IB Physics HL Theme B: The particulate nature of matter questions

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Question 1

Multiple choiceMedium1 mark

During a 10.0 min interval, a sample is melting at constant temperature. A 400 W heater is on for a total of 5.0 min; assume all energy supplied while it is on enters the sample. Which row gives the energy supplied by the heater and the particle behaviour during melting?

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Answer: B. The energy supplied is $E=Pt$ using the time for which the heater is on: $E=(400\ \text{W})(5.0\times60\ \text{s})=1.2\times10^5\ \text{J}=120\ \text{kJ}$ During melting, the energy changes particle behaviour and arrangement while the average kinetic energy, and therefore the temperature, remains constant.
A: treats 5.0 min as 5.0 s, giving 2.0 kJ.
C: calculates the energy correctly but applies the temperature-rise model instead of recognizing a phase change.
D: uses the full 10.0 min interval as the heater-on time, giving 240 kJ.
Remember: energy can be transferred during a phase change without increasing the temperature.

Question 2

Multiple choiceMedium1 mark

A methane source inventory records total emissions of $1.0\times10^5\ \text{kg day}^{-1}$ Wetlands are the only natural source in the inventory and emit methane at $1.5\times10^3\ \text{kg h}^{-1}$ all other emissions are human-made. What is the mass emitted by human-made sources per day?

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Answer: B. The human-made emissions are the total emissions minus the natural emissions. Wetlands emit $1.5\times10^3\times24=3.6\times10^4\ \text{kg day}^{-1}$ so the human-made emissions are $1.0\times10^5-3.6\times10^4=6.4\times10^4\ \text{kg day}^{-1}$

A: converts the wetland rate to a daily mass but gives the natural emissions alone.
C: uses the total daily emissions without subtracting the natural contribution.
D: adds the natural contribution to the total instead of subtracting it.

Remember: convert rates to matching time units before combining emissions.

Question 3

Multiple choiceEasy1 mark

A fixed amount of ideal gas changes between the two states shown. The labelled pressures are absolute; $p$ is a positive pressure and $V$ is a positive volume. Which expression gives $T_B/T_A$

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Answer: A. For a fixed amount of ideal gas, $PV/T=\text{constant}$ so $T_B/T_A=P_BV_B/(P_AV_A)$ Substituting the values shown gives $(3p)(2V)/[(2p)(V)]=3$ the temperature ratio is dimensionless.
B: reverses the state ratio, giving $T_A/T_B=1/3$
C: omits the volume ratio and uses pressure alone, giving $3/2$
D: omits the pressure ratio and uses volume alone, giving $2$
Remember: compare the products $PV$ when finding temperature ratios for a fixed amount of ideal gas.

Question 4

Multiple choiceMedium1 mark

A real isolated system consists of parts A and B. During an irreversible change, their entropy changes are $\Delta S_A=-x$ and $\Delta S_B=+y$ where $x$ and $y$ are positive. Which relation gives the entropy change of the whole system and the resulting constraint on $x$ and $y$

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Answer: B. The changes of the two parts add: $\Delta S_{\text{whole}}=\Delta S_A+\Delta S_B=-x+y=y-x$ Since the irreversible change occurs in a real isolated system, its total entropy increases; hence $y-x>0$ so $y>x$
A: treats the decrease $-x$ as a positive contribution.
C: reverses the subtraction, giving the wrong total change and constraint.
D: counts only the increase in part B and omits part A’s entropy change.
Remember: the entropy increase applies to the isolated whole, not necessarily to each part.

Question 5

Multiple choiceMedium1 mark

A metal wire is connected as shown. Its temperature is held constant for the first readings, then allowed to rise as the potential difference is increased. Which row describes the first readings and the resistance trend in the later readings?

	Behaviour of first readings	Resistance during later readings
A	Ohmic	Increases
B	Ohmic	Decreases
C	Non-ohmic	Increases
D	Non-ohmic	Decreases

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Answer: A. Resistance is $R=V/I$ For the fixed-temperature readings, $1.0/0.20=5.0\ \Omega$ and $2.0/0.40=5.0\ \Omega$ so current is proportional to potential difference and the wire is ohmic. For the later readings, $3.6/0.60=6.0\ \Omega$ and $5.6/0.80=7.0\ \Omega$ so resistance increases.

B: treats the falling gradient of the current-against-potential-difference graph, $I/V$ as resistance; its decrease corresponds to increasing $V/I$

C: uses the changing gradient in the later section to classify the fixed-temperature readings, which lie on a straight line through the origin.

D: combines those errors, applying the later section’s changing gradient to the first readings and reversing resistance with $I/V$

Remember: On an $I$ -against- $V$ graph, the gradient is $I/V$ whereas resistance is $V/I$

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