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Edexcel International A Level Physics · Unit 5 · Thermodynamics, Radiation, Oscillations and Cosmology

Internal energy and the ideal gas

40 free, high-quality past-paper-style questions on Internal energy and the ideal gas for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.40 new this month · 143 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1IALP-U5-2-A-01 Unit 5 · A2Easy1 mark

Which statement about the internal energy of a fixed mass of an ideal gas is correct?

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AnswerMarkNotes
In an ideal gas there are no forces between the molecules except during collisions, so the molecules have no potential energy. The internal energy is the total random kinetic energy of the molecules, which is proportional to the absolute (kelvin) temperature.✓ 1Answer A

Answer: A

Exam tip: Internal energy = sum of the randomly distributed kinetic and potential energies of the molecules. For an ideal gas the potential energy term is zero.

2IALP-U5-2-A-02 Unit 5 · A2Easy1 mark

The temperature of a gas rises from −73 °C to 127 °C.
By what factor does the mean kinetic energy of its molecules increase?

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AnswerMarkNotes
Convert to kelvin: −73 + 273 = 200 K and 127 + 273 = 400 K. Mean kinetic energy ∝ T, so the factor is 400 ÷ 200 = 2.0.✓ 1Answer C

Answer: C

Common error: Dividing Celsius temperatures. Mean kinetic energy is proportional to the absolute temperature, so convert to kelvin first.

3IALP-U5-2-A-03 Unit 5 · A2Easy1 mark

A sealed flask of volume 2.0 × 10−3 m3 contains air at a pressure of 1.0 × 105 Pa and a temperature of 27 °C. k = 1.38 × 10−23 J K−1.
How many molecules are in the flask?

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AnswerMarkNotes
N = pV/kT = (1.0 × 105 × 2.0 × 10−3) ÷ (1.38 × 10−23 × 300) = 4.8 × 1022.✓ 1Answer B

Answer: B

Exam tip: Before using pV = NkT, write every quantity in SI units: Pa, m3 and K.

4IALP-U5-2-A-07 Unit 5 · A2Easy1 mark

A block of ice at 0 °C is heated and melts, forming water at 0 °C.
Which row describes the changes in the mean kinetic energy and in the potential energy of the molecules during the melting?

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AnswerMarkNotes
The temperature stays at 0 °C, so the mean kinetic energy of the molecules is unchanged. The energy supplied separates the molecules against their attractive forces, so their potential energy increases and the internal energy increases.✓ 1Answer A

Answer: A

Exam tip: Constant temperature means constant mean kinetic energy. During a change of state, the energy supplied goes into molecular potential energy.

5IALP-U5-2-B-01 Unit 5 · A2Easy4 marks

A sealed plastic bottle of air is carried in a moving car from the coast to a mountain village. The temperature of the air in the bottle stays the same throughout. Treat the air as an ideal gas.

(a)

State what is meant by the internal energy of a body.

(2)
(b)

Explain why the internal energy of the air in the bottle does not change during the journey.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
the sum of the kinetic and potential energies of the molecules✓ 1
which are randomly distributed among the molecules✓ 1
Part (b)
an ideal gas has no forces between its molecules, so no molecular potential energy; its internal energy is the total random kinetic energy, which depends only on the temperature✓ 1
the kinetic and gravitational potential energy of the bottle as a whole are not part of internal energy, and the temperature is unchanged✓ 1

Common error: Counting the speed or height of the whole object as internal energy. Only the random, molecular-scale energies count.

6IALP-U5-2-B-02 Unit 5 · A2Easy5 marks

A constant-volume gas thermometer consists of a fixed mass of gas in a rigid bulb connected to a pressure gauge. The pressure of the gas is measured at two fixed points: 0 °C: 92.0 kPa; 100 °C: 125.7 kPa. The pressure varies linearly with temperature.

2040608010092.0100110125.7θ / °Cp / kPa
Figure: pressure of the gas at the two fixed points
(a)

State what is meant by absolute zero.

(1)
(b)

Use the readings to estimate the value of absolute zero in °C.

(2)
(c)

Calculate the pressure of the gas when the bulb is at −40 °C.

(2)
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AnswerMarkNotes
Part (a)
the lowest possible temperature, at which the molecules have minimum (kinetic) energy / at which the pressure of an ideal gas would be zero✓ 1
Part (b)
pressure rises by 125.7 − 92.0 = 33.7 kPa per 100 °C, i.e. 0.337 kPa per °C✓ 1
zero pressure at 0 − 92.0 ÷ 0.337 = −273 °C✓ 1
Part (c)
p ∝ T: T = 233 K, or 92.0 − 40 × 0.337✓ 1
p = 92.0 × 233 ÷ 273 = 78.5 kPa✓ 1

Model sketch

−300−200−10010020406080100120θ / °Cp / kPa−273 °C
Figure: extrapolating the line to zero pressure

Exam tip: A straight-line extrapolation to zero pressure needs two points: find the gradient first, then how far below 0 °C the line meets the axis.

7IALP-U5-2-FA-02 Unit 5 · A2Easy1 mark

Which graph shows how the mean kinetic energy of the molecules of an ideal gas varies with the Celsius temperature θ of the gas?

−273 0 θ / °C mean KE A −273 0 θ / °C mean KE B −273 0 θ / °C mean KE C −273 0 θ / °C mean KE D
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AnswerMarkNotes
½m⟨c2⟩ = 3/2kT, so the mean kinetic energy is proportional to the kelvin temperature T = θ + 273. Against θ this is a straight line that reaches zero at θ = −273 °C (absolute zero): graph A.✓ 1Answer A

Answer: A

Exam tip: A quantity proportional to T plotted against θ in °C gives a straight line that is not through the origin: it cuts the θ axis at −273 °C.

8IALP-U5-2-FA-08 Unit 5 · A2Easy1 mark

The graph shows the distribution of molecular speeds in two samples, X and Y, of the same gas.
Which statement is correct?

