Kinetic theory of gases — Paper 4 · A Level Physics 9702

2 The air in a car tyre has a constant volume of 3.1 ×10–2m3. The pressure of this air is 2.9 ×105Pa at a temperature of 17 °C. The air may be considered to be an ideal gas. (a) State what is meant by an ideal gas. … … … [2] (b) Calculate the amount of air, in mol, in the tyre. amount = ……………………………. mol [2] (c) The pressure in the tyre is to be increased using a pump. On each stroke of the pump, 0.012 mol of air is forced into the tyre. Calculate the number of strokes of the pump required to increase the pressure to 3.4 ×105Pa at a temperature of 27 °C. number = ……………………………. [3]

7 marks

Mark scheme: 2 (a) obeys the law pV = constant × T ………………………..……………….. M1 at all values of p, V and T ………………………………………………. A1 [2] (b) n = (2.9 × 105 × 3.1 × 10–2) / (8.31 × 290) …..………………..………... C1 = 3.73 mol ………………………………………………………………. A1 [2] 3 . 4 290 (c) at new pressure, nn = 3.73 × × 2 . 9 300 = 4.23 mol ….………………………………………. C1 change = 0.50 mol ……………………………………………………….… C1 number of strokes = 0.50 / 0.012 = 42 (must round up for mark) ……. A1 [3]

This question in 9702/41 Oct/Nov 2005

2 (a) The equation pV = constant × T relates the pressure p and volume V of a gas to its kelvin (thermodynamic) temperature T. State two conditions for the equation to be valid. 1. … … 2. … … [2] (b) A gas cylinder contains 4.00 ×104cm3 of hydrogen at a pressure of 2.50 ×107Pa and a temperature of 290 K. The cylinder is to be used to fill balloons. Each balloon, when filled, contains 7.24 ×103cm3 of hydrogen at a pressure of 1.85 ×105Pa and a temperature of 290 K. Calculate, assuming that the hydrogen obeys the equation in (a), (i) the total amount of hydrogen in the cylinder, amount = ……………………….. mol [3] (ii) the number of balloons that can be filled from the cylinder. number = ……………………….. [3]

8 marks

Mark scheme: 2 (a) e.g. fixed mass/ amount of gas ideal gas (any two, 1 each) B2 [2] (b) (i) n = pV / RT C1 = (2.5 × 107 × 4.00 × 104 x 10-6) / (8.31 × 290) C1 = 415 mol A1 [3] (ii) volume of gas at 1.85 × 105 Pa = (2.5 × 107 × 4.00 × 104) / (1.85 × 105) = 5.41 × 106 cm3 C1 so, 5.41 × 106 = 4.00 × 104 + 7.24 × 103 N C1 N = 741 A1 [3] (answer 740 or fails to allow for gas in cylinder, max 2/3)

This question in 9702/41 May/June 2006

2 (a) An amount of 1.00 mol of Helium-4 gas is contained in a cylinder at a pressure of 1.02 × 105 Pa and a temperature of 27 °C. (i) Calculate the volume of gas in the cylinder. volume = … m3 [2] (ii) Hence show that the average separation of gas atoms in the cylinder is approximately 3.4 × 10–9 m. [2] (b) Calculate (i) the gravitational force between two Helium-4 atoms that are separated by a distance of 3.4 × 10–9 m, force = … N [3] Examiner’s Use (ii) the ratio weight of a Helium-4 atom . gravitational force between two Helium-4 atoms with separation 3.4 × 10–9 m ratio = … [2] (c) Comment on your answer to (b)(ii) with reference to one of the assumptions of the kinetic theory of gases. … … … [2]

11 marks

Mark scheme: 2 (a) (i) pV = nRT V = (8.31 × 300)/(1.02 × 105) … C1 = 0.0244 m3 (if uses Celsius, then 0/2) … A1 [2] (ii) volume occupied by one atom = 0.0244 / (6.02 × 1023) = 4.06 × 10-26 m3 … M1 separation ≈ 3√(4.06 × 10-26) … A1 = 3.44 × 10-9 m … A0 [2] (b) (i) F = GMm / r2 … C1 = (6.67 × 10-11 × {4 × 1.66 × 10-27}2) / (3.44 × 10-9)2 … C1 = 2.49 × 10-46 N … A1 [3] (ii) ratio = (4 × 1.66 × 10-27 × 9.8) / 2.49 × 10-46 … C1 = 2.6 × 1020 … A1 [2] (c) assumption that forces between atoms are negligible … B1 comment e.g. ratio shows gravitational force to be very small e.g. force is very much less than weight e.g. if there are forces, they are not gravitational … B1 [2] GCE A/AS LEVEL – October/November 2007 9702 04

This question in 9702/41 Oct/Nov 2007

2 (a) Explain qualitatively how molecular movement causes the pressure exerted by a gas. For Examiner’s … Use … … … [3] (b) The density of neon gas at a temperature of 273 K and a pressure of 1.02 × 105 Pa is 0.900 kg m–3. Neon may be assumed to be an ideal gas. Calculate the root-mean-square (r.m.s.) speed of neon atoms at (i) 273 K, speed = … m s–1 [3] (ii) 546 K. speed = … m s–1 [2] (c) The calculations in (b) are based on the density for neon being 0.900 kg m–3. For Suggest the effect, if any, on the root-mean-square speed of changing the density at Examiner’s constant temperature. Use … … … [2]

10 marks

Mark scheme: 2 (a) molecule(s) rebound from wall of vessel / hits walls B1 change in momentum gives rise to impulse / force B1 either (many impulses) averaged to give constant force / pressure or the molecules are in random motion B1 [3] (b) (i) p = 1 ρ<c2> C1 3 1.02 × 105 = 1 × 0.900 × <c2> 3 <c2> = 3.4 × 105 C1 cRMS = 580 m s–1 A1 [3] (ii) either <c2> ∝ T or <c2> = 2 × 3.4 ×105 C1 cRMS = 830 m s–1 (allow 820) A1 [2] (c) cRMS depends on temperature (alone) B1 so no effect B1 [2] GCE A/AS LEVEL – May/June 2008 9702 04

This question in 9702/41 May/June 2008

4 (a) Write down an equation to represent the first law of thermodynamics in terms of the For heating q of a system, the work w done on the system and the increase ∆U in the Examiner’s internal energy. Use … [1] (b) The pressure of an ideal gas is decreased at constant temperature. Explain what change, if any, occurs in the internal energy of the gas. … … … … [3]

4 marks

Mark scheme: 4 (a) ∆U = q + w (allow correct word equation) B1 [1] (b) either kinetic energy constant because temperature constant M1 potential energy constant because no intermolecular forces M1 so no change in internal energy A1 [3] or kinetic energy and potential energy both constant (M1) so no change in internal energy (A1) reason for either constant k.e. or constant p.e. given (A1)

This question in 9702/41 Oct/Nov 2008

5 Two deuterium (21H) nuclei are travelling directly towards one another. When their separation For is large compared with their diameters, they each have speed v as illustrated in Fig. 5.1. Examiner’s Use v v deuterium deuterium nucleus nucleus Fig. 5.1 The diameter of a deuterium nucleus is 1.1 × 10–14 m. (a) Use energy considerations to show that the initial speed v of the deuterium nuclei must be approximately 2.5 × 106 m s–1 in order that they may come into contact. Explain your working. [3] (b) For a fusion reaction to occur, the deuterium nuclei must come into contact. Assuming that deuterium behaves as an ideal gas, deduce a value for the temperature of the deuterium such that the nuclei have an r.m.s. speed equal to the speed calculated in (a). temperature = … K [4] (c) Comment on your answer to (b). … … [1]

8 marks

Mark scheme: 5 (a) change/loss in kinetic energy = change/gain in electric potential energy B1 2 × ½mv2 = q2 / 4πε0r C1 2 × ½ × 2 × 1.67 × 10–27 × v2 = (1.6 × 10–19)2 / (4π × 8.85 × 10–12 × 1.1 × 10–14) M1 v = 2.5 × 106 m s–1 A0 [3] (b) pV = ½Nm<c2> and pV = NkT C1 ½ m<c2> = 3 2 kT (award 1 mark of first two if <c2> not used) C1 ½ × 2 × 1.67 × 10–27 × (2.5 × 106)2 = 3 2 × 1.38 × 10–23 × T C1 T = 5 × 108 K A1 [4] (c) e.g. this is very high temperature temperature found in stars (any sensible comment, 1 mark) (if T < 106 K, should comment that too low for fusion to occur) B1 [1]

This question in 9702/41 Oct/Nov 2008

2 An ideal gas occupies a container of volume 4.5 × 103 cm3 at a pressure of 2.5 × 105 Pa and For a temperature of 290 K. Examiner’s Use (a) Show that the number of atoms of gas in the container is 2.8 × 1023. [2] (b) Atoms of a real gas each have a diameter of 1.2 × 10–10 m. (i) Estimate the volume occupied by 2.8 × 1023 atoms of this gas. volume = … m3 [2] (ii) By reference to your answer in (i), suggest whether the real gas does approximate to an ideal gas. … … … [2]

6 marks

Mark scheme: 2 (a) either pV = NkT or pV = nRT and n = N / NA … C1 clear correct substitution e.g. 2.5 × 105 × 4.5 × 103 × 10-6 = N × 1.38 × 10-23 × 290 … M1 N = 2.8 × 1023 … A0 [2] (allow 1 mark for calculation of n = 0.467 mol) 4 (b) (i) volume = (1.2 × 10-10)3 × 2.8 × 1023 or πr3 × 2.8 × 1023 … C1 3 = 4.8 × 10-7 m3 2.53 × 10-7 m3 … A1 [2] (ii) either 4.5 × 103 cm3 >> 0.48 cm3 or ratio of volumes is about 10-4 … B1 justified because volume of molecules is negligible … B1 [2] [Total: 6] GCE A/AS LEVEL – October/November 2009 9702 41

This question in 9702/41 Oct/Nov 2009

2 (a) Some gas, initially at a temperature of 27.2 °C, is heated so that its temperature rises For to 38.8 °C. Examiner’s Calculate, in kelvin, to an appropriate number of decimal places, Use (i) the initial temperature of the gas, initial temperature = … K [2] (ii) the rise in temperature. rise in temperature = … K [1] (b) The pressure p of an ideal gas is given by the expression 1 2 p = 3ρ c where ρ is the density of the gas. (i) State the meaning of the symbol c 2 . … … [1] (ii) Use the expression to show that the mean kinetic energy <EK> of the atoms of an ideal gas is given by the expression 3 <EK> = 2 kT. Explain any symbols that you use. … … … … … [4] (c) Helium-4 may be assumed to behave as an ideal gas. For A cylinder has a constant volume of 7.8 × 103 cm3 and contains helium-4 gas at a Examiner’s pressure of 2.1 × 107 Pa and at a temperature of 290 K. Use Calculate, for the helium gas, (i) the amount of gas, amount = … mol [2] (ii) the mean kinetic energy of the atoms, mean kinetic energy = … J [2] (iii) the total internal energy. internal energy = … J [3]

15 marks

Mark scheme: 2 (a) (i) 27.2 + 273.15 or 27.2 + 273.2 C1 300.4 K A1 [2] (ii) 11.6 K A1 [1] (b) (i) (<c2> is the) mean / average square speed B1 [1] (ii) ρ = Nm/V with N explained B1 so, pV = 1/3 Nm<c2> B1 and pV = NkT with k explained B1 so mean kinetic energy / <EK> = ½m<c2> = 3/2 kT B1 [4] (c) (i) pV = nRT 2.1 × 107 × 7.8 × 10–3 = n × 8.3 × 290 C1 n = 68 mol A1 [2] (ii) mean kinetic energy = 3/2 kT = 3/2 × 1.38 × 10–23 × 290 C1 = 6.0 × 10–21 J A1 [2] (iii) realisation that total internal energy is the total kinetic energy C1 energy = 6.0 × 10–21 × 68 × 6.02 × 1023 C1 = 2.46 × 105 J A1 [3]

This question in 9702/41 May/June 2010

2 (a) (i) State the basic assumption of the kinetic theory of gases that leads to the conclusion For that the potential energy between the atoms of an ideal gas is zero. Examiner’s Use … … [1] (ii) State what is meant by the internal energy of a substance. … … … [2] (iii) Explain why an increase in internal energy of an ideal gas is directly related to a rise in temperature of the gas. … … … [2] (b) A fixed mass of an ideal gas undergoes a cycle PQRP of changes as shown in Fig. 2.1. 10 P 8 volume / 10–4 m3 6 4 2 Q R 0 0 5 10 15 20 25 30 pressure / 105 Pa Fig. 2.1 (i) State the change in internal energy of the gas during one complete cycle PQRP. For Examiner’s change = … J [1] Use (ii) Calculate the work done on the gas during the change from P to Q. work done = … J [2] (iii) Some energy changes during the cycle PQRP are shown in Fig. 2.2. work done on gas heating supplied increase in change / J to gas / J internal energy / J P Q … –600 … Q R 0 +720 … R P … +480 … Fig. 2.2 Complete Fig. 2.2 to show all of the energy changes. [3]

11 marks

Mark scheme: 2 (a) (i) no forces (of attraction or repulsion) between atoms / molecules / particles B1 [1] (ii) sum of kinetic and potential energy of atoms / molecules M1 due to random motion A1 [2] (iii) (random) kinetic energy increases with temperature M1 no potential energy (so increase in temperature increases internal energy) A1 [2] (b) (i) zero A1 [1] (ii) work done = p∆V C1 = 4.0 × 105 × 6 × 10–4 = 240 J (ignore any sign) A1 [2] (iii) change work done / J heating / J increase in internal energy / J P → Q +240 –600 –360 Q → R 0 +720 +720 R → P –840 +480 –360 (correct signs essential) (each horizontal line correct, 1 mark – max 3) B3 [3] GCE AS/A LEVEL – October/November 2010 9702 41

This question in 9702/41 Oct/Nov 2010

2 (a) (i) State the basic assumption of the kinetic theory of gases that leads to the conclusion For that the potential energy between the atoms of an ideal gas is zero. Examiner’s Use … … [1] (ii) State what is meant by the internal energy of a substance. … … … [2] (iii) Explain why an increase in internal energy of an ideal gas is directly related to a rise in temperature of the gas. … … … [2] (b) A fixed mass of an ideal gas undergoes a cycle PQRP of changes as shown in Fig. 2.1. 10 P 8 volume / 10–4 m3 6 4 2 Q R 0 0 5 10 15 20 25 30 pressure / 105 Pa Fig. 2.1 (i) State the change in internal energy of the gas during one complete cycle PQRP. For Examiner’s change = … J [1] Use (ii) Calculate the work done on the gas during the change from P to Q. work done = … J [2] (iii) Some energy changes during the cycle PQRP are shown in Fig. 2.2. work done on gas heating supplied increase in change / J to gas / J internal energy / J P Q … –600 … Q R 0 +720 … R P … +480 … Fig. 2.2 Complete Fig. 2.2 to show all of the energy changes. [3]

11 marks

Mark scheme: 2 (a) (i) no forces (of attraction or repulsion) between atoms / molecules / particles B1 [1] (ii) sum of kinetic and potential energy of atoms / molecules M1 due to random motion A1 [2] (iii) (random) kinetic energy increases with temperature M1 no potential energy (so increase in temperature increases internal energy) A1 [2] (b) (i) zero A1 [1] (ii) work done = p∆V C1 = 4.0 × 105 × 6 × 10–4 = 240 J (ignore any sign) A1 [2] (iii) change work done / J heating / J increase in internal energy / J P → Q +240 –600 –360 Q → R 0 +720 +720 R → P –840 +480 –360 (correct signs essential) (each horizontal line correct, 1 mark – max 3) B3 [3] GCE AS/A LEVEL – October/November 2010 9702 42

This question in 9702/42 Oct/Nov 2010

2 (a) State the basic assumptions of the kinetic theory of gases. For Examiner’s … Use … … … … … [4] (b) Use equations for the pressure of an ideal gas to deduce that the average translational kinetic energy <EK> of a molecule of an ideal gas is given by the expression 3 R <EK> = T 2 NA where R is the molar gas constant, NA is the Avogadro constant and T is the thermodynamic temperature of the gas. [3] 2 (c) A deuterium nucleus 1H and a proton collide. A nuclear reaction occurs, represented by the equation 2 1 3 1H + 1p 2He + c. (i) State and explain whether the reaction represents nuclear fission or nuclear fusion. … … … [2] (ii) For the reaction to occur, the minimum total kinetic energy of the deuterium nucleus For and the proton is 2.4 × 10–14 J. Examiner’s Assuming that a sample of a mixture of deuterium nuclei and protons behaves as Use an ideal gas, calculate the temperature of the sample for this reaction to occur. temperature = … K [3] (iii) Suggest why the assumption made in (ii) may not be valid. … … [1]

