Topic B. The particulate nature of matter HL TEST 1

Topic B — The Particulate Nature of Matter
HL Test 1

B.1–B.5 • 25 multiple-choice questions • 5 extended responses • 1 lab-style question

Test instructions

Total: 100 marks. Suggested time: 120 minutes. Suggested allocation: 35 minutes for Section A, 65 minutes for Section B, and 20 minutes for Section C.

Answer all questions. Each multiple-choice question has one correct answer and is worth one mark.
Show working and units for extended responses and the investigation. Use three significant figures for calculated results unless stated otherwise.
A calculator and the IB Physics data booklet may be used. This is an original topic test with Paper 1A-style multiple choice, Paper 2-style extended responses, and a Paper 1B-style investigation; the timings and mark allocation are set for this test.
For thermodynamics, use $Q=\Delta U+W$ : heat supplied to the gas and work done by the gas are positive.
Assume ideal-gas behaviour and negligible circuit lead resistance unless stated otherwise. Use absolute temperature in gas and radiation equations.
All answers and worked solutions are in the boxes at the bottom.

Constants

$R=8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$ ; $k_B=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}$ ; $\sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$ ; $b=2.90\times10^{-3}\,\mathrm{m\,K}$ .

Section A — Multiple choice [25 marks]

Question 1 [1 mark]

Which quantity is most directly associated with the absolute temperature of an ideal gas?

Total internal energy of the sample
Total molecular momentum of the sample
Mean intermolecular potential energy
Mean translational kinetic energy per molecule

Question 2 [1 mark]

A spherical planet has solar constant $1360\,\mathrm{W\,m^{-2}}$ and albedo $0.25$ . What is its absorbed solar power per unit area, averaged over its entire surface?

$85\,\mathrm{W\,m^{-2}}$
$255\,\mathrm{W\,m^{-2}}$
$340\,\mathrm{W\,m^{-2}}$
$1020\,\mathrm{W\,m^{-2}}$

Question 3 [1 mark]

A fixed amount of ideal gas initially has pressure $1.80\times10^5\,\mathrm{Pa}$ . Its volume doubles while its absolute temperature increases by a factor of $1.50$ . What is its final pressure?

$0.600\times10^5\,\mathrm{Pa}$
$1.20\times10^5\,\mathrm{Pa}$
$1.35\times10^5\,\mathrm{Pa}$
$5.40\times10^5\,\mathrm{Pa}$

Question 4 [1 mark]

A gas loses $80\,\mathrm{J}$ of heat while $220\,\mathrm{J}$ of work is done on it. What is its internal-energy change?

$+140\,\mathrm{J}$
$-140\,\mathrm{J}$
$+300\,\mathrm{J}$
$-300\,\mathrm{J}$

Question 5 [1 mark]

A uniform wire of resistance $R$ is stretched to three times its original length at constant volume. Its resistivity remains unchanged. What is its new resistance?

$R/9$
$R/3$
$3R$
$9R$

Question 6 [1 mark]

$0.200\,\mathrm{kg}$ of water at $80.0^\circ\mathrm{C}$ is mixed with $0.300\,\mathrm{kg}$ of water at $20.0^\circ\mathrm{C}$ . Neglect the container’s heat capacity and heat exchange with the surroundings. What is the final temperature?

$32.0^\circ\mathrm{C}$
$40.0^\circ\mathrm{C}$
$44.0^\circ\mathrm{C}$
$50.0^\circ\mathrm{C}$

Question 7 [1 mark]

Two black bodies have peak emission wavelengths of $0.50\,\mathrm{\mu m}$ and $10.0\,\mathrm{\mu m}$ . What is the temperature of the first divided by that of the second?

$0.050$
$0.500$
$20.0$
$400$

Question 8 [1 mark]

Helium and nitrogen gases are at the same temperature. Their molar masses are $4.0\,\mathrm{g\,mol^{-1}}$ and $28\,\mathrm{g\,mol^{-1}}$ . What is $v_{\mathrm{rms,He}}/v_{\mathrm{rms,N_2}}$ ?

$\sqrt7$
$1/\sqrt7$
$7$
$1$

Question 9 [1 mark]

The same ideal gas expands reversibly from the same initial state to the same larger final volume by either an isothermal or an adiabatic path. Which statement is correct?