0 200 400 600 800 1000 1200 1400 1600 number of molecules per unit speed interval speed / m s⁻¹ X Y
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AnswerMarkNotes
The molecules in Y are spread to higher speeds, so their r.m.s. speed and mean kinetic energy are greater: Y is at the higher temperature. The area under each curve represents the total number of molecules, and the two areas are equal, so the samples contain the same number of molecules. Y's peak is lower because the same number of molecules is spread over a wider range of speeds.✓ 1Answer D

Answer: D

Exam tip: On a speed-distribution graph, the area under the curve is the number of molecules. Heating the gas moves the peak to a higher speed and makes it lower and broader; the area stays the same.

9IALP-U5-2-FB-01 Unit 5 · A2Easy5 marks

A thermometer is left inside a freezer until its reading is steady. Figure 1 shows the thermometer. Treat the air in the freezer as an ideal gas. k = 1.38 × 10−23 J K−1

−30 −20 −10 0 10 20 30 θ / °C liquid
(a)

Read the temperature of the air in the freezer from Figure 1 and convert it to kelvin.

(2)
(b)

Calculate the mean kinetic energy of a molecule of the air in the freezer.

(2)
(c)

State the temperature, in °C, at which the mean kinetic energy of the air molecules would be twice its value in the freezer.

(1)
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AnswerMarkNotes
Part (a)
θ = −18 °C✓ 1
T = −18 + 273 = 255 K✓ 1
Part (b)
mean kinetic energy = 3/2kT = 1.5 × 1.38 × 10−23 × 255✓ 1
= 5.3 × 10−21 J (5.28 × 10−21 J)✓ 1
Part (c)
T = 2 × 255 = 510 K, so θ = 237 °C✓ 1

Common error: Doubling the Celsius reading (−18 °C → −36 °C). Mean kinetic energy is proportional to the kelvin temperature, so double the kelvin value and convert back.

10IALP-U5-2-FB-02 Unit 5 · A2Easy5 marks

A block of ice at −20 °C is heated by a 70 W heater. All the energy from the heater is transferred to the ice, which melts, and then to the water formed. Figure 1 shows how the temperature θ varies with time t.

0 1 2 3 4 5 6 7 8 9 10 11 12 −30 −20 −10 0 10 20 30 θ / °C t / min
(a)

Describe what happens to the mean kinetic energy and to the potential energy of the molecules between t = 1.0 min and t = 9.0 min.

(2)
(b)

Calculate the increase in the internal energy of the ice while it is melting.

(2)
(c)

Calculate the total increase in internal energy from t = 0 to t = 11.0 min.

(1)
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AnswerMarkNotes
Part (a)
the temperature is constant, so the mean kinetic energy of the molecules does not change✓ 1
the potential energy of the molecules increases, as they are separated against the attractive forces between them (bonds are broken)✓ 1
Part (b)
melting time = 9.0 − 1.0 = 8.0 min = 480 s✓ 1
ΔU = 70 × 480 = 3.4 × 104 J (33 600 J)✓ 1
Part (c)
70 × 11.0 × 60 = 4.6 × 104 J (46 200 J)✓ 1

Exam tip: During a change of state the temperature is constant, so the energy supplied goes into molecular potential energy. Internal energy still increases.

11IALP-U5-2-A-04 Unit 5 · A2Medium1 mark

A fixed mass of gas in a cylinder has a volume of 450 cm3, a pressure of 1.2 × 105 Pa and a temperature of 17 °C. A piston compresses it to a volume of 180 cm3, and its temperature rises to 77 °C.
What is the new pressure of the gas?

gas450 cm³1.2 × 10⁵ Pa17 °Cgas180 cm³p = ?77 °Cbeforeafter
Figure: the gas before and after compression (not to scale)
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AnswerMarkNotes
For a fixed mass, pV/T is constant: p2 = 1.2 × 105 × (450 ÷ 180) × (350 ÷ 290) = 3.6 × 105 Pa.✓ 1Answer C

Answer: C

Exam tip: Write p1V1/T1 = p2V2/T2, then check each ratio: a smaller volume and a higher temperature must both increase the pressure.

12IALP-U5-2-A-05 Unit 5 · A2Medium1 mark

A container holds a mixture of hydrogen molecules (mass 2.0 u) and oxygen molecules (mass 32 u) at the same temperature.
What is the ratio (r.m.s. speed of the hydrogen molecules) ÷ (r.m.s. speed of the oxygen molecules)?

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AnswerMarkNotes
At the same temperature the mean kinetic energy ½m⟨c2⟩ = 3/2kT is the same, so crms ∝ 1/√m. Ratio = √(32 ÷ 2.0) = 4.0.✓ 1Answer C

Answer: C

Exam tip: Same temperature means the same mean kinetic energy, not the same speed. The lighter molecules move faster.

13IALP-U5-2-A-06 Unit 5 · A2Medium1 mark

The r.m.s. speed of the molecules of a gas is 480 m s−1 at a temperature of 300 K.
At what temperature is the r.m.s. speed 960 m s−1?

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AnswerMarkNotes
crms = √(3kT/m), so T ∝ crms2. Doubling the r.m.s. speed needs 22 = 4 times the temperature: 4 × 300 = 1200 K.✓ 1Answer D

Answer: D

Common error: Doubling the temperature to double the speed. The temperature is proportional to the mean square speed, so the speed only rises as √T.

14IALP-U5-2-A-08 Unit 5 · A2Medium1 mark

In an experiment on a fixed mass of gas at constant temperature, the volume of air trapped in a uniform tube is found from V = πd2l/4. The internal diameter is d = 1.00 ± 0.02 cm and the length of the air column is l = 25.0 ± 0.1 cm.
What is the percentage uncertainty in V?

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AnswerMarkNotes
%d = 0.02 ÷ 1.00 × 100 = 2.0%, which is doubled because d is squared: 4.0%. %l = 0.1 ÷ 25.0 × 100 = 0.4%. %V = 4.0 + 0.4 = 4.4%.✓ 1Answer D

Answer: D

Exam tip: When quantities are multiplied, add their percentage uncertainties; a quantity raised to a power has its percentage uncertainty multiplied by that power.