13 marks

Mark scheme: 2 (a) atoms / molecules / particles behave as elastic (identical) spheres (1) volume of atoms / molecules negligible compared to volume of containing vessel (1) time of collision negligible to time between collisions (1) no forces of attraction or repulsion between atoms / molecules (1) atoms / molecules / particles are in (continuous) random motion (1) (any four, 1 each) B4 [4] (b) pV = 31 Nm<c2> and pV = nRT or pV = NkT B1 1 Nm<c2> = nRT or = NkT and <EK> = ½m<c2> B1 3 n = N/NA or k = R/NA B1 3 <EK> = × R/NA × T A0 [3] 2 (c) (i) reaction represents either build-up of nucleus from light nuclei or build-up of heavy nucleus from nuclei M1 so fusion reaction A1 [2] (ii) proton and deuterium nucleus will have equal kinetic energies B1 1.2 × 10–14 = 32 × 8.31 / (6.02 × 1023) × T C1 T = 5.8 × 108 K A1 [3] (use of E = 2.4 × 10–14 giving 1.16 × 109 K scores 1 mark) (iii) either inter-molecular / atomic / nuclear forces exist or proton and deuterium nucleus are positively charged / repel B1 [1] GCE A LEVEL – October/November 2010 9702 43

This question in 9702/43 Oct/Nov 2010

2 (a) One assumption of the kinetic theory of gases is that gas molecules behave as if they For are hard, elastic identical spheres. Examiner’s Use State two other assumptions of the kinetic theory of gases. 1. … … 2. … … [2] (b) Using the kinetic theory of gases, it can be shown that the product of the pressure p and the volume V of an ideal gas is given by the expression pV = 13Nm <c 2> where m is the mass of a gas molecule. (i) State the meaning of the symbol 1. N, … [1] 2. <c 2>. … [1] (ii) Use the expression to deduce that the mean kinetic energy <EK > of a gas molecule at temperature T is given by the equation <EK> = 32 kT where k is a constant. [2] (c) (i) State what is meant by the internal energy of a substance. For Examiner’s … Use … … [2] (ii) Use the equation in (b)(ii) to explain that, for an ideal gas, a change in internal energy ΔU is given by ΔU ∝ ΔT where ΔT is the change in temperature of the gas. … … … [2]

10 marks

Mark scheme: 2 (a) e.g. moving in random (rapid) motion of molecules/atoms/particles no intermolecular forces of attraction/repulsion volume of molecules/atoms/particles negligible compared to volume of container time of collision negligible to time between collisions (1 each, max 2) B2 [2] (b) (i) 1. number of (gas) molecules B1 [1] 2. mean square speed/velocity (of gas molecules) B1 [1] (ii) either pV = NkT or pV = nRT and links n and k and <EK> = ½m<c2> M1 3 clear algebra leading to <EK> = kT A1 [2] 2 (c) (i) sum of potential energy and kinetic energy of molecules/atoms/particles M1 reference to random (distribution) A1 [2] (ii) no intermolecular forces so no potential energy B1 (change in) internal energy is (change in) kinetic energy and this is proportional to (change in ) T B1 [2] GCE AS/A LEVEL – October/November 2011 9702 41

This question in 9702/41 Oct/Nov 2011

2 (a) One assumption of the kinetic theory of gases is that gas molecules behave as if they For are hard, elastic identical spheres. Examiner’s Use State two other assumptions of the kinetic theory of gases. 1. … … 2. … … [2] (b) Using the kinetic theory of gases, it can be shown that the product of the pressure p and the volume V of an ideal gas is given by the expression pV = 13Nm <c 2> where m is the mass of a gas molecule. (i) State the meaning of the symbol 1. N, … [1] 2. <c 2>. … [1] (ii) Use the expression to deduce that the mean kinetic energy <EK > of a gas molecule at temperature T is given by the equation <EK> = 32 kT where k is a constant. [2] (c) (i) State what is meant by the internal energy of a substance. For Examiner’s … Use … … [2] (ii) Use the equation in (b)(ii) to explain that, for an ideal gas, a change in internal energy ΔU is given by ΔU ∝ ΔT where ΔT is the change in temperature of the gas. … … … [2]

10 marks

Mark scheme: 2 (a) e.g. moving in random (rapid) motion of molecules/atoms/particles no intermolecular forces of attraction/repulsion volume of molecules/atoms/particles negligible compared to volume of container time of collision negligible to time between collisions (1 each, max 2) B2 [2] (b) (i) 1. number of (gas) molecules B1 [1] 2. mean square speed/velocity (of gas molecules) B1 [1] (ii) either pV = NkT or pV = nRT and links n and k and <EK> = ½m<c2> M1 3 clear algebra leading to <EK> = kT A1 [2] 2 (c) (i) sum of potential energy and kinetic energy of molecules/atoms/particles M1 reference to random (distribution) A1 [2] (ii) no intermolecular forces so no potential energy B1 (change in) internal energy is (change in) kinetic energy and this is proportional to (change in ) T B1 [2] GCE AS/A LEVEL – October/November 2011 9702 42

This question in 9702/42 Oct/Nov 2011

1 The planet Mars may be considered to be an isolated sphere of diameter 6.79 × 106 m with its mass of 6.42 × 1023 kg concentrated at its centre. A rock of mass 1.40 kg rests on the surface of Mars. For this rock, (a) (i) determine its weight, weight = … N [3] (ii) show that its gravitational potential energy is –1.77 × 107 J. [2] (b) Use the information in (a)(ii) to determine the speed at which the rock must leave the surface of Mars so that it will escape the gravitational attraction of the planet. speed = … m s–1 [3] (c) The mean translational kinetic energy <EK> of a molecule of an ideal gas is given by the For expression Examiner’s Use <EK> = 32kT where T is the thermodynamic temperature of the gas and k is the Boltzmann constant. (i) Determine the temperature at which the root-mean-square (r.m.s.) speed of hydrogen molecules is equal to the speed calculated in (b). Hydrogen may be assumed to be an ideal gas. A molecule of hydrogen has a mass of 2 u. temperature = … K [2] (ii) State and explain one reason why hydrogen molecules may escape from Mars at temperatures below that calculated in (i). … … … [2]

12 marks

Mark scheme: 1 (a) (i) weight = GMm/r 2 C1 = (6.67 × 10–11 × 6.42 × 1023 × 1.40)/(½ × 6.79 × 106)2 C1 = 5.20 N A1 [3] (ii) potential energy = –GMm/r C1 = –(6.67 × 10–11 × 6.42 × 1023 × 1.40)/(½ × 6.79 × 106) M1 = –1.77 × 107 J A0 [2] (b) either ½mv 2 = 1.77 × 107 C1 v 2 = (1.77 × 107 × 2)/1.40 C1 v = 5.03 × 103 m s–1 A1 or ½mv 2 = GMm/r (C1) v 2 = (2 × 6.67 x 10–11 × 6.42 × 1023)/(6.79 × 106/2) (C1) v = 5.02 × 103 m s–1 (A1) [3] 3 (c) (i) ½ × 2 × 1.66 × 10–27 × (5.03 × 103)2 = × 1.38 × 10–23 × T C1 2 T = 2030 K A1 [2] (ii) either because there is a range of speeds M1 some molecules have a higher speed A1 or some escape from point above planet surface (M1) so initial potential energy is higher (A1) [2]

This question in 9702/43 Oct/Nov 2011

2 (a) The kinetic theory of gases is based on some simplifying assumptions. For The molecules of the gas are assumed to behave as hard elastic identical spheres. Examiner’s State the assumption about ideal gas molecules based on Use (i) the nature of their movement, … … [1] (ii) their volume. … … … [2] (b) A cube of volume V contains N molecules of an ideal gas. Each molecule has a For Examiner’s component cX of velocity normal to one side S of the cube, as shown in Fig. 2.1. Use side S cx Fig. 2.1 The pressure p of the gas due to the component cX of velocity is given by the expression pV = NmcX2 where m is the mass of a molecule. Explain how the expression leads to the relation 1 pV = 3Nm<c2> where <c2> is the mean square speed of the molecules. [3] (c) The molecules of an ideal gas have a root-mean-square (r.m.s.) speed of 520 m s–1 at a temperature of 27 °C. Calculate the r.m.s. speed of the molecules at a temperature of 100 °C. r.m.s. speed = … m s–1 [3]

9 marks

Mark scheme: 2 (a) (i) either random motion or constant velocity until hits wall/other molecule B1 [1] (ii) (total) volume of molecules is negligible M1 compared to volume of containing vessel A1 or radius/diameter of a molecule is negligible (M1) compared to the average intermolecular distance (A1) [2] (b) either molecule has component of velocity in three directions or c2 = cX2 + cY2 + cZ 2 M1 random motion and averaging, so <cX2> = <cY2> = <cZ 2> M1 <c2> = 3<cX2> A1 so, pV = ⅓Nm<c2> A0 [3] (c) <c2> ∝ T or crms ∝ T C1 temperatures are 300 K and 373 K C1 crms = 580 m s–1 A1 [3] (Do not allow any marks for use of temperature in units of ºC instead of K) GCE AS/A LEVEL – May/June 2012 9702 41

This question in 9702/41 May/June 2012

2 (a) The kinetic theory of gases is based on some simplifying assumptions. For The molecules of the gas are assumed to behave as hard elastic identical spheres. Examiner’s State the assumption about ideal gas molecules based on Use (i) the nature of their movement, … … [1] (ii) their volume. … … … [2] (b) A cube of volume V contains N molecules of an ideal gas. Each molecule has a For Examiner’s component cX of velocity normal to one side S of the cube, as shown in Fig. 2.1. Use side S cx Fig. 2.1 The pressure p of the gas due to the component cX of velocity is given by the expression pV = NmcX2 where m is the mass of a molecule. Explain how the expression leads to the relation 1 pV = 3Nm<c2> where <c2> is the mean square speed of the molecules. [3] (c) The molecules of an ideal gas have a root-mean-square (r.m.s.) speed of 520 m s–1 at a temperature of 27 °C. Calculate the r.m.s. speed of the molecules at a temperature of 100 °C. r.m.s. speed = … m s–1 [3]

9 marks

Mark scheme: 2 (a) (i) either random motion or constant velocity until hits wall/other molecule B1 [1] (ii) (total) volume of molecules is negligible M1 compared to volume of containing vessel A1 or radius/diameter of a molecule is negligible (M1) compared to the average intermolecular distance (A1) [2] (b) either molecule has component of velocity in three directions or c2 = cX2 + cY2 + cZ 2 M1 random motion and averaging, so <cX2> = <cY2> = <cZ 2> M1 <c2> = 3<cX2> A1 so, pV = ⅓Nm<c2> A0 [3] (c) <c2> ∝ T or crms ∝ T C1 temperatures are 300 K and 373 K C1 crms = 580 m s–1 A1 [3] (Do not allow any marks for use of temperature in units of ºC instead of K) GCE AS/A LEVEL – May/June 2012 9702 43

This question in 9702/43 May/June 2012

2 A student suggests that, when an ideal gas is heated from 100 °C to 200 °C, the internal For energy of the gas is doubled. Examiner’s Use (a) (i) State what is meant by internal energy. … … … [2] (ii) By reference to one of the assumptions of the kinetic theory of gases and your answer in (i), deduce what is meant by the internal energy of an ideal gas. … … … … [3] (b) State and explain whether the student’s suggestion is correct. … … … [2]

7 marks

Mark scheme: 2 (a) (i) sum of potential energy and kinetic energy of atoms / molecules / particles M1 reference to random A1 [2] (ii) no intermolecular forces B1 no potential energy B1 internal energy is kinetic energy (of random motion) of molecules B1 [3] (reference to random motion here then allow back credit to (i) if M1 scored) (b) kinetic energy ∝ thermodynamic temperature B1 either temperature in Celsius, not kelvin so incorrect or temperature in kelvin is not doubled B1 [2]

This question in 9702/42 Oct/Nov 2012

1 An ideal gas has volume V and pressure p. For this gas, the product pV is given by the expression pV = 13Nm <c 2> where m is the mass of a molecule of the gas. (a) State the meaning of the symbol (i) N, … [1] (ii) <c 2>. … [1] (b) A gas cylinder of volume 2.1 × 104 cm3 contains helium-4 gas at pressure 6.1 × 105 Pa and temperature 12 °C. Helium-4 may be assumed to be an ideal gas. (i) Determine, for the helium gas, 1. the amount, in mol, amount = … mol [3] 2. the number of atoms. number = … [2] (ii) Calculate the root-mean-square (r.m.s.) speed of the helium atoms. For Examiner’s Use r.m.s. speed = … m s–1 [3]

10 marks

Mark scheme: 1 (a) (i) number of molecules B1 [1] (ii) mean square speed B1 [1] (b) (i) 1. pV = nRT C1 n = (6.1 × 105 × 2.1 × 104 × 10–6) / (8.31 × 285) C1 n = 5.4 mol A1 [3] 2. either N = nNA = 5.4 × 6.02 × 1023 C1 = 3.26 × 1024 A1 or pV = NkT N = (6.1 × 105 × 2.1 × 104 × 10–6) / (1.38 × 10–23 × 285) (C1) N = 3.26 × 1024 (A1) [2] (ii) either 6.1 × 105 × 2.1 × 10–2 = 1/3 × 3.25 × 1024 × 4 × 1.66 × 10–27× <c2> C1 <c2> = 1.78 × 106 C1 cRMS = 1.33 × 103 m s–1 A1 or 1/2 × 4 × 1.66 × 10–27 × <c2> = 3/2 × 1.38 × 10–23 × 285 (C1) <c2> = 1.78 × 106 (C1) cRMS = 1.33 × 103 m s–1 (A1) [3]

This question in 9702/43 Oct/Nov 2012

2 (a) State what is meant by an ideal gas. For Examiner’s … Use … … … [3] (b) Two cylinders A and B are connected by a tube of negligible volume, as shown in Fig. 2.1. cylinder A cylinder B tap T 2.5 × 103 cm3 3.4 × 105 Pa 1.6 × 103 cm3 300 K 4.9 × 105 Pa tube Fig. 2.1 Initially, tap T is closed. The cylinders contain an ideal gas at different pressures. (i) Cylinder A has a constant volume of 2.5 × 103 cm3 and contains gas at pressure 3.4 × 105 Pa and temperature 300 K. Show that cylinder A contains 0.34 mol of gas. [1] (ii) Cylinder B has a constant volume of 1.6 × 103 cm3 and contains 0.20 mol of gas. For When tap T is opened, the pressure of the gas in both cylinders is 3.9 × 105 Pa. Examiner’s No thermal energy enters or leaves the gas. Use Determine the final temperature of the gas. temperature = … K [2] (c) By reference to work done and change in internal energy, suggest why the temperature of the gas in cylinder A has changed. … … … … [3]

9 marks

Mark scheme: 2 (a) obeys the equation pV = constant × T or pV = nRT M1 p, V and T explained A1 at all values of p, V and T/fixed mass/n is constant A1 [3] (b) (i) 3.4 × 105 × 2.5 × 103 × 10–6 = n × 8.31 × 300 M1 n = 0.34 mol A0 [1] (ii) for total mass/amount of gas 3.9 × 105 × (2.5 + 1.6) × 103 × 10–6 = (0.34 + 0.20) × 8.31 × T C1 T = 360 K A1 [2] (c) when tap opened gas passed (from cylinder B) to cylinder A B1 work done on gas in cylinder A (and no heating) M1 so internal energy and hence temperature increase A1 [3] GCE AS/A LEVEL – May/June 2013 9702 41