Adiabatic expansion has the greater final temperature.
The two paths do the same work.
Isothermal expansion does more work than adiabatic expansion.
Adiabatic expansion absorbs more heat than isothermal expansion.

Question 10 [1 mark]

A $4.0\,\Omega$ resistor is in series with a parallel combination of $12\,\Omega$ and $6.0\,\Omega$ , connected to an ideal $12\,\mathrm{V}$ supply. What is the supply current?

$1.0\,\mathrm{A}$
$1.5\,\mathrm{A}$
$2.0\,\mathrm{A}$
$3.0\,\mathrm{A}$

Question 11 [1 mark]

A heater supplies $300\,\mathrm{W}$ to $0.150\,\mathrm{kg}$ of a solid already at its melting point. Its specific latent heat of fusion is $2.00\times10^5\,\mathrm{J\,kg^{-1}}$ . Neglect losses. How long does melting take?

$0.100\,\mathrm{s}$
$10.0\,\mathrm{s}$
$66.7\,\mathrm{s}$
$100\,\mathrm{s}$

Question 12 [1 mark]

Why does carbon dioxide absorb strongly in some thermal-infrared bands while molecular nitrogen does not absorb comparably through its ordinary vibrational modes?

Some carbon-dioxide vibrations change its electric dipole moment and have suitable transition energies.
Carbon dioxide absorbs all wavelengths equally while nitrogen reflects them all.
Carbon dioxide has a permanent dipole moment while nitrogen has no vibrational energy levels.
Nitrogen molecules cannot vibrate because they contain only two atoms.

Question 13 [1 mark]

The pressure of a fixed amount of ideal gas doubles in a rigid container. What happens to its mean translational kinetic energy per molecule?

It is unchanged.
It doubles.
It increases by a factor of four.
It halves.

Question 14 [1 mark]

A gas follows a clockwise rectangular cycle on a pressure–volume graph. The rectangle’s pressure difference is $1.50\times10^5\,\mathrm{Pa}$ and its volume difference is $2.00\times10^{-3}\,\mathrm{m^3}$ . What is the net work done by the gas?

$-300\,\mathrm{J}$
$0\,\mathrm{J}$
$+300\,\mathrm{J}$
$+600\,\mathrm{J}$

Question 15 [1 mark]

A cell has emf $9.0\,\mathrm{V}$ and internal resistance $1.0\,\Omega$ . It supplies an external resistor of $8.0\,\Omega$ . What is its terminal voltage?

$1.0\,\mathrm{V}$
$7.0\,\mathrm{V}$
$9.0\,\mathrm{V}$
$8.0\,\mathrm{V}$

Question 16 [1 mark]

The steady thermal conduction rate through a slab is $H$ . Its thermal conductivity is doubled and its thickness is tripled, with area and face temperatures unchanged. What is the new rate?

$2H/3$
$3H/2$
$6H$
$H/6$

Question 17 [1 mark]

A planet’s albedo rises from $0.20$ to $0.60$ . Its solar constant and effective emissivity remain unchanged. In a simple radiative-equilibrium model, what is $T_{\mathrm{new}}/T_{\mathrm{old}}$ ?

$1/2$
$1/\sqrt2$
$2^{-1/4}$
$2^{1/4}$

Question 18 [1 mark]

What is the mean translational kinetic energy of one ideal-gas molecule at $400\,\mathrm{K}$ ? Use $k_B=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}$ .

$2.76\times10^{-21}\,\mathrm{J}$
$8.28\times10^{-21}\,\mathrm{J}$
$5.52\times10^{-21}\,\mathrm{J}$
$1.66\times10^{-20}\,\mathrm{J}$

Question 19 [1 mark]

A Carnot engine operates between $327^\circ\mathrm{C}$ and $27^\circ\mathrm{C}$ . Using $T/\mathrm{K}=\theta/{}^\circ\mathrm{C}+273$ , what is its efficiency?