15IALP-U5-2-B-03 Unit 5 · A2Medium6 marks

A weather balloon is filled with helium at ground level, where the pressure is 101 kPa and the temperature is 15 °C. The volume of the balloon at ground level is 4.0 m3. k = 1.38 × 10−23 J K−1.

(a)

Calculate the number of helium atoms in the balloon.

(2)
(b)

The balloon rises to a height where the pressure is 25 kPa and the temperature is −45 °C. The balloon fabric is slack, so the pressure inside equals the pressure outside. Calculate the new volume of the balloon.

(3)
(c)

State one assumption you made in your calculation in (b).

(1)
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AnswerMarkNotes
Part (a)
N = pV/kT = (101 × 103 × 4.0) ÷ (1.38 × 10−23 × 288)✓ 1
= 1.0 × 1026 (1.02 × 1026)✓ 1
Part (b)
T2 = −45 + 273 = 228 K✓ 1
V2 = V1 × (p1/p2) × (T2/T1) = 4.0 × (101 ÷ 25) × (228 ÷ 288)✓ 1
= 13 m3 (12.8 m3)✓ 1
Part (c)
no helium escapes (the number of atoms is constant) / helium behaves as an ideal gas✓ 1

Exam tip: Fall in pressure and fall in temperature act in opposite directions on the volume. Check that your answer is larger than 4.0 m3 but smaller than 4.0 × 101 ÷ 25.

16IALP-U5-2-B-04 Unit 5 · A2Medium5 marks

A steel cylinder of volume 5.0 × 10−3 m3 contains 2.4 mol of argon at 20 °C. The Avogadro constant NA = 6.02 × 1023 mol−1. k = 1.38 × 10−23 J K−1. The mass of an argon atom is 40 u, where 1 u = 1.66 × 10−27 kg.

(a)

Calculate the pressure of the argon.

(2)
(b)

Argon is used from the cylinder until the pressure has fallen to 4.0 × 105 Pa, with the temperature unchanged. Calculate the number of argon atoms that have been used.

(2)
(c)

Calculate the mass of argon that has been used.

(1)
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AnswerMarkNotes
Part (a)
N = 2.4 × 6.02 × 1023 = 1.44 × 1024✓ 1
p = NkT/V = 1.44 × 1024 × 1.38 × 10−23 × 293 ÷ 5.0 × 10−3 = 1.2 × 106 Pa (1.17 × 106 Pa)✓ 1
Part (b)
at constant V and T, N ∝ p: atoms remaining = 1.44 × 1024 × 4.0 × 105 ÷ 1.17 × 106 = 4.9 × 1023✓ 1
atoms used = 1.44 × 1024 − 4.9 × 1023 = 9.5 × 1023✓ 1
Part (c)
9.5 × 1023 × 40 × 1.66 × 10−27 = 0.063 kg (allow error carried forward)✓ 1

Exam tip: In pV = NkT, N is a number of molecules (here atoms). Convert moles to a number with N = n × NA before you substitute.

17IALP-U5-2-B-05 Unit 5 · A2Medium7 marks

A student investigates how the pressure of a fixed mass of air depends on its volume. The air is trapped in a gas syringe connected by a very short tube to a pressure sensor and data logger. She pushes the plunger in, in steps, and records the volume from the syringe scale and the pressure. The room temperature is 21 °C. Her results are:
V / cm3: 60, 50, 40, 30, 24, 20 — p / kPa: 101, 121, 152, 202, 252, 303.
k = 1.38 × 10−23 J K−1.

010203040506070050100150200250300V / cm³p / kPa
Figure: the student's readings of pressure against volume
(a)

Explain why she should wait for about a minute after each movement of the plunger before recording the pressure.

(2)
(b)

Show that these results are consistent with Boyle's law.

(2)
(c)

Calculate the number of air molecules in the syringe.

(2)
(d)

Suggest one advantage of using a pressure sensor and data logger rather than a dial gauge.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
compressing the air does work on it, so its temperature rises✓ 1
waiting allows the air to return to room temperature, so the temperature is the same (constant) for every reading✓ 1
Part (b)
calculates pV for at least three readings, e.g. 101 × 60 = 6060, 152 × 40 = 6080, 303 × 20 = 6060 (kPa cm3)✓ 1
pV is constant (within about 1%), so p ∝ 1/V✓ 1
Part (c)
pV = 101 × 103 × 60 × 10−6 = 6.06 J and T = 294 K✓ 1
N = 6.06 ÷ (1.38 × 10−23 × 294) = 1.5 × 1021 (1.49 × 1021)✓ 1
Part (d)
higher resolution / no parallax error in reading a pointer / readings recorded automatically at a chosen moment✓ 1

Exam tip: To test an inverse proportion from a table, calculate the product of the two variables for several rows. If the products agree within the precision of the data, the relationship holds.

18IALP-U5-2-B-07 Unit 5 · A2Medium6 marks

Air breathed out of the lungs is at body temperature, 37 °C. An oxygen molecule has a mass of 32 u, where 1 u = 1.66 × 10−27 kg. k = 1.38 × 10−23 J K−1.

(a)

Explain what is meant by the root-mean-square (r.m.s.) speed of the molecules of a gas.

(1)
(b)

Calculate the r.m.s. speed of the oxygen molecules in the exhaled air.

(3)
(c)

Calculate the mean kinetic energy of an oxygen molecule in the exhaled air.

