A Level H2 Physics Thermal Physics Quiz

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

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

• Answer: The sum of the random kinetic energy [1] and potential energy [1] of the molecules/atoms within the system.
• Note: Must mention both KE and PE. "Random" is key for KE.

2. State two assumptions of the kinetic theory of gases regarding the motion of molecules. [2]

• Answer: (Any two of the following)
  • Molecules move in random directions / random motion. [1]
  • Collisions between molecules and with walls are perfectly elastic. [1]
  • Intermolecular forces are negligible except during collisions. [1]
  • The volume of the molecules is negligible compared to the volume of the container. [1]
  • Time of collision is negligible compared to time between collisions. [1]

3. Explain, in terms of molecular behavior, why the internal energy of an ideal gas depends only on its temperature. [2]

• Answer:
  • For an ideal gas, there are no intermolecular forces, so the potential energy is zero (or constant). [1]
  • Therefore, internal energy consists only of kinetic energy, which is directly proportional to the absolute temperature. [1]

4. A student claims that if the temperature of a gas increases from $20^\circ\text{C}$ to $40^\circ\text{C}$, the average kinetic energy of its molecules doubles. Explain why this statement is incorrect. [2]

• Answer:
  • Average kinetic energy is proportional to absolute temperature (Kelvin), not Celsius. [1]
  • $20^\circ\text{C} = 293 \text{ K}$ and $40^\circ\text{C} = 313 \text{ K}$. The ratio is $313/293 \approx 1.07$, not 2. [1]

5. Two objects, A and B, are in thermal contact. Object A is at $80^\circ\text{C}$ and Object B is at $20^\circ\text{C}$.

• (a) State the direction of net thermal energy flow. [1]
  • Answer: From A to B (or from hot to cold).
• (b) State the condition required for thermal equilibrium to be reached. [1]
  • Answer: When both objects are at the same temperature.

6. Calculate the number of moles of gas present. [3]

• Answer:
  • Use $pV = nRT$
  • $n = \frac{pV}{RT}$
  • $n = \frac{(1.5 \times 10^5)(2.0 \times 10^{-3})}{(8.31)(300)}$
  • $n = \frac{300}{2493} \approx 0.120 \text{ mol}$
• Marks: 1 for formula, 1 for substitution, 1 for answer ($0.12$ or $0.120$).

7. Calculate the total number of molecules in the gas. [2]

• Answer:
  • $N = n \times N_A$
  • $N = 0.120 \times 6.02 \times 10^{23}$
  • $N \approx 7.22 \times 10^{22}$ molecules
• Marks: 1 for method, 1 for answer.

8. Calculate the new temperature of the gas in Kelvin. [2]

• Answer:
  • At constant volume, $p \propto T$ (Pressure Law).
  • $\frac{p_1}{T_1} = \frac{p_2}{T_2}$
  • Since $p_2 = 2p_1$, then $T_2 = 2T_1$.
  • $T_2 = 2 \times 300 = 600 \text{ K}$.
• Marks: 1 for reasoning/ratio, 1 for answer.

9. Explain, using the kinetic theory of gases, why the pressure of the gas increases when it is heated at constant volume. [3]

• Answer:
  • Temperature increase means molecules have higher average kinetic energy / speed. [1]
  • Molecules collide with the walls more frequently. [1]
  • Each collision involves a greater change in momentum (greater force per collision). [1]
  • (Result: Greater average force per unit area = higher pressure).

10. Show that $p = \frac{1}{3} \rho \langle c^2 \rangle$. [2]

• Answer:
  • Start with $pV = \frac{1}{3} N m \langle c^2 \rangle$. [1]
  • Density $\rho = \frac{\text{total mass}}{V} = \frac{Nm}{V}$.
  • Rearrange equation: $p = \frac{1}{3} \frac{Nm}{V} \langle c^2 \rangle$.
  • Substitute $\rho$: $p = \frac{1}{3} \rho \langle c^2 \rangle$. [1]

11. Calculate the r.m.s. speed of nitrogen molecules at $300 \text{ K}$. [3]

• Answer:
  • Molar mass $M = 28.0 \text{ g mol}^{-1} = 0.028 \text{ kg mol}^{-1}$.
  • Formula: $\frac{1}{2} M \langle c^2 \rangle = \frac{3}{2} RT \Rightarrow c_{rms} = \sqrt{\frac{3RT}{M}}$
  • $c_{rms} = \sqrt{\frac{3 \times 8.31 \times 300}{0.028}}$
  • $c_{rms} = \sqrt{\frac{7479}{0.028}} = \sqrt{267107}$
  • $c_{rms} \approx 517 \text{ m s}^{-1}$
• Marks: 1 for formula, 1 for conversion of mass, 1 for answer.