$8.3\%$
$91.7\%$
$100\%$
$50.0\%$

Question 20 [1 mark]

A fixed resistor connects an ideal supply’s positive terminal to a junction. An NTC thermistor connects the junction to the negative terminal. Output is measured across the thermistor with negligible loading. What happens as the thermistor warms?

Output voltage decreases; circuit current increases.
Output voltage increases; circuit current increases.
Output voltage decreases; circuit current decreases.
Output voltage increases; circuit current decreases.

Question 21 [1 mark]

A black surface at $400\,\mathrm{K}$ is surrounded by large black surroundings at $300\,\mathrm{K}$ . What fraction of its emitted power is its net radiative power loss? Give the answer to three significant figures.

$0.250$
$0.316$
$0.684$
$1.00$

Question 22 [1 mark]

Greenhouse-gas concentration increases while the planetary temperature, solar input, and albedo are initially unchanged. Which sequence best describes the enhanced greenhouse effect?

Outgoing infrared initially increases; the surface cools.
Outgoing infrared initially decreases; warming increases emission until balance can be restored.
Incoming solar radiation becomes zero; the atmosphere heats indefinitely.
Albedo must immediately become one; the surface warms.

Question 23 [1 mark]

An insulated ideal gas is rapidly compressed. Which molecular description explains the resulting rise in pressure?

The molecules become heavier.
Molecules collide less frequently with walls and move more slowly.
The number of molecules increases during compression.
Number density increases and mean translational kinetic energy increases.

Question 24 [1 mark]

$600\,\mathrm{J}$ of heat transfers directly from a reservoir at $500\,\mathrm{K}$ to one at $300\,\mathrm{K}$ . Both temperatures remain constant. What is the total entropy change?

$+0.800\,\mathrm{J\,K^{-1}}$
$-0.800\,\mathrm{J\,K^{-1}}$
$+3.20\,\mathrm{J\,K^{-1}}$
$0\,\mathrm{J\,K^{-1}}$

Question 25 [1 mark]

Resistance is calculated using $R=V/I$ . The voltage has fractional uncertainty $2\%$ and current $3\%$ . What is the estimated maximum fractional uncertainty in resistance?

$1\%$
$3\%$
$5\%$
$6\%$

Section B — Extended responses [60 marks]

Question 26 — Thermal energy transfers [12 marks]

A well-insulated calorimeter of heat capacity $80.0\,\mathrm{J\,K^{-1}}$ contains $0.250\,\mathrm{kg}$ of water. Both are initially at $40.0^\circ\mathrm{C}$ . A $0.0500\,\mathrm{kg}$ sample of ice at $0.0^\circ\mathrm{C}$ is added. Use water’s specific heat capacity $4200\,\mathrm{J\,kg^{-1}\,K^{-1}}$ and ice’s specific latent heat of fusion $3.34\times10^5\,\mathrm{J\,kg^{-1}}$ .

(a) Calculate the energy required to melt all the ice. [2 marks]

(b) Show that all the ice can melt by calculating the energy released if the initial water and calorimeter cool to $0.0^\circ\mathrm{C}$ . [2 marks]

(c) Determine the final equilibrium temperature, including the calorimeter’s heat capacity. [3 marks]

(d) Explain microscopically why the ice temperature stays constant while it melts. [2 marks]

(e) In a repeat, the system loses $2.68\,\mathrm{kJ}$ to its surroundings before reaching equilibrium. Determine the new final temperature. [3 marks]

Question 27 — Greenhouse effect and planetary energy balance [12 marks]

A spherical planet receives a solar constant of $1400\,\mathrm{W\,m^{-2}}$ and has albedo $0.300$ . Initially model it as having uniform effective emitting temperature, emissivity $1.00$ , and no atmospheric greenhouse effect.

(a) Derive the average absorbed solar power per unit area of the whole planet and calculate its value. Explain the geometrical factor used. [3 marks]

(b) Calculate the radiative-equilibrium temperature. [2 marks]

(c) Calculate the wavelength of maximum thermal emission and identify the spectral region. [2 marks]

(d) Explain how an atmosphere containing greenhouse gases can give a higher surface temperature than this simple model predicts, while the planet remains in overall energy balance. [3 marks]

(e) In the original atmosphere-free model, the albedo decreases to $0.200$ with solar constant unchanged. Determine the increase in equilibrium temperature. [2 marks]

Question 28 — Gas laws and compression [12 marks]

A cylinder contains $0.200\,\mathrm{mol}$ of a monatomic ideal gas at $300\,\mathrm{K}$ and volume $4.00\times10^{-3}\,\mathrm{m^3}$ . A piston compresses it reversibly and adiabatically to half this volume. You may use $pV^{5/3}=\mathrm{constant}$ for this process.