(1)
(d)

Exhaled air also contains carbon dioxide molecules, of mass 44 u. State how the r.m.s. speed of the carbon dioxide molecules compares with that of the oxygen molecules, and why.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
the square root of the mean of the squares of the speeds of all the molecules✓ 1
Part (b)
T = 310 K and m = 32 × 1.66 × 10−27 = 5.31 × 10−26 kg✓ 1
crms = √(3kT/m) = √(3 × 1.38 × 10−23 × 310 ÷ 5.31 × 10−26)✓ 1
= 490 m s−1 (492 m s−1)✓ 1
Part (c)
3/2kT = 1.5 × 1.38 × 10−23 × 310 = 6.4 × 10−21 J (6.42 × 10−21 J)✓ 1
Part (d)
lower, because at the same temperature they have the same mean kinetic energy but a larger mass✓ 1

Common error: Using the mass in u, or in grams, in √(3kT/m). The mass must be the mass of one molecule in kg.

19IALP-U5-2-B-10 Unit 5 · A2Medium5 marks

A sealed glass lamp contains xenon gas. When the lamp is off the xenon is at 25 °C; when it is running the xenon reaches 600 °C. The mass of a xenon atom is 131 u, where 1 u = 1.66 × 10−27 kg. k = 1.38 × 10−23 J K−1.

(a)

Calculate the ratio (mean kinetic energy of a xenon atom when the lamp is running) ÷ (mean kinetic energy when it is off).

(1)
(b)

Calculate the r.m.s. speed of the xenon atoms when the lamp is running.

(2)
(c)

The volume of the lamp does not change. State and explain the factor by which the pressure of the xenon increases.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
873 K ÷ 298 K = 2.93✓ 1
Part (b)
m = 131 × 1.66 × 10−27 = 2.17 × 10−25 kg; crms = √(3 × 1.38 × 10−23 × 873 ÷ 2.17 × 10−25)✓ 1
= 408 m s−1✓ 1
Part (c)
2.93 (the same factor as the mean kinetic energy)✓ 1
with N and V constant, pV = NkT gives p ∝ T (in kelvin)✓ 1

Common error: Taking the ratio of Celsius temperatures (600 ÷ 25 = 24). Mean kinetic energy and pressure at constant volume are proportional to the kelvin temperature.

20IALP-U5-2-FA-01 Unit 5 · A2Medium1 mark

The bar charts show the number of molecules N and the Celsius temperature θ of four samples, W, X, Y and Z, of the same ideal gas.
Which sample has the greatest internal energy?

0 1 2 3 4 5 N / 10²³ −100 0 100 200 300 θ / °C 1.0 W 2.0 X 4.0 Y 3.0 Z 300 W 160 X −60 Y 27 Z sample sample
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AnswerMarkNotes
For an ideal gas the internal energy is the total random kinetic energy of the molecules, N × 3/2kT, so it is proportional to NT with T in kelvin. NT (in units of 1023 K): W 1.0 × 573 = 573; X 2.0 × 433 = 866; Y 4.0 × 213 = 852; Z 3.0 × 300 = 900. The greatest is Z.✓ 1Answer D

Answer: D

Common error: Comparing temperatures in °C. The mean kinetic energy of a molecule is proportional to the kelvin temperature, so convert every temperature before multiplying by the number of molecules.

21IALP-U5-2-FA-03 Unit 5 · A2Medium1 mark

The graph shows how the pressure p of a fixed mass of ideal gas varies with its volume V at two different constant temperatures. The lower curve is for a temperature of 300 K.
What is the temperature T of the upper curve?

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 0 50 100 150 200 250 300 350 400 p / kPa V / 10⁻³ m³ P Q 300 K T
Show mark scheme
AnswerMarkNotes
For a fixed mass, pV ∝ T. At P: pV = 100 × 103 × 3.0 × 10−3 = 300 J (at 300 K). At Q: pV = 300 × 103 × 1.5 × 10−3 = 450 J. T = 300 × 450 ÷ 300 = 450 K.✓ 1Answer C

Answer: C

Exam tip: Points on different isotherms usually have different volumes and pressures. Compare the product pV, not p or V alone.

22IALP-U5-2-FA-04 Unit 5 · A2Medium1 mark

A fixed mass of gas is trapped in a vertical cylinder of uniform cross-section by a piston that moves freely without friction, so the pressure of the gas stays constant. The cylinder stands in a water bath. The diagram shows the apparatus when the water is at 17 °C; the ruler measures the length of the gas column.
The water is heated to 107 °C and the gas reaches this temperature. What is the new length of the gas column?

water piston gas 0 2 4 6 8 10 12 14 16 18 20 cm thermometer 17 °C
Show mark scheme
AnswerMarkNotes
Read the length: 12.0 cm. At constant pressure V ∝ T, and V ∝ length for a uniform cylinder. Length = 12.0 × (107 + 273) ÷ (17 + 273) = 12.0 × 380 ÷ 290 = 15.7 cm.✓ 1Answer B

Answer: B

Common error: Using the ratio of Celsius temperatures, 107 ÷ 17. Charles's law needs kelvin temperatures.

23IALP-U5-2-FA-06 Unit 5 · A2Medium1 mark

The bar chart shows the speeds of the ten molecules in a model of a gas.
What is the root-mean-square (r.m.s.) speed of the molecules?

0 1 2 3 4 5 number of molecules speed / m s⁻¹ 100 200 300 400
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AnswerMarkNotes
⟨c2⟩ = (2 × 1002 + 4 × 2002 + 3 × 3002 + 1 × 4002) ÷ 10 = 610 000 ÷ 10 = 61 000 m2 s−2. crms = √61 000 = 247 m s−1.✓ 1Answer A

Answer: A

Exam tip: r.m.s. means: square every speed, take the mean over all the molecules, then take the square root. Each speed counts once for every molecule that has it.

24IALP-U5-2-FA-09 Unit 5 · A2Medium1 mark

The graph shows how the product pV varies with the kelvin temperature T for two samples, X and Y, of an ideal gas.
What is the ratio (number of molecules in X) ÷ (number of molecules in Y)?

0 100 200 300 400 500 0 2 4 6 8 10 pV / J T / K X Y
Show mark scheme
AnswerMarkNotes
pV = NkT, so the gradient of each line is Nk. Gradient of X = 6.0 ÷ 300 = 0.020 J K−1; gradient of Y = 2.0 ÷ 400 = 0.0050 J K−1. Ratio = 0.020 ÷ 0.0050 = 4.0.✓ 1Answer D

Answer: D

Exam tip: On a graph of pV against T, compare gradients, not single readings: the gradient is Nk, so it is proportional to the number of molecules.

