of vaporising water. This means less water is boiled for the same energy input, so the calculated $L_v = E/m$ is higher than the accepted value. [2 marks]
2. Water not exactly at boiling point: if the water is slightly below boiling point, some energy is used to raise the temperature to boiling point rather than vaporisation. This reduces the mass boiled, making the calculated $L_v$ higher. [2 marks]
3. Impurities in the water: dissolved substances can alter the boiling point and latent heat, usually making the experimental value lower or higher depending on the impurity. [2 marks]
4. Splashing/water loss: if water is lost as liquid droplets rather than vapour, the measured mass boiled is overestimated, making the calculated $L_v$ lower. [2 marks]

Award marks for plausible reasons with correct effect on the value.

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13. Explain, in terms of the kinetic theory, why the pressure of an ideal gas increases when its volume is reduced at constant temperature. [3 marks]

Answer:

• Pressure is due to the rate of change of momentum of molecules colliding with the walls. [1 mark]
• When volume is reduced, the number of molecules per unit volume increases, so the frequency of collisions with the walls increases. [1 mark]
• Since temperature (and thus average speed of molecules) is constant, the average force per collision remains the same, but more collisions per second lead to a greater total force and thus higher pressure. [1 mark]

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14. A fixed mass of an ideal gas is heated at constant pressure. Explain why the work done by the gas is given by $W = p \Delta V$. [2 marks]

Answer: Work done by a gas during expansion: $W = \int p \, dV$. [1 mark] At constant pressure, $p$ is constant, so $W = p \int dV = p \Delta V$. [1 mark]

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15. Distinguish between the specific heat capacity and the specific latent heat of a substance. [2 marks]

Answer: Specific heat capacity is the energy required to raise the temperature of 1 kg of a substance by 1 K (or 1 °C) without a change of state. [1 mark] Specific latent heat is the energy required to change the state of 1 kg of a substance at constant temperature. [1 mark]

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Section D: Conceptual Understanding & Application (10 marks)

16. A gas is compressed isothermally. State whether the following quantities are positive, negative, or zero: (a) the work done on the gas, (b) the heat transferred to the gas, (c) the change in internal energy of the gas. [3 marks]

Answer: (a) Positive (work is done ON the gas during compression). [1 mark] (b) Negative (heat is transferred FROM the gas to the surroundings to maintain constant temperature). [1 mark] (c) Zero (internal energy of an ideal gas depends only on temperature; isothermal means $\Delta T = 0$, so $\Delta U = 0$). [1 mark]

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17. Explain why the temperature of a gas rises during an adiabatic compression. [2 marks]

Answer: In an adiabatic process, $Q = 0$. [1 mark] Work is done on the gas ($W > 0$), so by the First Law, $\Delta U = W > 0$. The internal energy increases, and since internal energy is proportional to temperature for an ideal gas, the temperature rises. [1 mark]

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18. An ideal gas undergoes a cyclic process. State the net change in internal energy over one complete cycle. Justify your answer. [2 marks]

Answer: The net change in internal energy is zero. [1 mark] Internal energy is a state function; when the gas returns to its initial state (same temperature, pressure, volume), the internal energy must be the same as at the start. [1 mark]

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19. State the Second Law of Thermodynamics in terms of heat flow. [1 mark]

Answer: Heat cannot spontaneously flow from a colder body to a hotter body. [1 mark] Accept: It is impossible for heat to flow from a colder body to a hotter body without external work being done.

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20. A student claims that the efficiency of a heat engine can be 100% if all the heat from the hot reservoir is converted into work. Explain why this is not possible. [2 marks]

Answer: According to the Second Law of Thermodynamics, it is impossible to convert all heat from a hot reservoir completely into work without rejecting some heat to a cold reservoir. [1 mark] Some heat must always be rejected to a cold sink, so the efficiency is always less than 100%. [1 mark]

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END OF ANSWER KEY