(a) Calculate the initial pressure. [2 marks]

(b) Calculate the final pressure and temperature. [4 marks]

(c) Calculate the work done on the gas, stating the first-law argument used. [3 marks]

(d) Calculate the final pressure for an isothermal compression to the same final volume. Use molecular reasoning to explain why it differs from the adiabatic result. [3 marks]

Question 29 — Thermodynamic cycle and efficiency [12 marks]

A fixed sample of monatomic ideal gas follows the rectangular cycle A → B → C → D → A on a pressure–volume graph. The states are listed below. Each branch is quasistatic and either constant-pressure or constant-volume. For the gas, $U=\frac32pV$ .

States for Question 29
State	Pressure / Pa	Volume / m³
A	1.00 × 10⁵	2.00 × 10⁻³
B	3.00 × 10⁵	2.00 × 10⁻³
C	3.00 × 10⁵	5.00 × 10⁻³
D	1.00 × 10⁵	5.00 × 10⁻³

(a) Calculate the work done by the gas on each branch. [3 marks]

(b) Determine the net work and net heat supplied over one complete cycle, explaining the internal-energy change. [2 marks]

(c) Calculate the heat supplied on A → B and B → C, and hence the total heat input per cycle. [3 marks]

(d) Calculate the heat-engine efficiency. [2 marks]

(e) State the net entropy change of the gas over the cycle. Explain what the second law requires for the total entropy change of the gas and its surroundings. [2 marks]

Question 30 — Current, circuits, and a real source [12 marks]

A source has emf $15.0\,\mathrm{V}$ and internal resistance $1.00\,\Omega$ . Its external circuit consists of a $3.00\,\Omega$ resistor in series with a parallel combination of $6.00\,\Omega$ and $12.0\,\Omega$ . Assume ohmic resistors at constant temperature.

(a) Calculate the external equivalent resistance and source current. [3 marks]

(b) Calculate the voltage across the parallel combination and the current through each parallel resistor. [3 marks]

(c) Calculate the energy dissipated in the $12.0\,\Omega$ resistor in $120\,\mathrm{s}$ . [2 marks]

(d) The $12.0\,\Omega$ branch is disconnected. Determine the new terminal voltage of the source. [2 marks]

(e) Explain why the terminal voltage changes even though the emf remains the same. [2 marks]

Section C — Lab-style investigation [15 marks]

Question 31 — Electrical heating of a metal block [15 marks]

A student heats a metal block of mass $0.250\,\mathrm{kg}$ using an electrical heater at constant voltage $6.00\,\mathrm{V}$ and current $2.00\,\mathrm{A}$ . The following idealised central readings are obtained. Use the quoted uncertainties only in part (c).

Temperature measurements during heating
Time / s	Temperature / °C
0	20.0
60	23.6
120	27.2
180	30.8
240	34.4

(a) Describe an apparatus and procedure to obtain these data. Identify important controls. [3 marks]

(b) Calculate the temperature–time gradient and determine the apparent specific heat capacity, initially neglecting apparatus heat capacity and heat loss. [3 marks]

(c) Estimate the maximum absolute uncertainty in this apparent specific heat capacity using $V=(6.00\pm0.03)\,\mathrm{V}$ , $I=(2.00\pm0.01)\,\mathrm{A}$ , $m=(0.250\pm0.002)\,\mathrm{kg}$ , $\Delta t=(240\pm1)\,\mathrm{s}$ , and $\Delta T=(14.4\pm0.2)\,\mathrm{K}$ . Add fractional uncertainty contributions. [3 marks]

(d) The heater and probe have combined heat capacity $25.0\,\mathrm{J\,K^{-1}}$ and warm with the block. Neglect heat loss and uncertainty for this correction. Calculate the corrected specific heat capacity of the metal. [3 marks]