25IALP-U5-2-FA-10 Unit 5 · A2Medium1 mark

A gas molecule of mass 4.6 × 10−26 kg travelling at 500 m s−1 collides elastically with a flat wall, as shown in the diagram.
What is the magnitude of the change in momentum of the molecule?

wall normal molecule 500 m s⁻¹ 60°
Show mark scheme
AnswerMarkNotes
Only the component of velocity at right angles to the wall reverses. The path is at 60° to the wall, so this component is 500 sin 60° = 433 m s−1. Δp = 2mv sin 60° = 2 × 4.6 × 10−26 × 433 = 4.0 × 10−23 N s, directed away from the wall.✓ 1Answer C

Answer: C

Common error: Using the angle to the wall as if it were the angle to the normal. Draw the normal and find the component of velocity along it before doubling.

26IALP-U5-2-FB-03 Unit 5 · A2Medium6 marks

Figure 1 shows how the pressure p and volume V of a fixed mass of ideal gas change as the gas is taken from state A to state B and then to state C. The temperature of the gas at A is 300 K. k = 1.38 × 10−23 J K−1

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 40 80 120 160 200 240 280 p / kPa V / 10⁻³ m³ A B C
(a)

Show that the temperature of the gas at B is the same as at A.

(2)
(b)

Calculate the number of molecules of gas.

(2)
(c)

Calculate the temperature of the gas at C.

(1)
(d)

Calculate the ratio (r.m.s. speed of the molecules at C) ÷ (r.m.s. speed of the molecules at A).

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
pV at A = 240 × 103 × 1.0 × 10−3 = 240 J and at B = 80 × 103 × 3.0 × 10−3 = 240 J✓ 1
pV = NkT with N constant, so equal pV means equal T (300 K)✓ 1
Part (b)
N = pV/kT = 240 ÷ (1.38 × 10−23 × 300)✓ 1
= 5.8 × 1022 (5.80 × 1022)✓ 1
Part (c)
at constant volume p ∝ T: T = 300 × 160 ÷ 80 = 6.00 × 102 K (600 K)✓ 1
Part (d)
crms ∝ √T: √(600 ÷ 300) = 1.41✓ 1

Exam tip: Read the volume axis label: 1.0 on an axis marked V / 10−3 m3 means 1.0 × 10−3 m3, and kPa must become Pa before you use pV = NkT.

27IALP-U5-2-FB-05 Unit 5 · A2Medium6 marks

Figure 1 shows how the air temperature θ varies with height above sea level in the lower atmosphere. Treat the air as an ideal gas consisting of nitrogen molecules, each of mass 28 u. 1 u = 1.66 × 10−27 kg k = 1.38 × 10−23 J K−1

−70 −60 −50 −40 −30 −20 −10 0 10 20 0 2 4 6 8 10 12 14 16 height / km air temperature θ / °C
(a)

Use Figure 1 to find the kelvin temperature of the air at a height of 10 km.

(1)
(b)

Calculate the r.m.s. speed of the molecules at a height of 10 km.

(3)
(c)

The air pressure at a height of 10 km is 26 kPa. Calculate the number of molecules in each cubic metre of air at this height.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
θ = −50 °C, so T = 223 K✓ 1
Part (b)
m = 28 × 1.66 × 10−27 = 4.65 × 10−26 kg✓ 1
crms = √(3kT/m) = √(3 × 1.38 × 10−23 × 223 ÷ 4.65 × 10−26)✓ 1
= 446 m s−1✓ 1
Part (c)
N/V = p/kT = 26 × 103 ÷ (1.38 × 10−23 × 223)✓ 1
= 8.4 × 1024 m−3 (8.45 × 1024 m−3)✓ 1

Common error: Putting the Celsius reading (−50) into √(3kT/m). A negative number under the square root is a sure sign that the temperature has not been converted to kelvin.

28IALP-U5-2-FB-06 Unit 5 · A2Medium5 marks

Figure 1 shows a hand pump used to pump air into a rigid tank of volume 5.0 × 10−3 m3. On each stroke the pump fills with air at atmospheric pressure, 1.0 × 105 Pa, and then pushes all of this air through the one-way valve into the tank. The dimensions in Figure 1 are the internal diameter of the pump barrel and the internal length swept by the piston. Assume the temperature of the air stays at 20 °C. k = 1.38 × 10−23 J K−1

piston (fully drawn out) 40.0 cm 3.0 cm one-way valve rigid tank
(a)

Calculate the number of air molecules pushed into the tank on each stroke.

(2)
(b)

The tank initially contains air at 1.0 × 105 Pa. Calculate the pressure in the tank after 20 strokes.

(2)
(c)

In practice the temperature of the air in the tank rises during pumping. Suggest why.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
volume of pump = π × (0.015)2 × 0.400 = 2.83 × 10−4 m3✓ 1
N = pV/kT = 1.0 × 105 × 2.83 × 10−4 ÷ (1.38 × 10−23 × 293) = 7.0 × 1021✓ 1
Part (b)
molecules in tank: initially 1.0 × 105 × 5.0 × 10−3 ÷ (1.38 × 10−23 × 293) = 1.24 × 1023; after 20 strokes 1.24 × 1023 + 20 × 7.0 × 1021 = 2.64 × 1023✓ 1
p = NkT/V = 2.1 × 105 Pa✓ 1
Part (c)
work is done on the air as it is compressed, which increases its internal energy (the mean kinetic energy of the molecules)✓ 1

Exam tip: With a fixed temperature, N ∝ pV. Adding molecules to a rigid container raises the pressure in proportion to the total number of molecules.

29IALP-U5-2-FB-08 Unit 5 · A2Medium7 marks

A computer model of an ideal gas calculates the mean kinetic energy of the molecules at different kelvin temperatures. Figure 1 shows the results with a line of best fit.