12. Sketch Maxwell-Boltzmann distribution. [3]

• Answer:
  • Axes: y-axis = Number of molecules (or fraction), x-axis = Speed. [1]
  • Curve $T_1$: Starts at origin, rises to peak, tails off asymptotically to x-axis. [1]
  • Curve $T_2$: Peak is lower and shifted to the right (higher speed) compared to $T_1$. Area under both curves is equal. [1]

13. State the First Law of Thermodynamics. [2]

• Answer:
  • $\Delta U = Q + W$ (or $\Delta U = Q - W$ depending on convention, must define).
  • $\Delta U$: Change in internal energy. [0.5]
  • $Q$: Thermal energy supplied to the system. [0.5]
  • $W$: Work done on the system. [1]
  • (If using $\Delta U = Q - W$, $W$ is work done by the system).

14. Calculate the work done by the gas. [2]

• Answer:
  • $W = p \Delta V$
  • $W = 1.0 \times 10^5 \times (0.05 - 0.02)$
  • $W = 1.0 \times 10^5 \times 0.03 = 3000 \text{ J}$
• Marks: 1 for formula/sub, 1 for answer.

15. Calculate the change in internal energy. [2]

• Answer:
  • $\Delta U = Q - W_{by}$ (Using convention where $W_{by}$ is work done by gas)
  • $Q = +5000 \text{ J}$ (supplied)
  • $W_{by} = +3000 \text{ J}$
  • $\Delta U = 5000 - 3000 = +2000 \text{ J}$
• Marks: 1 for correct signs/logic, 1 for answer.

16. Isothermal expansion.

• (a) State what happens to internal energy. [1]
  • Answer: Internal energy remains constant ($\Delta U = 0$).
• (b) Explain why thermal energy must be supplied. [2]
  • Answer:
    • Gas does work during expansion ($W_{by} > 0$). [1]
    • Since $\Delta U = 0$, $Q = W_{by}$. Energy must be supplied as heat to compensate for the work done, keeping temperature constant. [1]

17. Distinguish between adiabatic and isothermal processes. [2]

• Answer:
  • Adiabatic: No thermal energy enters or leaves the system ($Q=0$). [1]
  • Isothermal: Temperature remains constant ($\Delta T = 0$, so $\Delta U = 0$ for ideal gas). [1]

18. Explain why adiabatic curve is steeper than isothermal on p-V diagram. [3]

• Answer:
  • In isothermal expansion, $T$ is constant, so $p$ decreases only due to volume increase ($p \propto 1/V$). [1]
  • In adiabatic expansion, gas does work at the expense of internal energy, so $T$ decreases. [1]
  • The drop in temperature causes an additional decrease in pressure, making the pressure drop faster for the same volume increase. [1]

19. Heat engine efficiency.

• (a) Calculate maximum theoretical efficiency. [2]
  • Answer:
    • $\eta = 1 - \frac{T_C}{T_H}$
    • $\eta = 1 - \frac{300}{600} = 1 - 0.5 = 0.5$ or $50\%$
• (b) Suggest one reason for lower actual efficiency. [1]
  • Answer: Friction / Heat loss to surroundings / Irreversible processes / Energy used to move engine parts.

20. Rapid compression.

• (a) State whether adiabatic or isothermal. [1]
  • Answer: Adiabatic.
• (b) Explain temperature change. [2]
  • Answer:
    • Work is done on the gas ($W > 0$). [1]
    • Since $Q \approx 0$, $\Delta U = W$. Internal energy increases, so temperature increases. [1]