(e) Explain the effect of unaccounted heat loss on the apparent specific heat capacity. Suggest one specific improvement and explain how it helps. [3 marks]

Answers and marking guidance

Section A — Multiple-choice answer box

Answer key: one mark per correct answer
Question	Answer	Question	Answer	Question	Answer	Question	Answer	Question	Answer
1	D	6	C	11	D	16	A	21	C
2	B	7	C	12	A	17	C	22	B
3	C	8	A	13	B	18	B	23	D
4	A	9	C	14	C	19	D	24	A
5	D	10	B	15	D	20	A	25	C

Show multiple-choice explanations

1. D For an ideal gas, $\overline{E_k}=\frac32k_BT$ . Temperature is associated with the mean translational kinetic energy, rather than the total energy of the sample.

2. B The intercepting area is $\pi R_p^2$ and total surface area is $4\pi R_p^2$ , so the mean absorbed flux is $(1-0.25)(1360)/4=255\,\mathrm{W\,m^{-2}}$ .

3. C For fixed gas amount, $pV/T$ is constant. Thus $p_2=p_1(1.50)/2=1.35\times10^5\,\mathrm{Pa}$ .

4. A Using work done by the gas as positive, $Q=-80\,\mathrm{J}$ and $W=-220\,\mathrm{J}$ . Hence $\Delta U=Q-W=+140\,\mathrm{J}$ .

5. D Constant volume means the cross-sectional area falls to $A/3$ . Since resistance is $\rho L/A$ , the resistance increases by $3\times3=9$ .

6. C Energy conservation gives $T_f=[(0.200)(80.0)+(0.300)(20.0)]/0.500=44.0^\circ\mathrm{C}$ .

7. C Wien’s law gives $\lambda_{\max}T=b$ . Therefore $T_1/T_2=\lambda_2/\lambda_1=10.0/0.50=20.0$ .

8. A At fixed temperature, $v_{\mathrm{rms}}\propto1/\sqrt M$ , so the ratio is $\sqrt{28/4}=\sqrt7$ .

9. C The adiabatically expanding gas cools. Its pressure is lower than that on the isothermal path at intermediate volumes, so the area under the pressure–volume curve, and therefore the work, is smaller.

10. B The parallel resistance is $4.0\,\Omega$ , making the total $8.0\,\Omega$ . Current is $12/8.0=1.5\,\mathrm{A}$ .

11. D $t=mL/P=(0.150)(2.00\times10^5)/300=100\,\mathrm{s}$ .

12. A Some carbon-dioxide vibrations produce a changing dipole moment and suitable infrared transitions. Nitrogen has vibrational levels, but its homonuclear vibration does not create a changing dipole moment for ordinary electric-dipole infrared absorption.

13. B At fixed volume and gas amount, $p\propto T$ , so temperature doubles. Mean translational kinetic energy is proportional to $T$ and therefore also doubles.

14. C The signed enclosed area is positive for a clockwise cycle: $W=\Delta p\Delta V=+300\,\mathrm{J}$ .

15. D Current is $9.0/(8.0+1.0)=1.0\,\mathrm{A}$ . Thus $V=\varepsilon-Ir=9.0-1.0=8.0\,\mathrm{V}$ .

16. A Conduction rate is proportional to $k/L$ . The new rate is $(2/3)H$ .

17. C $T\propto(1-\alpha)^{1/4}$ . Therefore $T_{\mathrm{new}}/T_{\mathrm{old}}=(0.40/0.80)^{1/4}=2^{-1/4}$ .

18. B $\overline{E_k}=\frac32k_BT=(1.5)(1.38\times10^{-23})(400)=8.28\times10^{-21}\,\mathrm{J}$ .

19. D Use kelvin: $T_h=600\,\mathrm{K}$ and $T_c=300\,\mathrm{K}$ . Thus $\eta=1-300/600=0.500$ .

20. A Warming lowers the NTC resistance, reducing total resistance and increasing current. The thermistor’s share of the total resistance falls, so its output voltage decreases.