This question in 9702/41 May/June 2013

2 (a) State what is meant by an ideal gas. For Examiner’s … Use … … … [3] (b) Two cylinders A and B are connected by a tube of negligible volume, as shown in Fig. 2.1. cylinder A cylinder B tap T 2.5 × 103 cm3 3.4 × 105 Pa 1.6 × 103 cm3 300 K 4.9 × 105 Pa tube Fig. 2.1 Initially, tap T is closed. The cylinders contain an ideal gas at different pressures. (i) Cylinder A has a constant volume of 2.5 × 103 cm3 and contains gas at pressure 3.4 × 105 Pa and temperature 300 K. Show that cylinder A contains 0.34 mol of gas. [1] (ii) Cylinder B has a constant volume of 1.6 × 103 cm3 and contains 0.20 mol of gas. For When tap T is opened, the pressure of the gas in both cylinders is 3.9 × 105 Pa. Examiner’s No thermal energy enters or leaves the gas. Use Determine the final temperature of the gas. temperature = … K [2] (c) By reference to work done and change in internal energy, suggest why the temperature of the gas in cylinder A has changed. … … … … [3]

9 marks

Mark scheme: 2 (a) obeys the equation pV = constant × T or pV = nRT M1 p, V and T explained A1 at all values of p, V and T/fixed mass/n is constant A1 [3] (b) (i) 3.4 × 105 × 2.5 × 103 × 10–6 = n × 8.31 × 300 M1 n = 0.34 mol A0 [1] (ii) for total mass/amount of gas 3.9 × 105 × (2.5 + 1.6) × 103 × 10–6 = (0.34 + 0.20) × 8.31 × T C1 T = 360 K A1 [2] (c) when tap opened gas passed (from cylinder B) to cylinder A B1 work done on gas in cylinder A (and no heating) M1 so internal energy and hence temperature increase A1 [3] GCE AS/A LEVEL – May/June 2013 9702 43

This question in 9702/43 May/June 2013

2 The product of the pressure p and the volume V of an ideal gas is given by the expression For 1 Examiner’s Use pV = 3Nm<c 2> where m is the mass of one molecule of the gas. (a) State the meaning of the symbol (i) N, … [1] (ii) <c 2>. … [1] (b) The product pV is also given by the expression pV = NkT. Deduce an expression, in terms of the Boltzmann constant k and the thermodynamic temperature T, for the mean kinetic energy of a molecule of the ideal gas. [2] (c) A cylinder contains 1.0 mol of an ideal gas. (i) The volume of the cylinder is constant. Calculate the energy required to raise the temperature of the gas by 1.0 kelvin. energy = … J [2] (ii) The volume of the cylinder is now allowed to increase so that the gas remains at constant pressure when it is heated. Explain whether the energy required to raise the temperature of the gas by 1.0 kelvin is now different from your answer in (i). … … … [2]

8 marks

Mark scheme: 2 (a) (i) N: (total) number of molecules B1 [1] (ii) <c2>: mean square speed/velocity B1 [1] (b) pV = ⅓Nm<c2> = NkT (mean) kinetic energy = ½ m<c2> C1 algebra clear leading to ½ m<c2> = (3/2)kT A1 [2] (c) (i) either energy required = (3/2) × 1.38 × 10–23 × 1.0 × 6.02 × 1023 C1 either energy required = 12.5 J (12J if 2 s.f.) A1 [2] or energy = (3/2) × 8.31 × 1.0 (C1) or energy = 12.5 J (A1) (ii) energy is needed to push back atmosphere/do work against atmosphere M1 so total energy required is greater A1 [2]

This question in 9702/41 Oct/Nov 2013

2 The product of the pressure p and the volume V of an ideal gas is given by the expression For 1 Examiner’s Use pV = 3Nm<c 2> where m is the mass of one molecule of the gas. (a) State the meaning of the symbol (i) N, … [1] (ii) <c 2>. … [1] (b) The product pV is also given by the expression pV = NkT. Deduce an expression, in terms of the Boltzmann constant k and the thermodynamic temperature T, for the mean kinetic energy of a molecule of the ideal gas. [2] (c) A cylinder contains 1.0 mol of an ideal gas. (i) The volume of the cylinder is constant. Calculate the energy required to raise the temperature of the gas by 1.0 kelvin. energy = … J [2] (ii) The volume of the cylinder is now allowed to increase so that the gas remains at constant pressure when it is heated. Explain whether the energy required to raise the temperature of the gas by 1.0 kelvin is now different from your answer in (i). … … … [2]

8 marks

Mark scheme: 2 (a) (i) N: (total) number of molecules B1 [1] (ii) <c2>: mean square speed/velocity B1 [1] (b) pV = ⅓Nm<c2> = NkT (mean) kinetic energy = ½ m<c2> C1 algebra clear leading to ½ m<c2> = (3/2)kT A1 [2] (c) (i) either energy required = (3/2) × 1.38 × 10–23 × 1.0 × 6.02 × 1023 C1 either energy required = 12.5 J (12J if 2 s.f.) A1 [2] or energy = (3/2) × 8.31 × 1.0 (C1) or energy = 12.5 J (A1) (ii) energy is needed to push back atmosphere/do work against atmosphere M1 so total energy required is greater A1 [2]

This question in 9702/42 Oct/Nov 2013

3 (a) State what is meant by an ideal gas. … … [1] (b) A storage cylinder for an ideal gas has a volume of 3.0 × 10−4 m3. The gas is at a temperature of 23 °C and a pressure of 5.0 × 107 Pa. (i) Show that the amount of gas in the cylinder is 6.1 mol. [2] (ii) The gas leaks slowly from the cylinder so that, after a time of 35 days, the pressure has reduced by 0.40%. The temperature remains constant. Calculate the average rate, in atoms per second, at which gas atoms escape from the cylinder. rate = … s−1 [4]

7 marks

Mark scheme: 3 (a) obeys the equation pV / T = constant B1 [1] (accept pV = nRT) (b) (i) pV = nRT C1 5.0 × 107 × 3.0 × 10–4 = n × 8.31 × 296 giving n = 6.1 mol A1 [2] (ii) pressure ∝ amount of substance loss = 0.40 / 100 × 6.1 mol = 0.0244 mol C1 = 0.0244 × 6.02 × 1023 (atoms) C1 = 1.47 × 1022 atoms C1 rate = (1.47 × 1022) / (35 × 24 × 60 × 60) = 4.9 × 1015 s–1 A1 [4]

This question in 9702/41 Oct/Nov 2014

3 (a) State what is meant by an ideal gas. … … [1] (b) A storage cylinder for an ideal gas has a volume of 3.0 × 10−4 m3. The gas is at a temperature of 23 °C and a pressure of 5.0 × 107 Pa. (i) Show that the amount of gas in the cylinder is 6.1 mol. [2] (ii) The gas leaks slowly from the cylinder so that, after a time of 35 days, the pressure has reduced by 0.40%. The temperature remains constant. Calculate the average rate, in atoms per second, at which gas atoms escape from the cylinder. rate = … s−1 [4]

7 marks

Mark scheme: 3 (a) obeys the equation pV / T = constant B1 [1] (accept pV = nRT) (b) (i) pV = nRT C1 5.0 × 107 × 3.0 × 10–4 = n × 8.31 × 296 giving n = 6.1 mol A1 [2] (ii) pressure ∝ amount of substance loss = 0.40 / 100 × 6.1 mol = 0.0244 mol C1 = 0.0244 × 6.02 × 1023 (atoms) C1 = 1.47 × 1022 atoms C1 rate = (1.47 × 1022) / (35 × 24 × 60 × 60) = 4.9 × 1015 s–1 A1 [4]

This question in 9702/42 Oct/Nov 2014

2 In a sample of gas at room temperature, five atoms have the following speeds: 1.32 × 103 m s–1 1.50 × 103 m s–1 1.46 × 103 m s–1 1.28 × 103 m s–1 1.64 × 103 m s–1. For these five atoms, calculate, to three significant figures, (a) the mean speed, mean speed = … m s–1 [1] (b) the mean-square speed, mean-square speed = … m2 s–2 [2] (c) the root-mean-square speed. root-mean-square speed = … m s–1 [1]

4 marks

Mark scheme: 2 (a) mean speed = 1.44 × 103 m s–1 A1 [1] (b) evidence of summing of individual squared speeds C1 mean square speed = 2.09 × 106 m2 s–2 A1 [2] (c) root-mean-square speed = 1.45 × 103 m s–1 A1 [1] (allow ECF from (b) but only if arithmetic error)

This question in 9702/42 May/June 2015

2 (a) State what is meant by an ideal gas. … … … [2] (b) The mean-square speed of the atoms of a fixed mass of an ideal gas at 32 °C is 1.9 × 106 m2 s–2. The gas is heated at constant volume to a temperature of 80 °C. Determine (i) the rise, in kelvin, of the temperature of the gas, temperature rise = … K [1] (ii) the root-mean-square (r.m.s.) speed of the atoms at 80 °C. r.m.s. speed = … m s–1 [3]

6 marks

Mark scheme: 2 (a) obeys the equation pV = nRT or pV / T = constant M1 all symbols explained; T in kelvin/thermodynamic temperature A1 [2] (b) (i) temperature rise = 48 K A1 [1] (ii) <c2> ∝ T or equivalent C1 <c2> = (353 / 305) × 1.9 × 106 C1 cr.m.s. = 1480 m s–1 A1 [3]

This question in 9702/41 Oct/Nov 2015

2 (a) State what is meant by an ideal gas. … … … [2] (b) The mean-square speed of the atoms of a fixed mass of an ideal gas at 32 °C is 1.9 × 106 m2 s–2. The gas is heated at constant volume to a temperature of 80 °C. Determine (i) the rise, in kelvin, of the temperature of the gas, temperature rise = … K [1] (ii) the root-mean-square (r.m.s.) speed of the atoms at 80 °C. r.m.s. speed = … m s–1 [3]

6 marks

Mark scheme: 2 (a) obeys the equation pV = nRT or pV / T = constant M1 all symbols explained; T in kelvin/thermodynamic temperature A1 [2] (b) (i) temperature rise = 48 K A1 [1] (ii) <c2> ∝ T or equivalent C1 <c2> = (353 / 305) × 1.9 × 106 C1 cr.m.s. = 1480 m s–1 A1 [3]

This question in 9702/42 Oct/Nov 2015

2 (a) An ideal gas is said to consist of molecules that are hard elastic identical spheres. State two further assumptions of the kinetic theory of gases. 1. … … 2. … … [2] (b) The number of molecules per unit volume in an ideal gas is n. If it is assumed that all the molecules are moving with speed v, the pressure p exerted by the gas on the walls of the vessel is given by 1 p = 3nmv2 where m is the mass of one molecule. Explain the reasoning by which this expression is modified to give the formula 1 p = 3nm<c2>. … … [1] (c) The density of an ideal gas is 1.2 kg m−3 at a pressure of 1.0 × 105 Pa and a temperature of 27 °C. (i) Calculate the root-mean-square (r.m.s.) speed of the molecules of the gas at 27 °C. r.m.s. speed = … m s−1 [3] (ii) Calculate the mean-square speed of the molecules at 207 °C. mean-square speed = … m2 s−2 [2]

8 marks

Mark scheme: 2 (a) total volume of molecules negligible compared to that of containing vessel no intermolecular forces molecules in random motion time of collision small compared with the time between collisions large number of molecules any two B2 [2] (b) in a real gas there is a range of velocities or must take the average of v2 B1 [1] 1 (c) (i) either p = ρ <c2> 3 1 or 1.0 × 105 = × 1.2 × <c2> C1 3 <c2> = 2.5 × 105 C1 cr.m.s. = 500 m s–1 A1 [3] (ii) T ∝ <c2> C1 <c2> = 2.5 × 105 × 480 / 300 = 4.0 × 105 m2 s–2 (allow ECF from (c)(i)) A1 [2]

This question in 9702/43 Oct/Nov 2015

2 (a) State (i) what is meant by internal energy, … … … [2] (ii) the basic assumption of the kinetic theory of gases that leads to the conclusion that there is zero potential energy between the molecules of an ideal gas. … … [1] (b) The pressure p and volume V of an ideal gas are related by 1 pV = Nm 〈c2〉 3 where N is the number of molecules, m is the mass of a molecule and 〈c2〉 is the mean-square speed of the molecules. Use this equation to show that the mean kinetic energy 〈EK〉 of a molecule is given by 3 〈EK〉 = kT 2 where k is the Boltzmann constant and T is the thermodynamic temperature. [3] (c) A cylinder contains 17 g of oxygen gas at a temperature of 12 °C. The mass of 1.0 mol of oxygen gas is 32 g. It may be assumed that the oxygen behaves as an ideal gas. Calculate, for the oxygen gas in the cylinder, (i) the mean kinetic energy of a molecule, mean kinetic energy = … J [2] (ii) the number of molecules, number = … [2] (iii) the total internal energy. internal energy = … J [1] [Total: 11]

11 marks

Mark scheme: 2 (a) (i) sum of kinetic and potential energy of atoms/molecules M1 reference to random (distribution) A1 [2] (ii) no forces (of attraction or repulsion) between molecules B1 [1] (b) pV = NkT or pV = nRT and R = kNA , n = N / NA B1 1/3 Nm<c 2> = NkT or 1/3 m<c 2> = kT B1 <EK> = 1/2 m<c 2> so <EK> = 3/2 kT B1 [3] (c) (i) <EK> = 3/2 × 1.38 × 10–23 × (273 + 12) C1 = 5.9 (5.90) × 10–21 J A1 [2] (use of T = 12 K not T = 285 K scores 0 / 2) (ii) number = (17 / 32) × 6.02 × 1023 C1 = 3.2 (3.20) × 1023 A1 [2] (iii) internal energy = 5.9 × 10–21 × 3.2 × 1023 = 1900 (1890) J A1 [1]

This question in 9702/42 Feb/March 2016

2 (a) An ideal gas is assumed to consist of atoms or molecules that behave as hard, identical spheres that are in continuous motion and undergo elastic collisions. State two further assumptions of the kinetic theory of gases. 1. … … 2. … … [2] (b) Helium-4 (42He) may be assumed to be an ideal gas. (i) Show that the mass of one atom of helium-4 is 6.6 × 10−24 g. [1] (ii) The mean kinetic energy EK of an atom of an ideal gas is given by the expression EK = 32 kT. Calculate the root-mean-square (r.m.s.) speed of a helium-4 atom at a temperature of 27 °C. r.m.s. speed = … m s−1 [3] [Total: 6]

6 marks

Mark scheme: 2 (a) e.g. time of collisions negligible compared to time between collisions no intermolecular forces (except during collisions) random motion (of molecules) large numbers of molecules (total) volume of molecules negligible compared to volume of containing vessel or average/mean separation large compared with size of molecules any two B2 [2] 2 (b) (i) mass = 4.0 / (6.02 × 10 23) = 6.6 × 10–24 g or mass = 4.0 × 1.66 × 10–27 × 103 = 6.6 × 10–24 g B1 [1] 3 1 (ii) kT = m <c 2> C1 2 2 3 1 × 1.38 × 10–23 × 300 = × 6.6 × 10–27 × <c 2> 2 2 <c 2> = 1.88 × 106 (m2 s–2) C1 r.m.s. speed = 1.4 × 103 m s–1 A1 [3]

This question in 9702/41 May/June 2016

2 (a) An ideal gas is assumed to consist of atoms or molecules that behave as hard, identical spheres that are in continuous motion and undergo elastic collisions. State two further assumptions of the kinetic theory of gases. 1. … … 2. … … [2] (b) Helium-4 (42He) may be assumed to be an ideal gas. (i) Show that the mass of one atom of helium-4 is 6.6 × 10−24 g. [1] (ii) The mean kinetic energy EK of an atom of an ideal gas is given by the expression EK = 32 kT. Calculate the root-mean-square (r.m.s.) speed of a helium-4 atom at a temperature of 27 °C. r.m.s. speed = … m s−1 [3] [Total: 6]

6 marks

Mark scheme: 2 (a) e.g. time of collisions negligible compared to time between collisions no intermolecular forces (except during collisions) random motion (of molecules) large numbers of molecules (total) volume of molecules negligible compared to volume of containing vessel or average/mean separation large compared with size of molecules any two B2 [2] 2 (b) (i) mass = 4.0 / (6.02 × 10 23) = 6.6 × 10–24 g or mass = 4.0 × 1.66 × 10–27 × 103 = 6.6 × 10–24 g B1 [1] 3 1 (ii) kT = m <c 2> C1 2 2 3 1 × 1.38 × 10–23 × 300 = × 6.6 × 10–27 × <c 2> 2 2 <c 2> = 1.88 × 106 (m2 s–2) C1 r.m.s. speed = 1.4 × 103 m s–1 A1 [3]