0 100 200 300 400 500 600 0 2 4 6 8 10 12 14 mean kinetic energy / 10⁻²¹ J T / K
(a)

Determine the gradient of the line.

(2)
(b)

The model uses ½m⟨c2⟩ = 3/2kT. Use your gradient to find the value of the Boltzmann constant k in the model.

(1)
(c)

Explain why the internal energy of the model gas is equal to the number of molecules multiplied by the mean kinetic energy of a molecule.

(1)
(d)

The model contains 2.0 × 1022 molecules. Use Figure 1 to calculate the internal energy of the model gas at 400 K.

(2)
(e)

Each model molecule has a mass of 6.6 × 10−27 kg. Calculate the r.m.s. speed of the molecules at 400 K.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
uses a large triangle, e.g. from (0, 0) to (580 K, 12.0 × 10−21 J)✓ 1
gradient = 12.0 × 10−21 ÷ 580 = 2.07 × 10−23 J K−1 (allow 2.03 to 2.11 × 10−23 J K−1)✓ 1
Part (b)
gradient = 3/2k, so k = ⅔ × 2.07 × 10−23 = 1.38 × 10−23 J K−1 (allow 1.35 to 1.41 × 10−23, consistent with the gradient)✓ 1
Part (c)
in an ideal gas there are no forces between the molecules (except in collisions), so they have no potential energy and the internal energy is the total random kinetic energy✓ 1
Part (d)
mean kinetic energy at 400 K = 8.3 × 10−21 J (allow 8.2 to 8.4)✓ 1
U = 2.0 × 1022 × 8.3 × 10−21 = 1.7 × 102 J (166 J; allow 164 to 168 J)✓ 1
Part (e)
⟨c2⟩ = 2 × 8.3 × 10−21 ÷ 6.6 × 10−27, so crms = 1.6 × 103 m s−1✓ 1

Exam tip: To find a gradient, use a triangle that covers at least half of the line and read its corners at gridline intersections; never use a single data point unless it lies on the line.

30IALP-U5-2-FB-10 Unit 5 · A2Medium6 marks

Figure 1 shows apparatus used to investigate how the pressure of a fixed mass of air depends on its volume at constant temperature. Air is trapped above oil in a glass tube of uniform cross-section. A pump changes the pressure on the oil, and the Bourdon gauge shows the pressure of the trapped air.

air oil 0 5 10 15 20 25 30 35 40 cm oil reservoir to pump 0 50 100 150 200 250 300 350 400 kPa Bourdon gauge
(a)

Record the length ℓ of the air column and the pressure p shown in Figure 1.

(2)
(b)

An earlier reading was ℓ = 35.0 cm at p = 121 kPa. Show that the two readings are consistent with Boyle's law.

(2)
(c)

Explain why the length of the air column can be used in place of its volume.

(1)
(d)

The uncertainty in judging the position of the oil surface is ±0.2 cm. Calculate the percentage uncertainty in ℓ.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
ℓ = 28.0 cm✓ 1
p = 150 kPa✓ 1
Part (b)
pℓ = 121 × 35.0 = 4.24 × 103 kPa cm and 150 × 28.0 = 4.20 × 103 kPa cm✓ 1
the products agree to within about 1%, which is within the uncertainty of the readings, so pℓ, and hence pV, is constant✓ 1
Part (c)
the tube has a uniform cross-sectional area A, so V = Aℓ and V ∝ ℓ✓ 1
Part (d)
0.2 ÷ 28.0 × 100 = 0.7% (0.71%)✓ 1

Exam tip: When two readings should give a constant product, calculate the percentage difference between the products and compare it with the percentage uncertainties in the readings.

31IALP-U5-2-A-09 Unit 5 · A2Hard1 mark

A sealed rigid container of an ideal gas is heated from 300 K to 600 K. The pressure doubles. The average momentum change of a molecule in each collision with a wall increases by a factor of √2.
By what factor does the number of collisions per second with a wall increase?

Show mark scheme
AnswerMarkNotes
Pressure ∝ (collisions per second) × (momentum change per collision). 2 = (factor for collisions) × √2, so the collision rate increases by 2 ÷ √2 = √2 = 1.4. This is consistent with crms ∝ √T: faster molecules cross the container more often.✓ 1Answer B

Answer: B

Exam tip: Heating a gas at constant volume raises the pressure for two reasons: the molecules hit the walls more often and harder. Each effect contributes a factor of √(T2/T1).

32IALP-U5-2-A-10 Unit 5 · A2Hard1 mark

Two flasks are joined by a thin tube, so the gas in them is at the same pressure. Flask X has a volume of 1.0 litre and is kept at 300 K. Flask Y has a volume of 2.0 litres and is kept at 450 K.
What fraction of the gas molecules is in flask Y?

XY1.0 litre2.0 litres300 K450 K
Figure: two flasks joined by a thin tube (not to scale)
Show mark scheme
AnswerMarkNotes
N = pV/kT with the same p, so N ∝ V/T. X: 1.0/300; Y: 2.0/450. Fraction in Y = (2.0/450) ÷ (1.0/300 + 2.0/450) = 0.57.✓ 1Answer B

Answer: B

Exam tip: When the pressure is the same, the number of molecules in each part is proportional to V/T. A hotter region holds fewer molecules per unit volume.

33IALP-U5-2-B-06 Unit 5 · A2Hard8 marks

A flask of volume 250 cm3 containing dry air is sealed, connected to a pressure gauge by a short narrow tube, and placed in a water bath. The water is heated and the pressure of the air is recorded at several temperatures:
θ / °C: 10, 30, 50, 70, 90 — p / kPa: 97.8, 104.7, 111.6, 118.5, 125.4.
k = 1.38 × 10−23 J K−1.

01020304050607080901009095100105110115120125130θ / °Cp / kPa
Figure: the pressure readings plotted against temperature
(a)

Calculate the gradient of the graph of p against θ, in kPa K−1.