21. C The fraction is $(T^4-T_s^4)/T^4=1-(300/400)^4=0.683594\approx0.684$ .

22. B At the initially unchanged temperature profile, greater infrared absorption reduces radiation escaping to space in affected bands. Incoming energy then exceeds outgoing energy; warming increases emission toward a new balance.

23. D Compression increases number density. Work done on the insulated gas raises internal energy and temperature, increasing mean translational kinetic energy; both effects raise pressure.

24. A $\Delta S=-600/500+600/300=-1.20+2.00=+0.800\,\mathrm{J\,K^{-1}}$ .

25. C For a maximum uncertainty estimate in a quotient, add the fractional uncertainties: 2% + 3% = 5%.

Section B — Extended-response answer box

Suggested marking points are included. Accept equivalent correct methods and explanations; carry unrounded values through calculations.

Question 26 — Thermal energy transfers: full solution [12 marks]

(a) [2 marks] $\displaystyle Q_{\mathrm{melt}}=mL=(0.0500)(3.34\times10^5)=1.67\times10^4\,\mathrm{J}$ Credit the latent-heat relation and substitution/result.

(b) [2 marks] $\displaystyle Q_{\mathrm{available}}=(m_wc+C)(40.0)=[(0.250)(4200)+80.0](40.0)=4.52\times10^4\,\mathrm{J}$ This exceeds $1.67\times10^4\,\mathrm{J}$ , so all the ice melts and the equilibrium temperature is above zero.

(c) [3 marks] Let $\theta_f$ be the final temperature in degrees Celsius. Heat lost by the original water and calorimeter equals melting energy plus warming of the meltwater: $\displaystyle [(0.250)(4200)+80.0](40.0-\theta_f)=16700+(0.0500)(4200)\theta_f$ $\displaystyle \theta_f=\frac{45200-16700}{1130+210}=21.3^\circ\mathrm{C}$ Credit a correct energy balance, inclusion of both water masses and calorimeter, and the result.

(d) [2 marks] Energy supplied during melting increases intermolecular potential energy as the solid structure changes. The mean molecular kinetic energy, and therefore temperature, does not increase during the phase change at constant pressure. Credit the distinction between potential and kinetic energy.

(e) [3 marks] Include the energy transferred to the surroundings: $\displaystyle 1130(40.0-\theta_f)=16700+210\theta_f+2680$ $\displaystyle \theta_f=\frac{45200-16700-2680}{1340}=19.3^\circ\mathrm{C}$ The final temperature falls by $2.00\,\mathrm{K}$ . All the ice still melts.

Question 27 — Greenhouse effect and planetary energy balance: full solution [12 marks]

(a) [3 marks] The intercepted solar power is $S\pi R_p^2$ . Multiply by the absorbed fraction $1-\alpha$ and divide by the emitting surface area $4\pi R_p^2$ : $\displaystyle F_{\mathrm{abs}}=\frac{(1-\alpha)S}{4}=\frac{(0.700)(1400)}4=245\,\mathrm{W\,m^{-2}}$ Credit the two areas, absorbed fraction, and calculation.

(b) [2 marks] At equilibrium, $F_{\mathrm{abs}}=\sigma T^4$ . $\displaystyle T=\left(\frac{245}{5.67\times10^{-8}}\right)^{1/4}=256\,\mathrm{K}$ This is the model’s effective emitting temperature.

(c) [2 marks] $\displaystyle \lambda_{\max}=\frac bT=\frac{2.90\times10^{-3}}{256.3869}=1.13\times10^{-5}\,\mathrm{m}=11.3\,\mathrm{\mu m}$ The emission peak is in the infrared.

(d) [3 marks] Greenhouse gases absorb some outgoing thermal infrared in bands associated with molecular transitions and emit infrared both upward and downward. Downward radiation supplies additional energy to the surface, so it can equilibrate at a higher temperature. At the top of the atmosphere, total outgoing radiation still equals absorbed solar radiation at equilibrium; surface temperature and effective emitting temperature need not be the same. Credit absorption/emission, surface warming, and top-of-atmosphere balance.