This question in 9702/43 May/June 2016

2 (a) The equation of state for an ideal gas of volume V at pressure p is pV = nRT where R is the molar gas constant. State what is meant by (i) the symbol n, … … [1] (ii) the symbol T. … … [1] (b) An ideal gas is held in a container of volume 2.4 × 103 cm3 at pressure 4.9 × 105 Pa. The temperature of the gas is 100 °C. Show that the number of molecules of the gas in the container is 2.3 × 1023. [3] (c) Use data from (b) to estimate the mean distance between molecules in the gas. mean distance = … cm [3] [Total: 8]

8 marks

Mark scheme: 2 (a) (i) number of moles/amount of substance B1 [1] (ii) kelvin temperature/absolute temperature/thermodynamic temperature B1 [1] (b) pV = nRT 4.9 × 105 × 2.4 × 103 × 10–6 = n × 8.31 × 373 B1 n = 0.38 (mol) C1 number of molecules or N = 0.38 × 6.02 × 1023 = 2.3 × 1023 A1 [3] or pV = NkT (C1) 4.9 × 105 × 2.4 × 103 × 10–6 = N × 1.38 × 10–23 × 373 (M1) number of molecules or N = 2.3 × 1023 (A1) (c) volume occupied by one molecule = (2.4 × 103) / (2.3 × 1023) C1 = 1.04 × 10–20 cm3 mean spacing = (1.04 × 10–20)1/3 C1 = 2.2 × 10–7 cm (allow 1 s.f.) A1 [3] (allow other dimensionally correct methods e.g. V = (4/3)πr3)

This question in 9702/42 Oct/Nov 2016

4 (a) Describe the motion of molecules in a gas, according to the kinetic theory of gases. … … … [2] (b) Describe what is observed when viewing Brownian motion that provides evidence for your answer in (a). … … … [2] (c) At a pressure of 1.05 × 105 Pa and a temperature of 27 °C, 1.00 mol of helium gas has a volume of 0.0240 m3. The mass of 1.00 mol of helium gas, assumed to be an ideal gas, is 4.00 g. (i) Calculate the root-mean-square (r.m.s.) speed of an atom of helium gas for a temperature of 27 °C. r.m.s. speed = … m s–1 [3] (ii) Using your answer in (i), calculate the r.m.s. speed of the atoms at 177 °C. r.m.s. speed = … m s–1 [3] [Total: 10]

10 marks

Mark scheme: 4(a) random/haphazard B1 constant velocity or speed in a straight line between collisions or distribution of speeds/different directions B1 4(b) (small) specks of light/bright specks/pollen grains/dust particles/smoke particles M1 moving haphazardly/randomly/jerky/in a zigzag fashion A1 4(c)(i) pV = ⅓ Nm〈c2〉 1.05 × 105 × 0.0240 = ⅓ × 4.00 × 10–3 × 〈c2〉 C1 〈c2〉 = 1.89 × 106 C1 or ½ m〈c2〉 = (3 / 2) kT 0.5 × (4.00 × 10–3 / 6.02 × 1023) × 〈c2〉 = 1.5 × 1.38 × 10–23 × 300 (C1) 〈c2〉 = 1.87 × 106 (C1) or nRT = ⅓ Nm〈c2〉 1.00 × 8.31 × 300 = ⅓ × 4.00 × 10–3 × 〈c2〉 (C1) 〈c2〉 = 1.87 × 106 (C1) cr.m.s. = 1.37 × 103 m s–1 A1 Question Answer Marks 4(c)(ii) 〈c2〉 ∝ T C1 〈c2〉 at 177 °C = 1.89 × 106 × (450 / 300) C1 cr.m.s. at 177 °C = 1.68 × 103 m s–1 A1

This question in 9702/41 May/June 2017

2 (a) The pressure p and volume V of an ideal gas are related to the density ρ of the gas by the expression 1 p = ρ 〈c 2〉. 3 (i) State what is meant by the symbol 〈c 2〉. … … [1] (ii) Use the expression to show that the mean kinetic energy EK of a gas molecule is given by 3 EK = 2 kT where k is the Boltzmann constant and T is the thermodynamic temperature. [3] (b) (i) An ideal gas containing 1.0 mol of molecules is heated at constant volume. Use the expression in (a)(ii) to show that the thermal energy required to raise the 3 temperature of the gas by 1.0 K has a value of R, where R is the molar gas constant. 2 [3] (ii) Nitrogen may be assumed to be an ideal gas. The molar mass of nitrogen gas is 28 g mol–1. Use the answer in (b)(i) to calculate a value for the specific heat capacity, in J kg–1 K–1, at constant volume for nitrogen. specific heat capacity = … J kg–1 K–1 [2] [Total: 9]

9 marks

Mark scheme: 2(a)(i) mean/average square speed/velocity B1 2(a)(ii) pV = NkT or pV = nRT B1 ρ = Nm / V or ρ = nNAm / V and k = nR / N B1 EK = ½ m〈c2〉 with algebra to (3 / 2)kT B1 2(b)(i) no (external) work done or ∆U = q or w = 0 B1 q = NA × (3 / 2)k × 1.0 M1 NAk = R so q = (3 / 2)R A1 2(b)(ii) specific heat capacity = {(3 / 2) × R} / 0.028 C1 = 450 J kg–1 K–1 A1

This question in 9702/42 May/June 2017

4 (a) Describe the motion of molecules in a gas, according to the kinetic theory of gases. … … … [2] (b) Describe what is observed when viewing Brownian motion that provides evidence for your answer in (a). … … … [2] (c) At a pressure of 1.05 × 105 Pa and a temperature of 27 °C, 1.00 mol of helium gas has a volume of 0.0240 m3. The mass of 1.00 mol of helium gas, assumed to be an ideal gas, is 4.00 g. (i) Calculate the root-mean-square (r.m.s.) speed of an atom of helium gas for a temperature of 27 °C. r.m.s. speed = … m s–1 [3] (ii) Using your answer in (i), calculate the r.m.s. speed of the atoms at 177 °C. r.m.s. speed = … m s–1 [3] [Total: 10]

10 marks

Mark scheme: 4(a) random/haphazard B1 constant velocity or speed in a straight line between collisions or distribution of speeds/different directions B1 4(b) (small) specks of light/bright specks/pollen grains/dust particles/smoke particles M1 moving haphazardly/randomly/jerky/in a zigzag fashion A1 4(c)(i) pV = ⅓ Nm〈c2〉 1.05 × 105 × 0.0240 = ⅓ × 4.00 × 10–3 × 〈c2〉 C1 〈c2〉 = 1.89 × 106 C1 or ½ m〈c2〉 = (3 / 2) kT 0.5 × (4.00 × 10–3 / 6.02 × 1023) × 〈c2〉 = 1.5 × 1.38 × 10–23 × 300 (C1) 〈c2〉 = 1.87 × 106 (C1) or nRT = ⅓ Nm〈c2〉 1.00 × 8.31 × 300 = ⅓ × 4.00 × 10–3 × 〈c2〉 (C1) 〈c2〉 = 1.87 × 106 (C1) cr.m.s. = 1.37 × 103 m s–1 A1 Question Answer Marks 4(c)(ii) 〈c2〉 ∝ T C1 〈c2〉 at 177 °C = 1.89 × 106 × (450 / 300) C1 cr.m.s. at 177 °C = 1.68 × 103 m s–1 A1

This question in 9702/43 May/June 2017

3 (a) (i) State what is meant by the internal energy of a system. … … … … [2] (ii) Explain why, for an ideal gas, the change in internal energy is directly proportional to the change in thermodynamic temperature of the gas. … … … … … [3] (b) A cylinder of volume 1.8 × 104 cm3 contains helium gas at pressure 6.4 × 106 Pa and temperature 25 °C. Helium gas may be considered to be an ideal gas consisting of single atoms. Calculate the number of helium atoms in the cylinder. number = … [3] [Total: 8]

8 marks

Mark scheme: 3(a)(i) sum of potential and kinetic energies (of molecules/atoms/particles) B1 (energy of) molecules/atoms/particles in random motion B1 3(a)(ii) (in ideal gas) no intermolecular forces so no potential energy B1 internal energy is (solely) kinetic energy (of particles) B1 (mean) kinetic energy (of particles) proportional to (thermodynamic) temperature of gas B1 3(b) pV = NkT C1 6.4 × 106 × 1.8 × 104 × 10–6 = N × 1.38 × 10–23 × 298 C1 or pV = nRT and N = n × NA (C1) 6.4 × 106 × 1.8 × 104 × 10–6 = n × 8.31 × 298 n = 46.5 (mol) N = 46.5 × 6.02 × 1023 (C1) N = 2.8 × 1025 A1

This question in 9702/41 May/June 2018

2 (a) Use one of the assumptions of the kinetic theory of gases to explain why the potential energy of the molecules of an ideal gas is zero. … … [1] (b) The average translational kinetic energy EK of a molecule of an ideal gas is given by the expression 1 3 EK = 2m 〈c2〉 = 2 kT where m is the mass of a molecule and k is the Boltzmann constant. State the meaning of the symbol (i) 〈c2〉, … [1] (ii) T. … [1] (c) A cylinder of constant volume 4.7 × 104 cm3 contains an ideal gas at pressure 2.6 × 105 Pa and temperature 173 °C. The gas is heated. The thermal energy transferred to the gas is 2900 J. The final temperature and pressure of the gas are T and p, as illustrated in Fig. 2.1. 4.7 × 104 cm3 4.7 × 104 cm3 2900 J 2.6 × 105 Pa p 173 °C T Fig. 2.1 (i) Calculate 1. the number N of molecules in the cylinder, N = … [3] 2. the increase in average kinetic energy of a molecule during the heating process. increase = … J [1] (ii) Use your answer in (i) part 2 to determine the final temperature T, in kelvin, of the gas in the cylinder. T = … K [3] [Total: 10]

10 marks

Mark scheme: 2(a) no intermolecular forces (so no potential energy) B1 2(b)(i) mean square speed (of molecule(s)) B1 2(b)(ii) kelvin/thermodynamic/absolute temperature B1 2(c)(i)1. pV = NkT C1 4.7 × 10–2 × 2.6 × 105 = N × 1.38 × 10–23 × 446 C1 or pV = nRT and N = nNA 4.7 × 10–2 × 2.6 × 105 = n × 8.31 × 446 n = 3.3 (mol) (C1) N = 3.3 × 6.02 × 1023 (C1) N = 2.0 × 1024 A1 2(c)(i)2. average increase = 2900 / (2.0 × 1024) = 1.5 × 10–21 J A1 2(c)(ii) ∆EK = (3/2)k (∆)T 1.5 × 10–21 = (3/2) × 1.38 × 10–23 × (∆)T C1 (∆)T in range 70–72 K C1 T = 173 + 273 + 70 = 520 K A1

This question in 9702/42 May/June 2018

3 (a) During melting, a solid becomes liquid with little or no change in volume. Use kinetic theory to explain why, during the melting process, thermal energy is required although there is no change in temperature. … … … … … [3] (b) An aluminium can of mass 160 g contains a mass of 330 g of warm water at a temperature of 38 °C, as illustrated in Fig. 3.1. ice warm water aluminium can Fig. 3.1 A mass of 48 g of ice at –18 °C is taken from a freezer and put in to the water. The ice melts and the final temperature of the can and its contents is 23 °C. Data for the specific heat capacity c of aluminium, ice and water are given in Fig. 3.2. c / J g–1 K–1 aluminium 0.910 ice 2.10 water 4.18 Fig. 3.2 Assuming no exchange of thermal energy with the surroundings, (i) show that the loss in thermal energy of the can and the warm water is 2.3 × 104 J, [2] (ii) use the information in (i) to calculate a value L for the specific latent heat of fusion of ice. L = … J g–1 [2] [Total: 7]

7 marks

Mark scheme: 3(a) (during melting,) bonds between atoms/molecules are broken B1 potential energy of atoms/molecules is increased B1 no/little work done so required input of energy is thermal B1 3(b)(i) (∆Q =) mc∆θ C1 loss = (160 × 0.910 × 15) + (330 × 4.18 × 15) = 2.3 × 104 J A1 3(b)(ii) 2.3 × 104 = (48 × 2.10 × 18) + 48L + (48 × 4.18 × 23) C1 48L = 1.66 × 104 L = 350 J g–1 A1

This question in 9702/42 May/June 2018

3 (a) (i) State what is meant by the internal energy of a system. … … … … [2] (ii) Explain why, for an ideal gas, the change in internal energy is directly proportional to the change in thermodynamic temperature of the gas. … … … … … [3] (b) A cylinder of volume 1.8 × 104 cm3 contains helium gas at pressure 6.4 × 106 Pa and temperature 25 °C. Helium gas may be considered to be an ideal gas consisting of single atoms. Calculate the number of helium atoms in the cylinder. number = … [3] [Total: 8]

8 marks

Mark scheme: 3(a)(i) sum of potential and kinetic energies (of molecules/atoms/particles) B1 (energy of) molecules/atoms/particles in random motion B1 3(a)(ii) (in ideal gas) no intermolecular forces so no potential energy B1 internal energy is (solely) kinetic energy (of particles) B1 (mean) kinetic energy (of particles) proportional to (thermodynamic) temperature of gas B1 3(b) pV = NkT C1 6.4 × 106 × 1.8 × 104 × 10–6 = N × 1.38 × 10–23 × 298 C1 or pV = nRT and N = n × NA (C1) 6.4 × 106 × 1.8 × 104 × 10–6 = n × 8.31 × 298 n = 46.5 (mol) N = 46.5 × 6.02 × 1023 (C1) N = 2.8 × 1025 A1

This question in 9702/43 May/June 2018

2 (a) State what is meant by an ideal gas. … … … [2] (b) An ideal gas comprised of single atoms is contained in a cylinder and has a volume of 1.84 × 10–2 m3 at a pressure of 2.12 × 107 Pa. The mass of gas in the cylinder is 3.20 kg. (i) Determine, to three significant figures, the root-mean-square (r.m.s.) speed of the atoms of the gas. r.m.s. speed = … m s–1 [3] (ii) The temperature of the gas in the cylinder is 22 °C. Determine, to three significant figures, 1. the amount, in mol, of the gas, amount = … mol [2] 2. the mass of one atom of the gas. mass = … kg [2] (c) Use your answer in (b)(ii) part 2 to determine the nucleon number A of an atom of the gas. A = … [1] [Total: 10]

10 marks

Mark scheme: 2(a) M1 symbols p,V and T explained A1 2(b)(i) pV = ⅓ Nm<c2> and M = Nm (and so) p = ⅓ρ <c2> C1 2.12 × 107 = ⅓ × [3.20 / (1.84 × 10–2)] × <c2> C1 cr.m.s. = 605 m s–1 A1 2(b)(ii) 1. pV = nRT and T = (22 + 273) K C1 n = (2.12 × 107 × 1.84 × 10–2) / (8.31 × 295) = 159 mol A1 2. mass = 3.20 / (159 × 6.02 × 1023) or mass = [2 × (3 / 2) × 1.38 × 10–23 × 295] / 6052 C1 mass = 3.34 × 10–26 kg A1 2(c) A = (3.34 × 10–26) / (1.66 × 10–27) = 20 A1

This question in 9702/42 Oct/Nov 2018

2 The pressure p of an ideal gas having density ρ is given by the expression p = 13 ρ〈c2〉. (a) State what is meant by: (i) an ideal gas … … … [2] (ii) the symbol 〈c2〉. … … [1] (b) A cylinder contains a fixed mass of a gas at a temperature of 120 °C. The gas has a volume of 6.8 × 10–3 m3 at a pressure 2.4 × 105 Pa. (i) Assuming the gas acts like an ideal gas, show that the number of atoms of gas in the cylinder is 3.0 × 1023. [3] (ii) Each atom of the gas, assumed to be a sphere, has a radius of 3.2 × 10–11 m. Use the answer in (i) to estimate the actual volume occupied by the gas atoms. volume = … m3 [2] (iii) One of the assumptions of the kinetic theory of gases is related to the volume of the atoms. State this assumption. Explain whether your answer in (ii) is consistent with this assumption. … … … … [2] [Total: 10]