(2)
(b)

Use your gradient to determine the temperature, in °C, at which the pressure would be zero.

(2)
(c)

The gradient of the graph is equal to Nk/V. Calculate the number of air molecules in the flask. The percentage uncertainty in the gradient is 2.0% and in the volume of the flask is 1.2%; calculate the percentage uncertainty in your value of N.

(3)
(d)

Suggest why the tube connecting the flask to the gauge should be short and narrow.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient = (125.4 − 97.8) ÷ (90 − 10)✓ 1
= 0.345 kPa K−1 (a change of 1 °C is a change of 1 K)✓ 1
Part (b)
θ = 10 − 97.8 ÷ 0.345✓ 1
= −273 °C (allow −274 °C)✓ 1
Part (c)
N = gradient × V/k = 345 × 2.50 × 10−4 ÷ 1.38 × 10−23✓ 1
= 6.25 × 1021✓ 1
percentage uncertainty = 2.0 + 1.2 = 3.2%✓ 1
Part (d)
the air in the tube is not at the temperature of the water bath, so its volume must be negligible compared with the flask✓ 1

Model sketch

−300−200−10010020406080100120θ / °Cp / kPa≈ −273 °C
Figure: the best-fit line extrapolated to zero pressure

Exam tip: A gradient read in kPa per °C is the same number in kPa per K, because the two scales have degrees of the same size. Only the zero is different.

34IALP-U5-2-B-08 Unit 5 · A2Hard8 marks

A gas is sealed in a rigid metal can. The can is placed in boiling water and the pressure of the gas increases.

(a)

State two assumptions made in the kinetic model of an ideal gas.

(2)
(b)

Explain, using the kinetic model of a gas, how the gas exerts a pressure on the walls of the can and why this pressure increases when the gas is heated.

(6)
Show mark scheme
AnswerMarkNotes
Part (a)
any two, one mark each: molecules are in random motion / the volume of the molecules is negligible compared with the volume of the container / there are no forces between molecules except during collisions / collisions are elastic / the duration of a collision is negligible compared with the time between collisions✓ 2
Part (b)
Indicative content, one mark for each point:—
molecules move randomly and collide with the walls✓ 1
in each collision a molecule's momentum changes (it rebounds), so the wall exerts a force on it✓ 1
by Newton's third law the molecule exerts an equal and opposite force on the wall; pressure is the total force per unit area of wall✓ 1
heating increases the mean kinetic energy, so the molecules move faster / the r.m.s. speed increases✓ 1
each collision involves a larger change of momentum✓ 1
and collisions with the walls are more frequent, since the volume is constant, so the rate of change of momentum per unit area, and the pressure, increase✓ 1

Exam tip: Build the explanation as a chain: collision → momentum change → force (Newton's second and third laws) → pressure = force/area. Then say what heating changes: harder collisions and more frequent collisions.

35IALP-U5-2-B-09 Unit 5 · A2Hard7 marks

The kinetic theory of gases models a gas as N molecules, each of mass m, moving at random in a cube of side L.

muLLface
Figure: one molecule moving at right angles to a face of the cube (not to scale)
(a)

A molecule moves with a velocity component u at right angles to one face of the cube and collides elastically with it. Show that the average force exerted on that face by this molecule is mu2/L.

(3)
(b)

State why ⟨u2⟩ = ⅓⟨c2⟩ for the molecules of the gas.

(1)
(c)

The theory leads to pV = ⅓Nm⟨c2⟩. Use this and pV = NkT to show that the mean kinetic energy of a molecule is proportional to the absolute temperature.

(1)
(d)

A sample of neon has a pressure of 2.0 × 105 Pa and a density of 1.6 kg m−3. Calculate the r.m.s. speed of its atoms.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
change of momentum in each collision = mu − (−mu) = 2mu✓ 1
time between collisions with the same face = 2L/u✓ 1
force = rate of change of momentum = 2mu ÷ (2L/u) = mu2/L✓ 1
Part (b)
the motion is random, so the mean square velocity components in the three perpendicular directions are equal, and c2 = u2 + v2 + w2✓ 1
Part (c)
⅓Nm⟨c2⟩ = NkT, so ½m⟨c2⟩ = 3/2kT, and 3/2k is a constant✓ 1
Part (d)
p = ⅓ρ⟨c2⟩, so ⟨c2⟩ = 3 × 2.0 × 105 ÷ 1.6 = 3.75 × 105 m2 s−2✓ 1
crms = 610 m s−1 (612 m s−1)✓ 1

Exam tip: In a “show that” derivation, write each physical step as a separate line with its reason: momentum change, time between collisions, then force as rate of change of momentum.

36IALP-U5-2-FA-05 Unit 5 · A2Hard1 mark

A student traps a fixed mass of air at 17 °C and measures its pressure p at different volumes V, keeping the temperature constant. The graph shows p against 1/V, with a line of best fit through the origin. P is a point on the line. k = 1.38 × 10−23 J K−1
How many air molecules are trapped?

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0 20 40 60 80 100 120 140 160 180 200 p / kPa (1/V) / cm⁻³ P
Show mark scheme
AnswerMarkNotes
Gradient = pV = NkT. From P: gradient = 150 kPa ÷ 0.050 cm−3 = 3000 kPa cm3 = 3.0 × 103 × 103 Pa × 10−6 m3 = 3.0 J. N = 3.0 ÷ (1.38 × 10−23 × 290) = 7.5 × 1020.✓ 1Answer B

Answer: B

Exam tip: Convert a gradient's units before using it: 1 kPa cm3 = 103 Pa × 10−6 m3 = 10−3 J.

37IALP-U5-2-FA-07 Unit 5 · A2Hard1 mark

The graph shows how the mean square speed ⟨c2⟩ of the molecules of a gas varies with the kelvin temperature T. P is a point on the line. k = 1.38 × 10−23 J K−1 1 u = 1.66 × 10−27 kg
What is the mass of one molecule of the gas?