(e) [2 marks] $\displaystyle T_2=\left(\frac{(0.800)(1400)}{4(5.67\times10^{-8})}\right)^{1/4}=265\,\mathrm{K}$ Using unrounded temperatures: $\displaystyle \Delta T=265.0902-256.3869=8.70\,\mathrm{K}$

Question 28 — Gas laws and compression: full solution [12 marks]

(a) [2 marks] $\displaystyle p_1=\frac{nRT_1}{V_1}=\frac{(0.200)(8.31)(300)}{4.00\times10^{-3}}=1.25\times10^5\,\mathrm{Pa}$ Retain $124650\,\mathrm{Pa}$ in subsequent calculations.

(b) [4 marks] $\displaystyle p_2=p_1\left(\frac{V_1}{V_2}\right)^{5/3}=(124650)(2^{5/3})=3.96\times10^5\,\mathrm{Pa}$ $\displaystyle T_2=\frac{p_2V_2}{nR}=\frac{(395739.1)(2.00\times10^{-3})}{(0.200)(8.31)}=476\,\mathrm{K}$ Credit the adiabatic relation and pressure, then the ideal-gas relation and temperature.

(c) [3 marks] Adiabatic means $Q=0$ . With $W$ positive for work done by the gas, $\Delta U=Q-W=-W$ . Therefore work done on the gas equals its internal-energy increase: $\displaystyle W_{\mathrm{on}}=\Delta U=\frac32nR(T_2-T_1)$ $\displaystyle W_{\mathrm{on}}=\frac32(0.200)(8.31)(476.2203-300)=439\,\mathrm{J}$

(d) [3 marks] For isothermal compression: $\displaystyle p_{2,\mathrm{iso}}=p_1\frac{V_1}{V_2}=(124650)(2)=2.49\times10^5\,\mathrm{Pa}$ Both compressions produce the same final number density. In the adiabatic case, work also increases mean translational kinetic energy, so molecules transfer more momentum to the walls and the final pressure is higher.

Question 29 — Thermodynamic cycle and efficiency: full solution [12 marks]

(a) [3 marks] Constant-volume branches do no boundary work. $\displaystyle W_{AB}=0,\qquad W_{BC}=(3.00\times10^5)(3.00\times10^{-3})=+900\,\mathrm{J}$ $\displaystyle W_{CD}=0,\qquad W_{DA}=(1.00\times10^5)(-3.00\times10^{-3})=-300\,\mathrm{J}$ Credit zero work on the vertical branches and the signed work on each horizontal branch.

(b) [2 marks] $\displaystyle W_{\mathrm{net}}=900-300=+600\,\mathrm{J}$ The gas returns to its initial state, so $\Delta U_{\mathrm{cycle}}=0$ and $Q_{\mathrm{net}}=W_{\mathrm{net}}=+600\,\mathrm{J}$ .

(c) [3 marks] Use $Q=\Delta U+W$ . $\displaystyle Q_{AB}=\frac32(p_BV_B-p_AV_A)=\frac32(600-200)=600\,\mathrm{J}$ $\displaystyle Q_{BC}=\frac32(1500-600)+900=2250\,\mathrm{J}$ $\displaystyle Q_h=600+2250=2850\,\mathrm{J}$ On C → D and D → A the gas rejects $1500\,\mathrm{J}$ and $750\,\mathrm{J}$ respectively, so these are not included as positive heat inputs.

(d) [2 marks] $\displaystyle \eta=\frac{W_{\mathrm{net}}}{Q_h}=\frac{600}{2850}=0.211=21.1\%$ Use total heat absorbed, not net heat, in the denominator.

(e) [2 marks] The gas returns to the same state, so its net entropy change is zero. The total entropy change of the gas and surroundings must be non-negative: zero for an entirely reversible cycle and positive if irreversible effects occur. Returning the gas to its initial state does not require the surroundings to return to their initial state.

Question 30 — Current, circuits, and a real source: full solution [12 marks]

(a) [3 marks] $\displaystyle R_{\parallel}=\frac{(6.00)(12.0)}{6.00+12.0}=4.00\,\Omega$ $\displaystyle R_{\mathrm{external}}=3.00+4.00=7.00\,\Omega$ $\displaystyle I=\frac{\varepsilon}{R_{\mathrm{external}}+r}=\frac{15.0}{8.00}=1.875\,\mathrm{A}\approx1.88\,\mathrm{A}$

(b) [3 marks] $\displaystyle V_{\parallel}=IR_{\parallel}=(1.875)(4.00)=7.50\,\mathrm{V}$ $\displaystyle I_6=\frac{7.50}{6.00}=1.25\,\mathrm{A},\qquad I_{12}=\frac{7.50}{12.0}=0.625\,\mathrm{A}$ The branch currents sum to $1.875\,\mathrm{A}$ .