10 marks

Mark scheme: 2(a)(i) gas obeys formula pV / T = constant M1 symbols V and T explained A1 2(a)(ii) mean-square-speed (of atoms / molecules) B1 2(b)(i) use of T = 393 C1 pV = nRT C1 2.4 × 105 × 6.8 × 10–3 = n × 8.31 × 393 and N = n × 6.02 × 1023 = 3.0 × 1023 A1 or pV = NkT (C1) 2.4 × 105 × 6.8 × 10–3 = N × 1.38 x 10–23 × 393 hence N = 3.0 × 1023 (A1) 2(b)(ii) volume of one atom = 4 / 3πr3 C1 volume occupied = 3.0 × 1023 × 4 / 3 × π × (3.2 × 10–11 )3 = 4 × 10–8 m3 A1 2(b)(iii) assumption: volume of atoms negligible compared to volume of container / cylinder B1 4 × 10–8 (m3) << 6.8 × 10–3 (m3) so yes B1

This question in 9702/42 Feb/March 2019

2 (a) The kinetic theory of gases is based on a number of assumptions about the molecules of a gas. State the assumption that is related to the volume of the molecules of the gas. … … … [2] (b) An ideal gas occupies a volume of 2.40 × 10–2 m3 at a pressure of 4.60 × 105 Pa and a temperature of 23 °C. (i) Calculate the number of molecules in the gas. number = … [3] (ii) Each molecule has a diameter of approximately 3 × 10–10 m. Estimate the total volume of the gas molecules. volume = … m3 [3] (c) By reference to your answer in (b)(ii), suggest why the assumption in (a) is justified. … … [1] [Total: 9]

9 marks

Mark scheme: 2(a) (total volume of molecules is) negligible M1 compared with volume occupied by the gas A1 2(b)(i) pV = NkT C1 4.60 × 105 × 2.40 × 10–2 = N × 1.38 × 10–23 × (273 + 23) C1 or pV = nRT (C1) 4.60 × 105 × 2.40 × 10–2 = n × 8.31 × (273 + 23) n = 4.49 (mol) N = nNA = 4.49 × 6.02 × 1023 (C1) N = 2.7 × 1024 A1 Question Answer Marks 2(b)(ii) volume of one atom = d3 (= 2.7 × 10–29 m3) C1 volume of all atoms = 2.7 × 10–29 × 2.7 × 1024 C1 = 7 × 10–5 m3 A1 or volume of one atom = (4 / 3)πr3 (= 1.41 × 10–29 m3) (C1) volume of all atoms = 2.7 × 1024 × 1.41 × 10–29 (C1) = 4 × 10–5 m3 (A1) 2(c) numerical comparison between answer to (b)(ii) and 2.4 × 10–2 (m3) showing (b)(ii) is much less than 2.4 × 10–2 (m3) B1

This question in 9702/41 Oct/Nov 2019

2 (a) Smoke particles are suspended in still air. Brownian motion of the smoke particles is seen through a microscope. Describe: (i) what is seen through the microscope … … [1] (ii) how Brownian motion provides evidence for the nature of the movement of gas molecules. … … … [2] (b) A fixed mass of an ideal gas has volume 2.40 × 103 cm3 at pressure 3.51 × 105 Pa and temperature 290 K. The gas is heated at constant volume until the temperature is 310 K at pressure 3.75 × 105 Pa, as illustrated in Fig. 2.1. 2.40 × 103 cm3 2.40 × 103 cm3 3.51 × 105 Pa 3.75 × 105 Pa 290 K 310 K Fig. 2.1 The quantity of thermal energy required to raise the temperature of 1.00 mol of the gas by 1.00 K at constant volume is 12.5 J. Calculate, to three significant figures: (i) the amount, in mol, of the gas amount = … mol [3] (ii) the thermal energy transfer during the change. energy transfer = … J [2] (c) For the change in the gas in (b), state: (i) the quantity of external work done on the gas work done = … J [1] (ii) the change in internal energy, with the direction of this change. change = … J direction … [2] [Total: 11]

11 marks

Mark scheme: 2(a)(i) specks of light moving haphazardly B1 2(a)(ii) (gas) molecules collide with (smoke) particles or random motion of the (gas) molecules M1 causes the (haphazard) motion of the smoke particles or causes the smoke particles to change direction A1 2(b)(i) pV = nRT C1 n = (3.51 × 105 × 2.40 × 10–3) / (8.31 × 290) or n = (3.75 × 105 × 2.40 × 10–3) / (8.31 × 310) C1 or pV = NkT (C1) n = (3.51 × 105 × 2.40 × 10–3) / (1.38 × 10–23 × 6.02 × 1023 × 290) or n = (3.75 × 105 × 2.40 × 10–3) / (1.38 × 10–23 × 6.02 × 1023 × 310) (C1) n = 0.350 mol or 0.349 mol A1 2(b)(ii) energy transfer = (0.349 or 0.35) × 12.5 × (310 – 290) C1 = 87.3 J or 87.5 J A1 2(c)(i) zero A1 2(c)(ii) 87.3 J or 87.5 J A1 increase B1

This question in 9702/42 Oct/Nov 2019

2 (a) The kinetic theory of gases is based on a number of assumptions about the molecules of a gas. State the assumption that is related to the volume of the molecules of the gas. … … … [2] (b) An ideal gas occupies a volume of 2.40 × 10–2 m3 at a pressure of 4.60 × 105 Pa and a temperature of 23 °C. (i) Calculate the number of molecules in the gas. number = … [3] (ii) Each molecule has a diameter of approximately 3 × 10–10 m. Estimate the total volume of the gas molecules. volume = … m3 [3] (c) By reference to your answer in (b)(ii), suggest why the assumption in (a) is justified. … … [1] [Total: 9]

9 marks

Mark scheme: 2(a) (total volume of molecules is) negligible M1 compared with volume occupied by the gas A1 2(b)(i) pV = NkT C1 4.60 × 105 × 2.40 × 10–2 = N × 1.38 × 10–23 × (273 + 23) C1 or pV = nRT (C1) 4.60 × 105 × 2.40 × 10–2 = n × 8.31 × (273 + 23) n = 4.49 (mol) N = nNA = 4.49 × 6.02 × 1023 (C1) N = 2.7 × 1024 A1 Question Answer Marks 2(b)(ii) volume of one atom = d3 (= 2.7 × 10–29 m3) C1 volume of all atoms = 2.7 × 10–29 × 2.7 × 1024 C1 = 7 × 10–5 m3 A1 or volume of one atom = (4 / 3)πr3 (= 1.41 × 10–29 m3) (C1) volume of all atoms = 2.7 × 1024 × 1.41 × 10–29 (C1) = 4 × 10–5 m3 (A1) 2(c) numerical comparison between answer to (b)(ii) and 2.4 × 10–2 (m3) showing (b)(ii) is much less than 2.4 × 10–2 (m3) B1

This question in 9702/43 Oct/Nov 2019

2 A large container of volume 85 m3 is filled with 110 kg of an ideal gas. The pressure of the gas is 1.0 × 105 Pa at temperature T. The mass of 1.0 mol of the gas is 32 g. (a) Show that the temperature T of the gas is approximately 300 K. [3] (b) The temperature of the gas is increased to 350 K at constant volume. The specific heat capacity of the gas for this change is 0.66 J kg−1 K−1. Calculate the energy supplied to the gas by heating. energy = … J [2] (c) Explain how movement of the gas molecules causes pressure in the container. … … … … … … … [3] (d) The temperature of a gas depends on the root-mean-square (r.m.s.) speed of its molecules. Calculate the ratio: r.m.s. speed of gas molecules at 350 K . r.m.s. speed of gas molecules at 300 K ratio = … [2] [Total: 10]

10 marks

Mark scheme: 2(a) n = 110 / 0.032 or 110000 / 32 or 3440 C1 pV = nRT C1 T = (1.0 × 105 × 85) / (8.31 × (110 / 0.032)) = 300 K A1 2(b) E = mcΔθ = 110 × 0.66 × 50 C1 = 3600 J A1 2(c) Any 3 from: • molecule collides with wall • momentum of molecule changes during collision (with wall) • force on molecule so force on wall • many forces act over surface area of container exerting a pressure B3 2(d) KE ∝ T v ∝ √T C1 ratio = √(350 / 300) = 1.1 A1

This question in 9702/42 Feb/March 2020

2 (a) A square box of volume V contains N molecules of an ideal gas. Each molecule has mass m. Using the kinetic theory of ideal gases, it can be shown that, if all the molecules are moving with speed v at right angles to one face of the box, the pressure p exerted on the face of the box is given by the expression pV = Nmv 2. (equation 1) This expression leads to the formula 1 p = 3ρ 〈c 2〉 (equation 2) for the pressure p of an ideal gas, where ρis the density of the gas and 〈c 2〉 is the mean‑square speed of the molecules. Explain how each of the following terms in equation 2 is derived from equation 1: ρ : … … 1 : … 3 … 〈c 2〉: … … [4] (b) An ideal gas has volume, pressure and temperature as shown in Fig. 2.1. volume 6.0 × 10–3 m3 pressure 3.0 × 105 Pa temperature 17 °C Fig. 2.1 The mass of the gas is 20.7 g. Calculate the mass of one molecule of the gas. mass = … g [4] [Total: 8]

8 marks

Mark scheme: 2(a) ρ: Nm / V ⅓: molecules move in three dimensions (not one) so ⅓ in any (one) direction B1 <c2>: molecules have different speeds so take average M1 of (speed)2 A1 2(b) pV = NkT C1 N = (3.0 × 105 × 6.0 × 10–3) / (1.38 × 10–23 × 290) C1 = 4.5 × 1023 C1 mass = 20.7 / (4.5 × 1023) = 4.6 × 10–23 g A1

This question in 9702/42 May/June 2020

2 A fixed mass of an ideal gas is at a temperature of 21 °C. The pressure of the gas is 2.3 × 105 Pa and its volume is 3.5 × 10–3 m3. (a) (i) Calculate the number N of molecules in the gas. N = … [2] (ii) The mass of one molecule of the gas is 40 u. Determine the root-mean-square (r.m.s.) speed of the gas molecules. r.m.s. speed = … m s–1 [2] (b) The temperature of the gas is increased by 84 °C. Calculate the value of the ratio new r.m.s. speed of molecules original r.m.s. speed of molecules . ratio = … [2] [Total: 6]

6 marks

Mark scheme: 2(a)(i) pV NkT = or pV nRT = and A N nN = 5 3 23 2.3 10 3.5 10 1.38 10 294 N − − × × × = × × = 2.0 × 1023 A1 2(a)(ii) 2 1 3 pV Nmc = 5 3 2 23 27 3 2.3 10 3.5 10 2.0 10 40 1.66 10 c − − × × × × = × × × × = 182 000 r.m.s. speed = 430 m s–1 C1 or 2 3 12 2 mc kT = A1 23 2 27 3 1.38 10 294 40 1.66 10 c − − × × × = × × = 183 000 (C1) r.m.s.speed = 430 m s–1 (A1) Question Answer Marks 2(b) ( ) 23 23 2 23 27 3 2.0 10 1.38 10 294 84 2.0 10 40 1.66 10 c − − × × × × × + = × × × × 2 236000 c = 485 c = C1 485 1.1 430 ratio   = =     A1 OR v T ∝ or 2 v T ∝ (C1) 273 21 84 273 21 ratio + + = + or 378 294 ratio = 1.1 (A1)

This question in 9702/42 Feb/March 2021

3 (a) Using a simple kinetic model of matter, describe the structure of a solid. … … … [2] (b) The specific latent heat of vaporisation is much greater than the specific latent heat of fusion for the same substance. Explain this, in terms of the spacing of molecules. … … … [1] (c) A heater supplies energy at a constant rate to 0.045 kg of a substance. The variation with time of the temperature of the substance is shown in Fig. 3.1. The substance is perfectly insulated from its surroundings. 80 Q 60 temperature / °C 40 20 0 P –20 –40 –60 –80 –100 –120 0 1 2 3 4 5 6 7 8 9 10 time / min Fig. 3.1 (i) Determine the temperature at which the substance melts. temperature = … °C [1] (ii) The power of the heater is 150 W. Use data from Fig. 3.1 to calculate, in kJ kg–1, the specific latent heat of vaporisation L of the substance. L = … kJ kg–1 [3] (iii) Suggest what can be deduced from the fact that section Q on the graph is less steep than section P. … … [1] [Total: 8]

8 marks

Mark scheme: 3(a) Any 2 from: • particles / atoms / molecules / ions (very) close together / touching • regular, repeating pattern • vibrate about a fixed point B2 3(b) (much) greater increase in spacing of molecules (for vaporisation compared with fusion) B1 3(c)(i) –100 °C B1 Question Answer Marks 3(c)(ii) time = 8.5 – 3.0 = 5.5 min C1 Pt = mL energy = power × time = 150 × 5.5 × 60 = 49 500 J E L m = 49 500 0.045 = C1 1 1100 kJ kg − = A1 3(c)(iii) gas has a higher specific heat capacity (than liquid) B1

This question in 9702/42 Feb/March 2021

2 An ideal gas is contained in a cylinder by means of a movable frictionless piston, as illustrated in Fig. 2.1. cylinder movement of piston piston gas molecule Fig. 2.1 Initially, the gas has a volume of 1.8 × 10−3 m3 at a pressure of 3.3 × 105 Pa and a temperature of 310 K. (a) Show that the number of gas molecules in the cylinder is 1.4 × 1023. [2] (b) Use kinetic theory to explain why, when the piston is moved so that the gas expands, this causes a decrease in the temperature of the gas. … … … … [3] (c) The gas expands so that its volume increases to 2.4 × 10−3 m3 at a pressure of 2.3 × 105 Pa and a temperature of 288 K, as shown in Fig. 2.2. 1.8 × 10−3 m3 2.4 × 10−3 m3 3.3 × 105 Pa 2.3 × 105 Pa 310 K 288 K Fig. 2.2 (i) The average translational kinetic energy EK of a molecule of an ideal gas is given by 3 EK = kT 2 where k is the Boltzmann constant and T is the thermodynamic temperature. Calculate the increase in internal energy ΔU of the gas during the expansion. ΔU = … J [3] (ii) The work done by the gas during the expansion is 76 J. Use your answer in (i) to explain whether thermal energy is transferred to or from the gas during the expansion. … … … [2] [Total: 10]

10 marks

Mark scheme: 2(a) pV = NkT C1 N = (1.8 × 10–3 × 3.3 × 105) / (1.38 × 10–23 × 310) = 1.4 × 1023 A1 or pV = nRT and nNA = N (C1) N = (1.8 × 10–3 × 3.3 × 105 × 6.02 × 1023) / (8.31 × 310) = 1.4 × 1023 (A1) 2(b) speed of molecule decreases on impact with moving piston B1 mean square speed (directly) proportional to (thermodynamic) temperature or mean square speed (directly) proportional to kinetic energy (of molecules) or kinetic energy (of molecules) (directly) proportional to (thermodynamic) temperature B1 kinetic energy (of molecules) decreases (so temperature decreases) B1 2(c)(i) ΔU = 3/2 × k × ΔT × N C1 = 3/2 × 1.38 × 10–23 × (288 – 310) × 1.4 × 1023 C1 = – 64 J A1 2(c)(ii) decrease in internal energy is less than work done by gas M1 (thermal energy is) transferred to the gas (during the expansion) A1

This question in 9702/41 May/June 2021

2 An ideal gas is contained in a cylinder by means of a movable frictionless piston, as illustrated in Fig. 2.1. cylinder movement of piston piston gas molecule Fig. 2.1 Initially, the gas has a volume of 1.8 × 10−3 m3 at a pressure of 3.3 × 105 Pa and a temperature of 310 K. (a) Show that the number of gas molecules in the cylinder is 1.4 × 1023. [2] (b) Use kinetic theory to explain why, when the piston is moved so that the gas expands, this causes a decrease in the temperature of the gas. … … … … [3] (c) The gas expands so that its volume increases to 2.4 × 10−3 m3 at a pressure of 2.3 × 105 Pa and a temperature of 288 K, as shown in Fig. 2.2. 1.8 × 10−3 m3 2.4 × 10−3 m3 3.3 × 105 Pa 2.3 × 105 Pa 310 K 288 K Fig. 2.2 (i) The average translational kinetic energy EK of a molecule of an ideal gas is given by 3 EK = kT 2 where k is the Boltzmann constant and T is the thermodynamic temperature. Calculate the increase in internal energy ΔU of the gas during the expansion. ΔU = … J [3] (ii) The work done by the gas during the expansion is 76 J. Use your answer in (i) to explain whether thermal energy is transferred to or from the gas during the expansion. … … … [2] [Total: 10]