0 100 200 300 400 500 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 <c²> / 10⁶ m² s⁻² T / K P
Show mark scheme
AnswerMarkNotes
½m⟨c2⟩ = 3/2kT, so m = 3kT/⟨c2⟩ = 3 × 1.38 × 10−23 × 400 ÷ 2.5 × 106 = 6.62 × 10−27 kg. In u: 6.62 × 10−27 ÷ 1.66 × 10−27 = 4.0 u.✓ 1Answer B

Answer: B

Exam tip: Read the axis label carefully: the graph shows ⟨c2⟩ in units of 106 m2 s−2, so the reading at P is 2.5 × 106 m2 s−2 and needs no squaring.

38IALP-U5-2-FB-04 Unit 5 · A2Hard7 marks

A student investigates how the pressure p of a fixed mass of air depends on its volume V at a constant temperature of 25 °C. She plots lg (p / kPa) against lg (V / cm3), as shown in Figure 1, and draws a line of best fit. k = 1.38 × 10−23 J K−1

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 lg (p / kPa) lg (V / cm³)
(a)

Determine the gradient of the line.

(2)
(b)

Explain how the gradient shows that the results obey Boyle's law.

(1)
(c)

Use the line to show that pV is about 3.2 × 103 kPa cm3.

(2)
(d)

Calculate the number of air molecules trapped.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
uses two well-separated points on the line, e.g. (1.00, 2.50) and (2.00, 1.50)✓ 1
gradient = (1.50 − 2.50) ÷ (2.00 − 1.00) = −1.0 (allow −0.98 to −1.02)✓ 1
Part (b)
if p = AVn then lg p = n lg V + lg A, so the gradient is n; n = −1 means p ∝ 1/V (pV constant)✓ 1
Part (c)
on the line lg p + lg V = 3.50 (e.g. at lg V = 1.50, lg p = 2.00), so lg (pV) = 3.50✓ 1
pV = 103.50 = 3.16 × 103 kPa cm3✓ 1
Part (d)
pV = 3.16 × 103 × 103 Pa × 10−6 m3 = 3.16 J and T = 298 K✓ 1
N = 3.16 ÷ (1.38 × 10−23 × 298) = 7.7 × 1020✓ 1

Exam tip: A log–log graph turns p = AVn into a straight line: the gradient gives the power n, and any point on the line gives the constant.

39IALP-U5-2-FB-07 Unit 5 · A2Hard7 marks

A sealed metal can of volume 4.0 × 10−4 m3 contains a gas. The can is heated slowly and the pressure p of the gas is recorded at different kelvin temperatures T. At 400 K a safety valve opens briefly, lets some gas escape and closes again. Figure 1 shows the results. The volume of the can does not change. k = 1.38 × 10−23 J K−1

0 100 200 300 400 500 0 50 100 150 200 250 300 350 400 p / kPa T / K valve opens
(a)

Show that, between 280 K and 400 K, the pressure of the gas is proportional to its kelvin temperature.

(2)
(b)

Calculate the number of gas molecules in the can before the valve opens.

(2)
(c)

Calculate the fraction of the molecules that escaped through the valve.

(2)
(d)

Explain, in terms of the molecules, why the pressure fell when the gas escaped, although the temperature did not change.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
p/T at two points, e.g. 240 ÷ 300 = 0.80 kPa K−1 and 320 ÷ 400 = 0.80 kPa K−1✓ 1
p/T is constant (the line extends through the origin), so p ∝ T✓ 1
Part (b)
N = pV/kT = 320 × 103 × 4.0 × 10−4 ÷ (1.38 × 10−23 × 400)✓ 1
= 2.3 × 1022 (2.32 × 1022)✓ 1
Part (c)
after the valve closes, p/T = 300 ÷ 500 = 0.60 kPa K−1; at constant volume N ∝ p/T✓ 1
fraction remaining = 0.60 ÷ 0.80 = 0.75, so fraction escaped = 0.25✓ 1
Part (d)
the mean kinetic energy (speed) of the molecules is unchanged, but there are fewer molecules in the same volume, so there are fewer collisions with each unit area of wall per second and a smaller rate of change of momentum (force) per unit area✓ 1

Exam tip: At constant volume, pV = NkT gives p/T = Nk/V. The gradient of a p–T line through the origin is a direct measure of the number of molecules.

40IALP-U5-2-FB-09 Unit 5 · A2Hard8 marks

Figure 1 shows a gas in a cylinder of uniform cross-section fitted with a pressure gauge, before and after the piston is pushed in slowly. The temperature of the gas stays at 20 °C throughout and no gas escapes. Treat the gas as ideal.

before 110 kPa 0 5 10 15 20 25 30 cm after (slow compression) ? 0 5 10 15 20 25 30 cm
(a)

Use Figure 1 to determine the pressure of the gas after the compression.

(1)
(b)

Explain, using the kinetic model of a gas, why the pressure of the gas changes by this factor.

(6)
(c)

The piston is now locked in place and the gas is heated to 80 °C. Calculate the new pressure of the gas.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
the length of the gas column, and so the volume, halves (30.0 cm → 15.0 cm) at constant temperature, so p = 2 × 110 = 220 kPa✓ 1
Part (b)
Indicative content, one mark for each point:—
the temperature is constant, so the mean kinetic energy and the r.m.s. speed of the molecules are unchanged✓ 1
so the average change of momentum in each collision with a wall is unchanged✓ 1
the same number of molecules now occupies half the volume, so the number of molecules per unit volume doubles✓ 1
so the number of collisions per second with each unit area of wall doubles✓ 1
the rate of change of momentum, and so the force, on each unit area of wall doubles✓ 1
pressure = force ÷ area, so the pressure doubles: p ∝ 1/V, Boyle's law✓ 1
Part (c)
at constant volume p ∝ T: p = 220 × 353 ÷ 293 = 265 kPa✓ 1

Exam tip: In a kinetic-theory explanation, say what stays the same (speed, momentum change per collision) as well as what changes (collision rate per unit area). Both are needed to get the factor right.