(c) [2 marks] $\displaystyle E=I_{12}^2R_{12}t=(0.625)^2(12.0)(120)=563\,\mathrm{J}$ Accept $562.5\,\mathrm{J}$ before rounding.

(d) [2 marks] The external resistance is now $3.00+6.00=9.00\,\Omega$ . $\displaystyle I_{\mathrm{new}}=\frac{15.0}{9.00+1.00}=1.50\,\mathrm{A}$ $\displaystyle V_{\mathrm{new}}=\varepsilon-I_{\mathrm{new}}r=15.0-1.50=13.5\,\mathrm{V}$

(e) [2 marks] The original terminal voltage was $15.0-1.875=13.125\,\mathrm{V}$ . Disconnecting a branch raises external resistance and reduces current, so the internal drop $Ir$ decreases and terminal voltage rises. Emf is the energy supplied per unit charge by the source and is fixed in this model.

Section C — Investigation answer box

Question 31 — Full solution and evaluation [15 marks]

(a) [3 marks] Use a metal block of measured mass with close-fitting holes for an electrical heater and temperature probe. Insulate the block and use thermal paste for good contact. Connect an ammeter in series with the heater and a voltmeter across it. Record temperature at regular time intervals while monitoring voltage and current. Keep heater power approximately constant, use the same apparatus throughout, and allow for temperature uniformity; repeat the experiment to check reproducibility. Credit apparatus/meter positions, procedure, and controls.

(b) [3 marks] The temperature rise is $14.4\,\mathrm{K}$ in $240\,\mathrm{s}$ : $\displaystyle \frac{\Delta T}{\Delta t}=0.0600\,\mathrm{K\,s^{-1}}$ Heater power is $P=VI=(6.00)(2.00)=12.0\,\mathrm{W}$ . $\displaystyle c_{\mathrm{app}}=\frac{P}{m(\Delta T/\Delta t)}=\frac{12.0}{(0.250)(0.0600)}=800\,\mathrm{J\,kg^{-1}\,K^{-1}}$

(c) [3 marks] Use $c_{\mathrm{app}}=VI\Delta t/(m\Delta T)$ . $\displaystyle \frac{\delta c}{c}\approx\frac{0.03}{6.00}+\frac{0.01}{2.00}+\frac1{240}+\frac{0.002}{0.250}+\frac{0.2}{14.4}=0.0361$ $\displaystyle \delta c\approx(800)(0.0361)=29\,\mathrm{J\,kg^{-1}\,K^{-1}}$ With sensible uncertainty rounding, report approximately $(800\pm30)\,\mathrm{J\,kg^{-1}\,K^{-1}}$ . The supplied $0.2\,\mathrm{K}$ is already the uncertainty in the temperature difference.

(d) [3 marks] Electrical energy warms both block and apparatus: $\displaystyle VI\Delta t=(mc+C_{\mathrm{apparatus}})\Delta T$ $\displaystyle c=\frac{VI\Delta t/\Delta T-C_{\mathrm{apparatus}}}{m}=\frac{(6.00)(2.00)(240)/14.4-25.0}{0.250}$ $\displaystyle c=700\,\mathrm{J\,kg^{-1}\,K^{-1}}$ Credit the two heat stores, substitution, and result.

(e) [3 marks] If energy escapes to the surroundings, the observed temperature rise is smaller for the same electrical input. Assigning all input energy to the block therefore overestimates its specific heat capacity. Add effective insulation around the block to reduce heat loss, or measure the heat-loss rate near the relevant temperatures and include it in the energy balance. Credit the direction of bias and an improvement with its reason.

Topic B HL Test 1 • Original IB-style assessment • Total: 100 marks

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Topic B. The particulate nature of matter HL TEST 1