10 marks

Mark scheme: 2(a) pV = NkT C1 N = (1.8 × 10–3 × 3.3 × 105) / (1.38 × 10–23 × 310) = 1.4 × 1023 A1 or pV = nRT and nNA = N (C1) N = (1.8 × 10–3 × 3.3 × 105 × 6.02 × 1023) / (8.31 × 310) = 1.4 × 1023 (A1) 2(b) speed of molecule decreases on impact with moving piston B1 mean square speed (directly) proportional to (thermodynamic) temperature or mean square speed (directly) proportional to kinetic energy (of molecules) or kinetic energy (of molecules) (directly) proportional to (thermodynamic) temperature B1 kinetic energy (of molecules) decreases (so temperature decreases) B1 2(c)(i) ΔU = 3/2 × k × ΔT × N C1 = 3/2 × 1.38 × 10–23 × (288 – 310) × 1.4 × 1023 C1 = – 64 J A1 2(c)(ii) decrease in internal energy is less than work done by gas M1 (thermal energy is) transferred to the gas (during the expansion) A1

This question in 9702/43 May/June 2021

3 (a) Define specific heat capacity. … … … [2] (b) A sealed container of fixed volume V contains N molecules, each of mass m, of an ideal gas at pressure p. (i) State an expression, in terms of V, N, p and the Boltzmann constant k, for the thermodynamic temperature T of the gas. … [1] (ii) Show that the mean translational kinetic energy EK of a molecule of the gas is given by 3 EK = kT. 2 [2] (iii) Explain why the internal energy of the gas is equal to the total kinetic energy of the molecules. … … … [2] (c) The gas in (b) is supplied with thermal energy Q. (i) Explain, with reference to the first law of thermodynamics, why the increase in internal energy of the gas is Q. … … … [2] (ii) Use the expression in (b)(ii) and the information in (c)(i) to show that the specific heat capacity c of the gas is given by 3k c = . 2m [2] (d) The container in (b) is now replaced with one that does not have a fixed volume. Instead, the gas is able to expand, so that the pressure of the gas remains constant as thermal energy is supplied. Suggest, with a reason, how the specific heat capacity of the gas would now compare with the value in (c)(ii). … … … … [2] [Total: 13]

13 marks

Mark scheme: 3(a) (thermal) energy per unit mass (to cause temperature change) B1 (thermal) energy per unit change in temperature B1 3(b)(i) (T =) pV / Nk B1 3(b)(ii) (pV =) NkT = ⅓Nm<c2> or pV = NkT and pV = ⅓Nm<c2> M1 leading to ½m<c2> = (3/2)kT and ½m<c2> = EK A1 3(b)(iii) internal energy = ΣEK (of molecules) + ΣEP (of molecules) or no forces between molecules B1 potential energy of molecules is zero B1 3(c)(i) increase in internal energy = Q + work done B1 constant volume so no work done B1 3(c)(ii) c = Q / NmΔT C1 = [N × (3/2)kΔT] / (NmΔT) = 3k / 2m A1 3(d) (as it expands) gas does work (against the atmosphere/external pressure) B1 for same temperature rise) more (thermal) energy needed, so larger specific heat capacity B1

This question in 9702/41 Oct/Nov 2021

3 (a) One of the assumptions of the kinetic theory of gases is that all collisions involving molecules of the gas are elastic. (i) State what is meant by an elastic collision. … … [1] (ii) State two other assumptions of the kinetic theory of gases. 1. … … 2. … … [2] (b) A molecule of an ideal gas has mass m and is contained in a cubic box of side length L. The molecule is moving with velocity u towards the face of the box that is shaded in Fig. 3.1. L molecule u Fig. 3.1 The molecule collides elastically with the shaded face and the face opposite to it alternately. Deduce expressions, in terms of m, u and L, for: (i) the magnitude of the change in momentum of the molecule on colliding with a face change in momentum = … [1] (ii) the time between consecutive collisions of the molecule with the shaded face time = … [1] (iii) the average force exerted by the molecule on the shaded face force = … [1] (iv) the pressure on the shaded face if the force in (iii) is exerted over the whole area of the face. pressure = … [1] (c) When the model described in (b) is extended to three dimensions, and to a gas containing N molecules, each of mass m, travelling with mean-square speed 〈c2〉, it can be shown that 1 pV = 3 Nm〈c2〉 where p is the pressure exerted by the gas and V is the volume of the gas. Use this expression, together with the equation of state of an ideal gas, to show that the average translational kinetic energy EK of a molecule of an ideal gas is given by 3 EK = 2 kT where T is the thermodynamic temperature of the gas and k is the Boltzmann constant. [2] (d) The mass of a hydrogen molecule is 3.34 × 10–27 kg. Use the expression for EK in (c) to determine the root-mean-square (r.m.s.) speed of a molecule of hydrogen gas at 25 °C. r.m.s. speed = … m s–1 [2] [Total: 11]

11 marks

Mark scheme: 3(a)(i) no loss of kinetic energy B1 3(a)(ii) • molecules have negligible volume (compared with gas/container) • no forces between molecules (except during collisions) • molecules are in random motion • collisions are instantaneous Any two points, 1 mark each B2 3(b)(i) 2mu A1 3(b)(ii) 2L / u A1 3(b)(iii) force = change in momentum / time = 2mu / (2L / u) = mu2 / L A1 3(b)(iv) pressure = force / area = (mu2 / L) / L2 = mu2 / L3 A1 3(c) pV = NkT C1 NkT = ⅓Nm<c2> leading to ½m<c2> = (3/2)kT and ½m<c2> = EK A1 3(d) ½ × 3.34 × 10–27 × <c2> = (3/2) × 1.38 × 10–23 × (25 + 273) C1 r.m.s. speed = 1.9 × 103 m s–1 A1

This question in 9702/42 Oct/Nov 2021

3 (a) Define specific heat capacity. … … … [2] (b) A sealed container of fixed volume V contains N molecules, each of mass m, of an ideal gas at pressure p. (i) State an expression, in terms of V, N, p and the Boltzmann constant k, for the thermodynamic temperature T of the gas. … [1] (ii) Show that the mean translational kinetic energy EK of a molecule of the gas is given by 3 EK = kT. 2 [2] (iii) Explain why the internal energy of the gas is equal to the total kinetic energy of the molecules. … … … [2] (c) The gas in (b) is supplied with thermal energy Q. (i) Explain, with reference to the first law of thermodynamics, why the increase in internal energy of the gas is Q. … … … [2] (ii) Use the expression in (b)(ii) and the information in (c)(i) to show that the specific heat capacity c of the gas is given by 3k c = . 2m [2] (d) The container in (b) is now replaced with one that does not have a fixed volume. Instead, the gas is able to expand, so that the pressure of the gas remains constant as thermal energy is supplied. Suggest, with a reason, how the specific heat capacity of the gas would now compare with the value in (c)(ii). … … … … [2] [Total: 13]

13 marks

Mark scheme: 3(a) (thermal) energy per unit mass (to cause temperature change) B1 (thermal) energy per unit change in temperature B1 3(b)(i) (T =) pV / Nk B1 3(b)(ii) (pV =) NkT = ⅓Nm<c2> or pV = NkT and pV = ⅓Nm<c2> M1 leading to ½m<c2> = (3/2)kT and ½m<c2> = EK A1 3(b)(iii) internal energy = ΣEK (of molecules) + ΣEP (of molecules) or no forces between molecules B1 potential energy of molecules is zero B1 3(c)(i) increase in internal energy = Q + work done B1 constant volume so no work done B1 3(c)(ii) c = Q / NmΔT C1 = [N × (3/2)kΔT] / (NmΔT) = 3k / 2m A1 3(d) (as it expands) gas does work (against the atmosphere/external pressure) B1 for same temperature rise) more (thermal) energy needed, so larger specific heat capacity B1

This question in 9702/43 Oct/Nov 2021

3 A fixed mass of an ideal gas is initially at a temperature of 17 °C. The gas has a volume of 0.24 m3 and a pressure of 1.2 × 105 Pa. (a) (i) State what is meant by an ideal gas. … … … [2] (ii) Calculate the amount n of gas. n = … mol [2] (b) The gas undergoes three successive changes, as shown in Fig. 3.1. 4.0 C pressure / 105 Pa 3.0 2.0 B A 1.0 0 0.04 0.08 0.12 0.16 0.20 0.24 0.28 volume / m3 Fig. 3.1 The initial state is represented by point A. The gas is cooled at constant pressure to point B by the removal of 48.0 kJ of thermal energy. The gas is then heated at constant volume to point C. Finally, the gas expands at constant temperature back to its original pressure and volume at point A. During this expansion, the gas does 31.6 kJ of work. (i) Show that the magnitude of the work done during the change AB is 19.2 kJ. [2] (ii) Complete Table 3.1 to show the work done on the gas, the thermal energy supplied to the gas and the increase in internal energy of the gas, for each of the changes AB, BC and CA. Table 3.1 work done thermal energy increase in internal change on gas / kJ supplied to gas / kJ energy of gas / kJ AB – 48.0 BC CA – 31.6 [5] [Total: 11]

11 marks

Mark scheme: 3(a)(i) M1 where p = pressure, V = volume, T = thermodynamic temperature A1 3(a)(ii) T = (273 + 17) K C1 n = pV / RT = (1.2  105  0.24) / [8.31  (273 + 17)] = 12 mol A1 3(b)(i) work done = pV C1 = 1.2  105  (0.24 – 0.08) = 19200 J (= 19.2 kJ) A1 3(b)(ii) AB work done correct (19.2) A1 BC work done correct (0) A1 CA increase in internal energy correct (0) and CA thermal energy correct (31.6) A1 AB increase in internal energy calculated correctly from work done – 48.0 A1 BC increase in internal energy correctly calculated so the final column adds up to zero and BC thermal energy same as increase in internal energy (Fully correct table: AB 19.2 – 48.0 –28.8 BC 0 28.8 28.8 CA –31.6 31.6 0 ) A1

This question in 9702/41 May/June 2022

3 A fixed mass of an ideal gas is initially at a temperature of 17 °C. The gas has a volume of 0.24 m3 and a pressure of 1.2 × 105 Pa. (a) (i) State what is meant by an ideal gas. … … … [2] (ii) Calculate the amount n of gas. n = … mol [2] (b) The gas undergoes three successive changes, as shown in Fig. 3.1. 4.0 C pressure / 105 Pa 3.0 2.0 B A 1.0 0 0.04 0.08 0.12 0.16 0.20 0.24 0.28 volume / m3 Fig. 3.1 The initial state is represented by point A. The gas is cooled at constant pressure to point B by the removal of 48.0 kJ of thermal energy. The gas is then heated at constant volume to point C. Finally, the gas expands at constant temperature back to its original pressure and volume at point A. During this expansion, the gas does 31.6 kJ of work. (i) Show that the magnitude of the work done during the change AB is 19.2 kJ. [2] (ii) Complete Table 3.1 to show the work done on the gas, the thermal energy supplied to the gas and the increase in internal energy of the gas, for each of the changes AB, BC and CA. Table 3.1 work done thermal energy increase in internal change on gas / kJ supplied to gas / kJ energy of gas / kJ AB – 48.0 BC CA – 31.6 [5] [Total: 11]

11 marks

Mark scheme: 3(a)(i) M1 where p = pressure, V = volume, T = thermodynamic temperature A1 3(a)(ii) T = (273 + 17) K C1 n = pV / RT = (1.2  105  0.24) / [8.31  (273 + 17)] = 12 mol A1 3(b)(i) work done = pV C1 = 1.2  105  (0.24 – 0.08) = 19200 J (= 19.2 kJ) A1 3(b)(ii) AB work done correct (19.2) A1 BC work done correct (0) A1 CA increase in internal energy correct (0) and CA thermal energy correct (31.6) A1 AB increase in internal energy calculated correctly from work done – 48.0 A1 BC increase in internal energy correctly calculated so the final column adds up to zero and BC thermal energy same as increase in internal energy (Fully correct table: AB 19.2 – 48.0 –28.8 BC 0 28.8 28.8 CA –31.6 31.6 0 ) A1

This question in 9702/43 May/June 2022

3 (a) The equation of state for an ideal gas can be written as pV = NkT. State the meaning of each of the symbols in this equation. p: … V: … N: … k: … T: … [3] (b) Use the equation in (a) to show that the average translational kinetic energy EK of a molecule of an ideal gas is given by 3 EK = 2 kT. [2] (c) The mass of an oxygen molecule is 5.31 × 10–26 kg. Assume that oxygen behaves as an ideal gas. (i) Use the equation in (b) to determine the root-mean-square (r.m.s.) speed u of an oxygen molecule at 23 °C. u = … m s–1 [3] (ii) A fixed mass of oxygen gas at initial pressure P is sealed in a cylindrical container by a movable piston at one end, as shown in Fig. 3.1. oxygen piston cylinder Fig. 3.1 The temperature of the gas is 23 °C. The piston is slowly moved into the cylinder so that the oxygen gas is compressed. At all times, the gas and the container remain in thermal equilibrium with the surroundings. On Fig. 3.2, sketch the variation with pressure of the r.m.s. speed of the oxygen molecules as the pressure increases. r.m.s. speed u 0 P pressure Fig. 3.2 [2] [Total: 10]

10 marks

Mark scheme: 3(a) p = pressure (of gas), V = volume (of gas) and k = Boltzmann constant B1 N = number of molecules B1 T = thermodynamic temperature B1 3(b) (pV = NkT and pV = ⅓Nm<c2> leading to) NkT = ⅓Nm<c2> M1 algebra leading to (3/2)kT = ½m<c2> and use of ½m<c2> = EK leading to (3/2)kT = EK A1 3(c)(i) T = 296 K C1 ½m<c2> = (3/2)kT C1 ½  5.31  10–26  u2 = (3/2)  1.38  10–23  296 u = 480 m s–1 A1 3(c)(ii) line passing through (P, u) B1 horizontal straight line B1

This question in 9702/42 Oct/Nov 2022

2 (a) State what is meant by an ideal gas. … … … [2] (b) A fixed amount of helium gas is sealed in a container. The helium gas has a pressure of 1.10 × 105 Pa, and a volume of 540 cm3 at a temperature of 27 °C. The volume of the container is rapidly decreased to 30.0 cm3. The pressure of the helium gas increases to 6.70 × 106 Pa and its temperature increases to 742 °C, as illustrated in Fig. 2.1. initial state final state 1.10 × 105 Pa 6.70 × 106 Pa 540 cm3 30.0 cm3 27 °C 742 °C Fig. 2.1 No thermal energy enters or leaves the helium gas during this process. (i) Show that the helium gas behaves as an ideal gas. [2] (ii) The first law of thermodynamics may be expressed as ΔU = q + W. Use the first law of thermodynamics to explain why the temperature of the helium gas increases. … … … … … [2] (iii) The average translational kinetic energy EK of a molecule of an ideal gas is given by 3 EK = kT 2 where k is the Boltzmann constant and T is the thermodynamic temperature. Calculate the change in the total kinetic energy of the molecules of the helium gas. change in kinetic energy = … J [3] (c) The mass of nitrogen gas in another container is 24.0 g at a temperature of 27 °C. The gas is cooled to its boiling point of –196 °C. Assume all the gas condenses to a liquid. For this change the specific heat capacity of nitrogen gas is 1.04 kJ kg–1 K–1. The specific latent heat of vaporisation of nitrogen is 199 kJ kg–1. Determine the thermal energy, in kJ, removed from the nitrogen gas. energy = … kJ [3] [Total: 12]

12 marks

Mark scheme: 2(a) gas for which pV  T M1 where T is thermodynamic temperature A1 2(b)(i) evidence of two temperature conversions between C and K B1 two calculations shown, one for each state e.g. A1 1.10  105  540  10 −6 6.70  10 6  30  10 −6 = 0.198 and = 0.198 ( 273 + 27 ) ( 273 + 742 ) 2(b)(ii) work is done on the gas M1 internal energy increases (so temperature increases) A1 2(b)(iii) pV = NkT e.g. C1 1.10  10 5  540  10 −6 N = 1.38  10 −23  300 = 1.435  1022 Ek = (3 / 2) kTN 1.10  10 5  540  10 −6 C1 = (3 / 2)  1.38  1023  (742 – 27)  1.38  10 −23  300 = 212 J A1 2(c) E = mc and E = mL C1 = (27 + 196) or 223 C1 E = 0.0240  1.04  (27 + 196) + 0.0240  199 A1 = 10.3 kJ

