Home › Free Resources › Physics Worksheets › IB Physics – Theme B: The Particulate Nature of Matter
Free Physics Worksheet

IB Physics – Theme B: The Particulate Nature of Matter

Thermal energy transfers, gas laws, thermodynamics (HL), current and circuits.

WhatsApp Mr. AhmadBook a trial lesson (50% off)Free worksheets
12 questions  ·  39 marks  ·  answers included  ·  IB Physics SL & HL
Download the PDFAsk about a question

This worksheet covers:

  • Practice questions covering thermal energy transfers
  • The gas laws
  • Thermodynamics (some parts HL only, marked)
  • Current and circuits from IB Physics Theme B

A calculator is required; take the specific heat capacity of water as $4180\,\text{J kg}^{-1}\text{K}^{-1}$, the specific latent heat of fusion of ice as $334000\,\text{J kg}^{-1}$, and the molar gas constant as $R = 8.31\,\text{J mol}^{-1}\text{K}^{-1}$.

Questions

Section A – Thermal Energy Transfers

1

A $0.50\,\text{kg}$ sample of water is heated from $15\,^\circ\text{C}$ to $40\,^\circ\text{C}$. Calculate the thermal energy required, in kJ.

Show answer
$Q = mc\Delta T = 0.50 \times 4180 \times 25 = 52250\,\text{J} = 52.3\,\text{kJ}$.
[3]
2

Calculate the energy required to melt $0.20\,\text{kg}$ of ice at $0\,^\circ\text{C}$, giving your answer in kJ.

Show answer
$Q = mL_f = 0.20 \times 334000 = 66800\,\text{J} = 66.8\,\text{kJ}$.
[2]
3

$0.20\,\text{kg}$ of ice at $0\,^\circ\text{C}$ is converted completely into water at $20\,^\circ\text{C}$. Calculate the total thermal energy required, in kJ.

Show answer
Energy to melt: $66800\,\text{J}$. Energy to heat water from $0\,^\circ\text{C}$ to $20\,^\circ\text{C}$: $0.20 \times 4180 \times 20 = 16720\,\text{J}$. Total $= 83520\,\text{J} = 83.5\,\text{kJ}$.
[4]
4

Using the particle (kinetic) model of matter, explain why the temperature of a substance remains constant while it is melting, even though thermal energy is continuously being supplied.

Show answer
During melting, the energy supplied is used to break the intermolecular bonds holding particles in fixed positions in the solid lattice (increasing potential energy), rather than to increase the average kinetic energy of the particles, so the temperature (which depends on average kinetic energy) stays constant.
[3]

Section B – Gas Laws

Calculator required
5

A gas sample of $2.0\,\text{mol}$ occupies a volume of $0.050\,\text{m}^3$ at a temperature of $300\,\text{K}$. Use the ideal gas equation $pV = nRT$ to calculate the pressure of the gas.

Show answer
$p = \dfrac{nRT}{V} = \dfrac{2.0 \times 8.31 \times 300}{0.050} = 9.97\times10^{4}\,\text{Pa}$.
[3]
6

A fixed mass of gas at constant volume has pressure $1.0\times10^{5}\,\text{Pa}$ at a temperature of $300\,\text{K}$. The gas is heated to $450\,\text{K}$. Calculate the new pressure.

Show answer
$\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2} \Rightarrow P_2 = 1.0\times10^{5} \times \dfrac{450}{300} = 1.5\times10^{5}\,\text{Pa}$.
[3]
7

Using the kinetic model of an ideal gas, explain why the pressure of a fixed mass of gas increases when its temperature is increased at constant volume.

Show answer
Increasing temperature increases the average kinetic energy (and hence speed) of the gas molecules. This means molecules collide with the container walls more frequently and with greater momentum change per collision, increasing the average force per unit area, i.e. the pressure.
[3]
8

A gas occupies a volume of $2.0\times10^{-3}\,\text{m}^3$ at a pressure of $1.5\times10^{5}\,\text{Pa}$. It is compressed at constant temperature until its volume is $1.2\times10^{-3}\,\text{m}^3$. Calculate the new pressure.

Show answer
By Boyle’s law, $P_1V_1 = P_2V_2 \Rightarrow P_2 = \dfrac{1.5\times10^{5}\times 2.0\times10^{-3}}{1.2\times10^{-3}} = 2.5\times10^{5}\,\text{Pa}$.
[3]

Section C – Thermodynamics (HL) and Circuits

Q9 and Q12 are HL only
9

(HL) In one cycle, $500\,\text{J}$ of thermal energy is added to a gas, and the gas does $200\,\text{J}$ of work on its surroundings as it expands. Use the first law of thermodynamics to find the change in internal energy of the gas.

Show answer
$\Delta U = Q – W = 500 – 200 = 300\,\text{J}$ (the internal energy increases by $300\,\text{J}$).
[3]
10

Two resistors, $R_1 = 6\,\Omega$ and $R_2 = 3\,\Omega$, are available.(a) Find the equivalent resistance when they are connected in series.(b) Find the equivalent resistance when they are connected in parallel.

Show answer
(a) $R_{\text{series}} = 6+3 = 9\,\Omega$.(b) $R_{\text{parallel}} = \dfrac{6\times3}{6+3} = 2.0\,\Omega$.
[4]
11

A battery of EMF $12\,\text{V}$ and internal resistance $0.5\,\Omega$ is connected to an external resistor of $5.5\,\Omega$.(a) Calculate the current in the circuit.(b) Calculate the terminal potential difference of the battery.

Show answer
(a) $I = \dfrac{\varepsilon}{R+r} = \dfrac{12}{5.5+0.5} = 2.0\,\text{A}$.(b) $V = \varepsilon – Ir = 12 – (2.0)(0.5) = 11\,\text{V}$.
[4]
12

(HL) A heat engine takes in $800\,\text{J}$ of thermal energy from a hot reservoir each cycle and expels $500\,\text{J}$ to a cold reservoir.(a) Calculate the useful work done by the engine per cycle.(b) Calculate the efficiency of the engine, giving your answer as a percentage.

Show answer
(a) $W = Q_H – Q_C = 800-500 = 300\,\text{J}$.(b) Efficiency $= \dfrac{W}{Q_H} = \dfrac{300}{800} = 0.375 = 37.5\%$.
[4]

Related worksheets

More worksheets in IB Physics (SL & HL):

IB Physics – Theme A: Space, Time and MotionIB Physics – Theme C: Wave BehaviourIB Physics – Theme D: FieldsIB Physics – Theme E: Nuclear and Quantum PhysicsIB Physics – Data-Based Questions (Paper 1B)

All physics worksheets  ·  Full resource library  ·  Physics tuition with Mr. Ahmad

Get new physics worksheets by email

One short email when new worksheets are added. No spam, unsubscribe any time.



    Book a trial lesson (50% off)

    Try a full first lesson with Mr. Ahmad at 50% off before you commit to anything. It is a real lesson, not a sales call: bring a topic the student is stuck on and we will work through it together.

    • Online from anywhere in the world, or in person across Dubai
    • You get honest feedback on where the student stands and what to fix first
    • No contract — if it is not the right fit, that is fine

    Prefer to talk? Message on WhatsApp and you will hear back within 24 hours.



      Stuck on this topic?

      Book a trial lesson with Mr. Ahmad at 50% off and bring this worksheet with you — online worldwide or in person in Dubai.

      WhatsApp nowBook a trial lesson (50% off)
      Scroll to Top