This question in 9702/42 Feb/March 2023

4 (a) State two of the basic assumptions of the kinetic theory of gases. 1 … … 2 … … [2] (b) An ideal gas has amount of substance n. The gas is initially in state X, with pressure 2p and volume V. The gas is cooled at constant volume to state Y, with pressure p. The gas is then heated at constant pressure to state Z, with volume 2V. Finally, the gas returns at constant temperature to state X. (i) Determine an expression for the temperature T of the gas in state X, in terms of n, p and V. Identify any other symbols that you use. [2] (ii) On Fig. 4.1, sketch the variation with volume of pressure for the gas as the gas undergoes the three changes. The state X is labelled. Label states Y and Z. 2p X pressure p 0 0 V 2V volume Fig. 4.1 [3] (iii) During the change of state from Y to Z, the increase in internal energy of the gas is U. During the change of state from Z to X, the work done on the gas is W. Complete Table 4.1 to indicate, for each of the three changes of state, the increase in internal energy of the gas, the thermal energy transferred to the gas and the work done on the gas, in terms of p, V, U and W. Table 4.1 increase in internal thermal energy change work done on gas energy of gas transferred to gas X to Y Y to Z +U Z to X +W [5] [Total: 12]

12 marks

Mark scheme: 4(a) particles are in (continuous) random motion  particles have negligible volume (compared with the gas)  negligible forces between particles (except during collisions)  (all) collisions (perfectly) elastic  time of collision negligible (in comparison with time between collisions) Any two points, 1 mark each B2 4(b)(i) (general starting equation) pV = nRT C1 T = (2pV / nR) where R is the (molar) gas constant A1 4(b)(ii) sketch: straight vertical line XY from (V, 2p) to (V, p) B1 straight horizontal line YZ from (V, p) to (2V, p) B1 curve with gradient increasing from Z to X from (2V, p) to (V, 2p) B1 4(b)(iii) XY work done on gas correct (= 0) B1 ZX increase in internal energy correct (= 0) B1 YZ work done on gas correct (= –pV) B1 XY increase in internal energy such that the increase in internal energy column adds up to zero B1 all three thermal energies transferred such that U = q + w in each row (completely correct answer: change U Q w X to Y Y to Z Z to X –U [ +U ] 0 –U U + pV –W 0 –pV [ +W ] ) B1

This question in 9702/41 May/June 2023

2 (a) (i) State what is meant by an ideal gas. … … … [2] (ii) State the temperature, in degrees Celsius, of absolute zero. temperature = … °C [1] (b) A sealed vessel contains a mass of 0.0424 kg of an ideal gas at 227 °C. The pressure of the gas is 1.37 × 105 Pa and the volume of the gas is 0.640 m3. Calculate: (i) the number of molecules of the gas in the vessel number of molecules = … [3] (ii) the mass of one molecule of the gas mass = … kg [1] (iii) the root-mean-square (r.m.s.) speed v of the molecules of the gas. v = … m s–1 [3] (c) The gas in (b) is now cooled gradually to absolute zero. On Fig. 2.1, sketch the variation with thermodynamic temperature T of the r.m.s. speed of the molecules of the gas. v r.m.s. speed 0 0 500 T / K Fig. 2.1 [2] [Total: 12]

12 marks

Mark scheme: 2(a)(i) M1 where T is thermodynamic temperature A1 2(a)(ii) temperature = –273.15 °C A1 2(b)(i) pV = NkT C1 N = (1.37  105  0.640) / (1.38  10–23  (227 + 273)) C1 = 1.27  1025 A1 2(b)(ii) mass = 0.0424 / (1.27  1025) = 3.34  10–27 kg A1 2(b)(iii) ½m<c2> = (3 / 2)kT C1 3.34  10–27  v2 = 3  1.38  10–23  500 C1 v = 2490 m s–1 A1 or pV = ⅓(Nm) <c2> and Nm = mass of gas (C1) 0.0424  v2 = 3  1.37  105  0.640 (C1) v = 2490 m s–1 (A1) 2(c) sketch: line from (0, 0) to (500, v) B1 line with decreasing positive gradient throughout B1

This question in 9702/42 May/June 2023

4 (a) State two of the basic assumptions of the kinetic theory of gases. 1 … … 2 … … [2] (b) An ideal gas has amount of substance n. The gas is initially in state X, with pressure 2p and volume V. The gas is cooled at constant volume to state Y, with pressure p. The gas is then heated at constant pressure to state Z, with volume 2V. Finally, the gas returns at constant temperature to state X. (i) Determine an expression for the temperature T of the gas in state X, in terms of n, p and V. Identify any other symbols that you use. [2] (ii) On Fig. 4.1, sketch the variation with volume of pressure for the gas as the gas undergoes the three changes. The state X is labelled. Label states Y and Z. 2p X pressure p 0 0 V 2V volume Fig. 4.1 [3] (iii) During the change of state from Y to Z, the increase in internal energy of the gas is U. During the change of state from Z to X, the work done on the gas is W. Complete Table 4.1 to indicate, for each of the three changes of state, the increase in internal energy of the gas, the thermal energy transferred to the gas and the work done on the gas, in terms of p, V, U and W. Table 4.1 increase in internal thermal energy change work done on gas energy of gas transferred to gas X to Y Y to Z +U Z to X +W [5] [Total: 12]

12 marks

Mark scheme: 4(a) particles are in (continuous) random motion  particles have negligible volume (compared with the gas)  negligible forces between particles (except during collisions)  (all) collisions (perfectly) elastic  time of collision negligible (in comparison with time between collisions) Any two points, 1 mark each B2 4(b)(i) (general starting equation) pV = nRT C1 T = (2pV / nR) where R is the (molar) gas constant A1 4(b)(ii) sketch: straight vertical line XY from (V, 2p) to (V, p) B1 straight horizontal line YZ from (V, p) to (2V, p) B1 curve with gradient increasing from Z to X from (2V, p) to (V, 2p) B1 4(b)(iii) XY work done on gas correct (= 0) B1 ZX increase in internal energy correct (= 0) B1 YZ work done on gas correct (= –pV) B1 XY increase in internal energy such that the increase in internal energy column adds up to zero B1 all three thermal energies transferred such that U = q + w in each row (completely correct answer: change U Q w X to Y Y to Z Z to X –U [ +U ] 0 –U U + pV –W 0 –pV [ +W ] ) B1

This question in 9702/43 May/June 2023

3 (a) The product pV for an ideal gas is given by 1 pV = Nm〈c2〉 3 where p is the pressure of the gas and V is the volume of the gas. (i) State the meaning of the symbols N, m and 〈c2〉 in this equation. N: … m: … 〈c2〉: … [3] (ii) Use the equation of state for an ideal gas to show that the average translational kinetic energy EK of a molecule of the gas at thermodynamic temperature T is given by 3 EK = kT. 2 [2] (b) The surface of a star consists mainly of a gas that may be assumed to be ideal. The molecules of the gas have a root-mean-square (r.m.s.) speed of 9300 m s–1. The mass of a molecule of the gas is 3.34 × 10–27 kg. Determine, to three significant figures, the temperature of the surface of the star. temperature = … K [2] (c) The radiant flux intensity of the radiation from the star in (b) is 2.52 × 10–8 W m–2 when observed at a distance of 4.16 × 1016 m from the star. (i) Calculate the luminosity of the star. Give a unit with your answer. luminosity = … unit … [2] (ii) Determine the radius of the star. radius = … m [2] (d) The gas at the surface of a star has a very high pressure. Use the basic assumptions of the kinetic theory to suggest why, in practice, a gas at the surface of a star is unlikely to behave as an ideal gas. … … … … [2] [Total: 13]

13 marks

Mark scheme: 3(a)(i) N: number of molecules (of the gas) B1 m: mass of one molecule (of the gas) B1 <c2>: mean square speed (of molecules) B1 3(a)(ii) pV = NkT M1 NkT = ⅓Nm<c2> and EK = ½m<c2> leading to EK = (3/2) kT A1 3(b) ½  3.34  10–27  93002 = (3/2)  1.38  10–23  T C1 T = 6980 K A1 3(c)(i) L = F  4d2 C1 L = 2.52  10–8  4  (4.16  1016)2 A1 = 5.48  1026 W 3(c)(ii) L = 4r2T4 C1 5.48  1026 = 4  5.67  10–8  r2  69804 r = 5.69  108 m A1 3(d) (very high pressure so) molecules are (very) close together (not just ‘nearer’) B1 forces between molecules are not negligible B1 or volume of molecules not negligible compared with gas volume

This question in 9702/41 Oct/Nov 2023

2 (a) (i) Define gravitational potential at a point. … … … [2] (ii) The Moon may be considered to be an isolated uniform sphere of mass 7.3 × 1022 kg and radius 1.7 × 106 m. Calculate the gravitational potential at the surface of the Moon. Give a unit with your answer. gravitational potential = … unit … [2] (b) An isolated uniform spherical planet has gravitational potential φ at its surface. A particle of mass m is projected vertically upwards from the surface. The particle is given just enough kinetic energy to travel to an infinite distance away from the planet, escaping from the gravitational pull of the planet, without any additional work being done on it. (i) Determine an expression, in terms of m and φ, for the gravitational potential energy EP of the particle at the surface of the planet. EP = … [1] (ii) Show that the speed v at which the particle is projected upwards from the surface of the planet is given by v = –2φ. [2] (c) A particle is moving upwards at the surface of the Moon. Use your answer in (a)(ii) and the expression in (b)(ii) to determine the minimum speed of this particle that will result in it escaping from the gravitational pull of the Moon. speed = … m s–1 [1] (d) Hydrogen may be assumed to be an ideal gas. The mass of a hydrogen molecule is 3.34 × 10–27 kg. Calculate the root-mean-square (r.m.s.) speed of a hydrogen molecule in hydrogen gas that is at a temperature of 400 K. r.m.s. speed = … m s–1 [3] (e) The surface of the Moon reaches temperatures of approximately 400 K when in direct sunlight. Use your answers in (c) and (d) to suggest a reason why the Moon does not have an atmosphere consisting of hydrogen. … … [1] [Total: 12]

12 marks

Mark scheme: 2(a)(i) work done per unit mass B1 work (done) moving mass from infinity (to the point) B1 2(a)(ii)  = –GM / r C1 = – (6.67  10–11  7.3  1022) / (1.7  106) = – 2.9  106 J kg–1 A1 2(b)(i) EP = m B1 2(b)(ii) ½mv2 + m= 0 M1 correct algebra leading to v = √(–2) A1 2(c) speed = √(2  2.9  106) A1 = 2400 m s–1 2(d) ½m<c2> = (3/2)kT C1 3.34  10–27  <c2> = 3  1.38  10–23  400 C1 cr.m.s. = 2200 m s–1 A1 2(e) r.m.s. speed is an average so many molecules have speeds greater than the escape speed B1 or there is a distribution of molecular speeds (around the r.m.s. value) so many molecules have speeds greater than the escape speed

This question in 9702/42 Oct/Nov 2023

3 (a) The product pV for an ideal gas is given by 1 pV = Nm〈c2〉 3 where p is the pressure of the gas and V is the volume of the gas. (i) State the meaning of the symbols N, m and 〈c2〉 in this equation. N: … m: … 〈c2〉: … [3] (ii) Use the equation of state for an ideal gas to show that the average translational kinetic energy EK of a molecule of the gas at thermodynamic temperature T is given by 3 EK = kT. 2 [2] (b) The surface of a star consists mainly of a gas that may be assumed to be ideal. The molecules of the gas have a root-mean-square (r.m.s.) speed of 9300 m s–1. The mass of a molecule of the gas is 3.34 × 10–27 kg. Determine, to three significant figures, the temperature of the surface of the star. temperature = … K [2] (c) The radiant flux intensity of the radiation from the star in (b) is 2.52 × 10–8 W m–2 when observed at a distance of 4.16 × 1016 m from the star. (i) Calculate the luminosity of the star. Give a unit with your answer. luminosity = … unit … [2] (ii) Determine the radius of the star. radius = … m [2] (d) The gas at the surface of a star has a very high pressure. Use the basic assumptions of the kinetic theory to suggest why, in practice, a gas at the surface of a star is unlikely to behave as an ideal gas. … … … … [2] [Total: 13]

13 marks

Mark scheme: 3(a)(i) N: number of molecules (of the gas) B1 m: mass of one molecule (of the gas) B1 <c2>: mean square speed (of molecules) B1 3(a)(ii) pV = NkT M1 NkT = ⅓Nm<c2> and EK = ½m<c2> leading to EK = (3/2) kT A1 3(b) ½  3.34  10–27  93002 = (3/2)  1.38  10–23  T C1 T = 6980 K A1 3(c)(i) L = F  4d2 C1 L = 2.52  10–8  4  (4.16  1016)2 A1 = 5.48  1026 W 3(c)(ii) L = 4r2T4 C1 5.48  1026 = 4  5.67  10–8  r2  69804 r = 5.69  108 m A1 3(d) (very high pressure so) molecules are (very) close together (not just ‘nearer’) B1 forces between molecules are not negligible B1 or volume of molecules not negligible compared with gas volume

This question in 9702/43 Oct/Nov 2023

3 (a) (i) State what is meant by an ideal gas. … … … [2] (ii) Use one of the basic assumptions of the kinetic theory to explain what can be deduced about the potential energy associated with the random motion of molecules in an ideal gas. … … … [2] (b) A sample of 0.26 m3 of an ideal gas is at pressure 2.0 × 105 Pa and temperature 290 K. Determine: (i) the number N of molecules of the gas N = … [2] (ii) the average translational kinetic energy EK of one molecule of the gas EK = … J [2] (iii) the internal energy of the gas. Explain your reasoning. internal energy = … J [2] (c) The volume V of the gas in (b) is now varied, keeping its pressure constant. On Fig. 3.1, sketch the variation with V of the internal energy U of the gas. U 0 0 V Fig. 3.1 [2] [Total: 12]

12 marks

Mark scheme: 3(a)(i) M1 where T is thermodynamic temperature A1 3(a)(ii) no intermolecular forces B1 (so) potential energy is zero B1 3(b)(i) pV = NkT C1 N = (2.0  105  0.26) / (1.38  10–23  290) = 1.3  1025 A1 3(b)(ii) EK = (3/2) kT C1 EK = (3/2)  1.38  10–23  290 = 6.0  10–21 J A1 3(b)(iii) internal energy = total KE + PE of molecules or PE = 0 so internal energy = total KE of molecules B1 internal energy = 1.3  1025  6.0  10–21 = 7.8  104 J A1 3(c) straight line with positive gradient B1 line passing through the origin B1

This question in 9702/41 May/June 2024

2 (a) With reference to thermal energy, state what is meant by two objects being in thermal equilibrium. … … … [1] (b) Two cylinders X and Y each contain a sample of an ideal gas. The samples are in thermal equilibrium with each other. X has a volume of 0.0260 m3 and contains 0.740 mol of gas at a pressure of 1.20 × 105 Pa. Y has a volume of 0.0430 m3 and contains gas at a pressure of 2.90 × 105 Pa. Data for the two cylinders are shown in Fig. 2.1. X Y 0.740 mol 1.20 × 105 Pa 2.90 × 105 Pa 0.0260 m3 0.0430 m3 Fig. 2.1 (i) Show that the temperature of the gas in X is 234 °C. [3] (ii) Determine the number N of molecules of the gas in Y. Explain your reasoning. N = … [3] (iii) The gas in X consists of molecules that each have a mass that is four times the mass of a molecule of the gas in Y. Explain how the root-mean-square (r.m.s.) speed of the molecules in X compares with the r.m.s. speed of the molecules in Y. … … … … … [3] [Total: 10]

10 marks

Mark scheme: 2(a) (if in thermal contact) no net transfer of (thermal) energy (between them) B1 2(b)(i) pV = nRT C1 T = (1.20  105  0.0260) / (0.740  8.31) ( = 507 K) M1 temperature = 507 – 273 = 234 °C A1 2(b)(ii) thermal equilibrium so temperatures (of X and Y) are equal B1 pV = NkT C1 N = (2.90  105  0.0430) / (1.38  10–23  507) = 1.78  1024 A1 2(b)(iii)  (molecular) kinetic energy is proportional to temperature or kinetic energy (of molecules) is same in both cylinders  kinetic energy proportional to mass  mean-square speed or temperature proportional to mass  mean-square speed or r.m.s. speed proportional to √(temperature / mass)  mean-square speed inversely proportional to mass or r.m.s. speed inversely proportional to √(mass) Any two bulleted points, 1 mark each B2 r.m.s. speed (of molecules) in X is half r.m.s. speed (of molecules) in Y B1

This question in 9702/42 May/June 2024

3 (a) (i) State what is meant by an ideal gas. … … … [2] (ii) Use one of the basic assumptions of the kinetic theory to explain what can be deduced about the potential energy associated with the random motion of molecules in an ideal gas. … … … [2] (b) A sample of 0.26 m3 of an ideal gas is at pressure 2.0 × 105 Pa and temperature 290 K. Determine: (i) the number N of molecules of the gas N = … [2] (ii) the average translational kinetic energy EK of one molecule of the gas EK = … J [2] (iii) the internal energy of the gas. Explain your reasoning. internal energy = … J [2] (c) The volume V of the gas in (b) is now varied, keeping its pressure constant. On Fig. 3.1, sketch the variation with V of the internal energy U of the gas. U 0 0 V Fig. 3.1 [2] [Total: 12]

12 marks

Mark scheme: 3(a)(i) M1 where T is thermodynamic temperature A1 3(a)(ii) no intermolecular forces B1 (so) potential energy is zero B1 3(b)(i) pV = NkT C1 N = (2.0  105  0.26) / (1.38  10–23  290) = 1.3  1025 A1 3(b)(ii) EK = (3/2) kT C1 EK = (3/2)  1.38  10–23  290 = 6.0  10–21 J A1 3(b)(iii) internal energy = total KE + PE of molecules or PE = 0 so internal energy = total KE of molecules B1 internal energy = 1.3  1025  6.0  10–21 = 7.8  104 J A1 3(c) straight line with positive gradient B1 line passing through the origin B1

This question in 9702/43 May/June 2024

4 (a) State three of the basic assumptions of the kinetic theory of gases. 1 … … 2 … … 3 … … [3] (b) Explain how molecular movement causes the pressure exerted by a gas. … … … … … [3] (c) Fig. 4.1 shows the variation with thermodynamic temperature T of the mean‑square speeds 〈c2〉 for two gases X and Y. 6 〈c 2〉 / 106 m2 s–2 X 4 Y 2 0 0 100 200 300 400 T / K Fig. 4.1 Fig. 4.2 shows the variation with T of the product pV for samples of the two gases, where p is the pressure of the gas and V is the volume of the gas. 3 Y pV / 103 J 2 1 X 0 0 100 200 300 400 T / K Fig. 4.2 State three conclusions about the gases and their samples that may be drawn from Fig. 4.1 and Fig. 4.2. The conclusions may be qualitative or quantitative. Use the space below for any working that you need. 1 … … 2 … … 3 … … [3] [Total: 9]

9 marks

Mark scheme: 4(a) • molecules are in (constant) random motion B3 • (all) collisions between molecules are (perfectly) elastic • no forces between molecules (except during collisions) • volume of molecules is negligible (compared with volume of gas) • collisions involving molecules are instantaneous Any three points, 1 mark each 4(b) • molecules collide with (walls of) container B3 • momentum of molecule changes during collision (with walls) • change in momentum is caused by force on molecule by wall • molecule experiences force from wall so molecule exerts force on wall • many molecules exerting force across the area of the wall leads to pressure (on the wall) Any three points, 1 mark each 4(c) Any three bulleted points from: B3 • both gases are ideal Up to 2 points from: • mass of one molecule of gas X is 3.3  10–27 kg • mass of one molecule of gas Y is 6.6  10–27 kg • mass of one molecule of gas Y is double mass of one molecule of gas X Up to 2 points from: • sample of X contains 0.27 mol / 1.6  1023 molecules • sample of Y contains 0.81 mol / 4.9  1023 molecules • sample of Y contains treble the amount of gas / number of molecules as sample of X Up to 2 points from: • mass of gas X is 5.4  10–4 kg • mass of gas Y is 3.2  10–3 kg • mass of gas Y is six times mass of gas X

This question in 9702/42 Oct/Nov 2024

3 (a) Two metal cuboids P and Q are in thermal contact with each other. (i) P and Q are in thermal equilibrium. State what is meant by the term thermal equilibrium. … … … [2] (ii) Data for P and Q are given in Table 3.1. Table 3.1 P Q specific heat capacity / J kg–1 K–1 390 910 mass / kg 0.54 0.37 P and Q are initially both at the same temperature. P is supplied with 24 kJ of thermal energy. After some time, P and Q are once again both at the same temperature as each other. P and Q are perfectly insulated from the surroundings. Determine the change in temperature ΔT of Q. ΔT = … K [3] (b) Nitrogen may be assumed to be an ideal gas. A fixed amount of nitrogen gas is contained at a constant pressure of 1.6 × 105 Pa. The variation of the volume V of the gas with the temperature θ of the gas is shown in Fig. 3.1. 0.4 V / m3 0.3 0.2 0.1 0 0 100 200 300 θ / °C Fig. 3.1 (i) The temperature of the nitrogen gas is increased from 0 °C to 210 °C. Determine the work done on the gas. work done = … J [3] (ii) Determine the number N of molecules of nitrogen gas. N = … [2] (iii) The mass of a nitrogen molecule is 4.7 × 10–26 kg. Calculate the root‑mean‑square (r.m.s.) speed of a nitrogen molecule at 210 °C. r.m.s. speed = … m s–1 [2] [Total: 12]

12 marks

Mark scheme: 3(a)(i) (P and Q are at the) same temperature B1 no net transfer of thermal energy (between P and Q) B1 3(a)(ii) Q = mcT C1 24  103 = (0.54  390  T) + (0.37  910  T) C1 T = 44 K A1 3(b)(i) work done = pV C1 = (1.6  105)  (0.18 – 0.32) C1 = –2.2  104 J A1 3(b)(ii) pV = NkT C1 N = (1.6  105  0.18) / (1.38  10–23  273) A1 = 7.6  1024 3(b)(iii) ½m<c2> = (3 / 2)kT C1 r.m.s. speed = √[(3  1.38  10–23  (210 + 273) / (4.7  10–26)] A1 = 650 m s–1

This question in 9702/42 Feb/March 2025

4 (a) The equation of state for an ideal gas may be written as pVA = NBT where p is the pressure of the gas, V is the volume of the gas, A is the Avogadro constant, B is another constant and N is the number of molecules of the gas. (i) State the meaning, in the equation, of the symbol T. … [1] (ii) Identify the constant B. … [1] (b) The product pV for an ideal gas is also given by pV = 1 Nm 〈c 2〉. 3 (i) State the meanings, in this equation, of the symbols m and 〈c 2〉. m: … 〈c 2〉: … [2] (ii) Use the equations in (a) and (b) to derive an expression, in terms of A, B and T, for the mean kinetic energy EK of a molecule of the gas. EK = … [2] (c) On Fig. 4.1, sketch the variation with T of the root-mean-square (r.m.s.) speed of the molecules of an ideal gas. r.m.s. speed 0 0 T Fig. 4.1 [2] [Total: 8]

8 marks

Mark scheme: 4(a)(i) thermodynamic temperature B1 4(a)(ii) molar gas constant B1 4(b)(i) m: mass of one molecule (of the gas) B1 〈c2〉: mean-square speed (of molecules) B1 4(b)(ii) 1 M1 NBT / A = Nm〈c2〉 3 1 A1 clear use of EK = m〈c2〉 leading to EK = 3BT / 2A 2 4(c) line with positive gradient passing through the origin B1 smooth curve with decreasing positive gradient B1

This question in 9702/42 May/June 2025

3 (a) State what is meant by an ideal gas. … … … [2] (b) An ideal gas at a pressure of 1.6 × 105 Pa has a density of 1.9 kg m–3. (i) Show that the root-mean-square (r.m.s.) speed of molecules of this gas is approximately 500 m s–1. [3] (ii) One molecule of the gas has a mass of 4.7 × 10–26 kg. Determine the thermodynamic temperature of the gas. temperature = … K [2] (c) Calculate the internal energy U of 6.0 mol of the gas in (b). Explain your reasoning. U = … J [3] [Total: 10]

10 marks

Mark scheme: 3(a) (gas that obeys) pV  T (at all values of p, V and T) M1 where T is thermodynamic temperature A1 3(b)(i) pV = ⅓ Nm〈c2〉 and Nm / V =  C1 (p = ⅓ 〈c2〉 ) r.m.s. speed = √〈c2〉 C1 1.6  105 = ⅓  1.9 × 〈c2〉 leading to r.m.s. speed = 500 m s–1 A1 or r.m.s. speed = √(3  1.6  105 / 1.9) = 500 m s–1 3(b)(ii) (pV =) ⅓ Nm〈c2〉 = NkT C1 (so) ½ m〈c2〉 = (3 / 2) kT ½  4.7  10–26  5032 = (3 / 2)  1.38  10–23  T A1 T = 290 K or pV = NkT and Nm / V =  (C1) (so) T = pm / k T = 1.6  105  4.7  10–26 / (1.9  1.38  10–23) (A1) = 290 K 3(c) potential energy (of molecules) is zero B1 U = N  ½ m〈c2〉 = 6.0  6.02  1023  ½  4.7  10–26  5032 C1 or U = N  (3 / 2) kT = 6.0  6.02  1023  (3 / 2)  1.38  10–23  287 or U = n  (3 / 2) RT = 6.0  (3 / 2)  8.31  287 U = 21000 J A1

This question in 9702/44 May/June 2025

4 (a) State the value of absolute zero on: (i) the Celsius temperature scale temperature = … °C [1] (ii) the thermodynamic temperature scale. Give a unit with your answer. temperature = … unit … [1] (b) A sample contains a fixed amount of gas. The gas has pressure p, volume V and thermodynamic temperature T. Fig. 4.1 shows the variation of pV with kT for the sample, where k is the Boltzmann constant. 300 pV / J 200 100 0 0 2 4 6 8 kT / 10–21 J Fig. 4.1 (i) State what is indicated about the nature of the gas from the variation shown in Fig. 4.1. … [1] (ii) Determine the number N of molecules of the gas in the sample. N = … [2] (iii) Use your answer in (b)(ii) to determine the amount n of gas in the sample. n = … mol [1] (c) The root-mean-square (r.m.s.) speed of the molecules of the gas is 1900 m s–1 when pV is equal to 270 J. Determine the mass, in u, of one molecule of the gas, where u is the unified atomic mass unit. mass = … u [4] [Total: 10]

10 marks

Mark scheme: 4(a)(i) temperature = –273.15 °C A1 4(a)(ii) temperature = 0 K A1 4(b)(i) gas is ideal B1 4(b)(ii) pV = NkT C1 N = 270 / (8.0  10–21) A1 = 3.4  1022 4(b)(iii) n = (3.4  1022) / (6.02  1023) A1 = 0.056 mol 4(c) ½ m<c2> = (3 / 2) kT C1 ½  m  19002 = 1.5  8.0  10–21 C1 (m = 6.65  10–27 kg) m = (6.65  10–27) / (1.66  10–27) C1 = 4.0 u A1

This question in 9702/41 Oct/Nov 2025

2 (a) State Newton’s law of gravitation. … … … [2] (b) One of the basic assumptions of the kinetic theory of gases is that there are no forces exerted between the molecules of the gas except during collisions. State two other basic assumptions of the kinetic theory of gases. 1 … … 2 … … [2] (c) Hydrogen gas consists of molecules that each have a mass of 3.34 × 10–27 kg. Hydrogen may be considered to be an ideal gas. A spherical balloon contains 0.0160 mol of hydrogen gas at a temperature of 282 K. At this temperature, the volume of gas in the balloon is 1.87 × 10– 4 m3. (i) Determine the pressure of the gas. pressure = … Pa [2] (ii) Estimate the average separation of the hydrogen molecules in the gas. average separation = … m [2] (d) (i) Use your answer in (c)(ii) to calculate the average gravitational force between adjacent molecules in hydrogen gas. average force = … N [2] (ii) By considering the weight of a molecule, suggest with a reason whether your answer in (d)(i) is consistent with the assumption of the kinetic theory of gases that there are no forces exerted between molecules. … … … [1] [Total: 11]

11 marks

Mark scheme: 2(a) (gravitational) force is (directly) proportional to product of masses B1 force (between point masses) is inversely proportional to the square of their separation B1 2(b) Any two points from: B2 • molecules are in continuous random motion • molecules have negligible volume compared with volume of gas • collisions (involving molecules) are (perfectly) elastic • collisions (of molecules) are instantaneous 2(c)(i) pV = nRT C1 p = (0.0160  8.31  282) / (1.87  10–4) A1 = 2.01  105 Pa 2(c)(ii) number of molecules = 0.0160  6.02  1023 C1 separation = 3√[(1.87  10–4) / (0.0160  6.02  1023)] A1 = 2.7  10–9 m (allow any answer that is 3 10–9 m to one significant figure) 2(d)(i) F = 6.67  10–11  (3.34  10–27)2 / (2.7  10–9)2 C1 = 1.0  10–46 N A1 2(d)(ii) numerical comparison between 10–46 N (F) and 10–26 N (the weight of molecule) leading to a conclusion that the assumption B1 is supported

This question in 9702/42 Oct/Nov 2025

4 (a) State the value of absolute zero on: (i) the Celsius temperature scale temperature = … °C [1] (ii) the thermodynamic temperature scale. Give a unit with your answer. temperature = … unit … [1] (b) A sample contains a fixed amount of gas. The gas has pressure p, volume V and thermodynamic temperature T. Fig. 4.1 shows the variation of pV with kT for the sample, where k is the Boltzmann constant. 300 pV / J 200 100 0 0 2 4 6 8 kT / 10–21 J Fig. 4.1 (i) State what is indicated about the nature of the gas from the variation shown in Fig. 4.1. … [1] (ii) Determine the number N of molecules of the gas in the sample. N = … [2] (iii) Use your answer in (b)(ii) to determine the amount n of gas in the sample. n = … mol [1] (c) The root-mean-square (r.m.s.) speed of the molecules of the gas is 1900 m s–1 when pV is equal to 270 J. Determine the mass, in u, of one molecule of the gas, where u is the unified atomic mass unit. mass = … u [4] [Total: 10]

10 marks

Mark scheme: 4(a)(i) temperature = –273.15 °C A1 4(a)(ii) temperature = 0 K A1 4(b)(i) gas is ideal B1 4(b)(ii) pV = NkT C1 N = 270 / (8.0  10–21) A1 = 3.4  1022 4(b)(iii) n = (3.4  1022) / (6.02  1023) A1 = 0.056 mol 4(c) ½ m<c2> = (3 / 2) kT C1 ½  m  19002 = 1.5  8.0  10–21 C1 (m = 6.65  10–27 kg) m = (6.65  10–27) / (1.66  10–27) C1 = 4.0 u A1

This question in 9702/43 Oct/Nov 2025

2 (a) The equation of state for an ideal gas may be expressed as pV = NkT. (i) State the meaning of each of the symbols in this equation. p: … V: … N: … k: … T: … [3] (ii) Using the equation of state, derive an expression for the average translational kinetic energy EK of a particle in the gas in terms of some or all of N, k and T. EK = … [2] (b) A molecule of hydrogen gas consists of two hydrogen atoms, each of nucleon number 1. A molecule of oxygen gas consists of two oxygen atoms, each of nucleon number 16. Assume that hydrogen and oxygen both behave as ideal gases. A sample of hydrogen gas is at the same temperature as a sample of oxygen gas. For the two samples, determine the ratio root-mean-square (r.m.s.) speed of hydrogen molecules . root-mean-square (r.m.s.) speed of oxygen molecules ratio = … [2] [Total: 7]

7 marks

Mark scheme: 2(a)(i) p = pressure (of gas), V = volume (of gas) and k = Boltzmann constant B1 N = number of molecules (in the gas) B1 T = thermodynamic temperature (of gas) B1 2(a)(ii) (pV =) NkT = ⅓Nm<c2> M1 EK = ½m<c2> = ½  3kT = (3 / 2)kT A1 2(b) (same temperature so) (½)m<c2> must be same for both gases C1 ratio = √(mO / mH) = √(32 / 2) A1 = 4.0

This question in 9702/44 Oct/Nov 2025