Internal energy — Paper 4 · A Level Physics 9702

2 (a) State what is meant by the internal energy of a gas. For Examiner’s … Use … … [2] (b) The first law of thermodynamics may be represented by the equation U = q + w. State what is meant by each of the following symbols. + U … +q … +w … [3] (c) An amount of 0.18 mol of an ideal gas is held in an insulated cylinder fitted with a piston, as shown in Fig. 2.1. piston gas insulated cylinder Fig. 2.1 Atmospheric pressure is 1.0 × 105 Pa. The volume of the gas is suddenly increased from 1.8 × 103 cm3 to 2.1 × 103 cm3. For the expansion of the gas, (i) calculate the work done by the gas and hence show that the internal energy changes by 30 J, [3] (ii) determine the temperature change of the gas and state whether the change is an For increase or a decrease. Examiner’s Use change = … K … [3]

11 marks

Mark scheme: 2 (a) sum of kinetic and potential energies of molecules / particles / atoms … M1 random (distribution) … A1 [2] (b) +∆U: increase in internal energy … B1 +q: heating of / heat supplied to system … B1 +w: work done on system … B1 [3] (c) (i) work done = p∆V … C1 = 1.0 × 105 × (2.1 – 1.8) × 10-3 = 30 J … M1 w = 30 J, q = 0 so ∆U = 30 J … A1 [3] (ii) these three marks were removed, as insufficient data was given in the question. [Total: 8] GCE A/AS LEVEL – October/November 2009 9702 42

This question in 9702/42 Oct/Nov 2009

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

3 (a) State what is meant by the internal energy of a system. For Examiner’s … Use … … [2] (b) State and explain qualitatively the change, if any, in the internal energy of the following systems: (i) a lump of ice at 0 °C melts to form liquid water at 0 °C, … … … … [3] (ii) a cylinder containing gas at constant volume is in sunlight so that its temperature rises from 25 °C to 35 °C. … … … … [3]

8 marks

Mark scheme: 3 (a) sum of potential energy and kinetic energy of atoms/molecules/particles M1 reference to random (distribution) A1 [2] (b) (i) as lattice structure is ‘broken’/bonds broken/forces between molecules reduced (not molecules separate) B1 no change in kinetic energy, potential energy increases M1 internal energy increases A1 [3] (ii) either molecules/atoms/particles move faster/ <c2> is increasing or kinetic energy increases with temperature (increases) B1 no change in potential energy, kinetic energy increases M1 internal energy increases A1 [3]

This question in 9702/42 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 9702/41 Oct/Nov 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

2 (a) (i) State what is meant by the internal energy of a system. For Examiner’s … Use … … [2] (ii) Explain why, for an ideal gas, the internal energy is equal to the total kinetic energy of the molecules of the gas. … … … [2] (b) The mean kinetic energy <EK> of a molecule of an ideal gas is given by the expression <EK> = 32kT where k is the Boltzmann constant and T is the thermodynamic temperature of the gas. A cylinder contains 1.0 mol of an ideal gas. The gas is heated so that its temperature changes from 280 K to 460 K. (i) Calculate the change in total kinetic energy of the gas molecules. change in energy = … J [2] (ii) During the heating, the gas expands, doing 1.5 × 103 J of work. For State the first law of thermodynamics. Use the law and your answer in (i) to Examiner’s determine the total energy supplied to the gas. Use … … total energy = … J [3]

9 marks

Mark scheme: 2 (a) (i) sum of kinetic and potential energies of the molecules M1 reference to random distribution A1 [2] (ii) for ideal gas, no intermolecular forces M1 so no potential energy (only kinetic) A1 [2] (b) (i) either change in kinetic energy = 3/2 × 1.38 × 10–23 × 1.0 × 6.02 × 1023 × 180 C1 = 2240 J A1 [2] or R = kNA energy = 3/2 × 1.0 × 8.31 × 180 (C1) = 2240 J (A1) (ii) increase in internal energy = heat supplied + work done on system B1 2240 = energy supplied – 1500 C1 energy supplied = 3740 J A1 [3]

This question in 9702/43 Oct/Nov 2013

2 A constant mass of an ideal gas has a volume of 3.49 × 103 cm3 at a temperature of 21.0 °C. When the gas is heated, 565 J of thermal energy causes it to expand to a volume of 3.87 × 103 cm3 at 53.0 °C. This is illustrated in Fig. 2.1. 3.49 × 103 cm3 3.87 × 103 cm3 565 J 21.0 °C 53.0 °C Fig. 2.1 (a) Show that the initial and final pressures of the gas are equal. [2] (b) The pressure of the gas is 4.20 × 105 Pa. For this heating of the gas, (i) calculate the work done by the gas, work done = … J [2] (ii) use the first law of thermodynamics and your answer in (i) to determine the change in internal energy of the gas. change in internal energy = … J [2] (c) Explain why the change in kinetic energy of the molecules of this ideal gas is equal to the change in internal energy. … … … … [3]

9 marks

Mark scheme: 2 (a) use of kelvin temperatures B1 both values of (V / T) correct (11.87), V / T is constant so pressure is constant M1 [2] (allow use of n R1. Do not allow other values of n.) (b) (i) work done R p∆V R 4.2 × 105 × (3.87 – 3.49) × 103 × 10–6 C1 = = = = = R 160 J A1 [2] (do not allow use of V instead of ∆V) (ii) increase / change in internal energy R heating of system N work done on system C1 R 565 – 160 = = = = R 405 J A1 [2] (c) internal energy R sum of kinetic energy and potential energy / EK N EP B1 no intermolecular forces M1 no potential energy (so ∆U R ∆EK) A1 [3]

This question in 9702/42 May/June 2014

2 (a) State what is meant by internal energy. … … … [2] (b) The variation with volume V of the pressure p of an ideal gas as it undergoes a cycle ABCA of changes is shown in Fig. 2.1. 4.0 p / 105 Pa B 3.5 3.0 2.5 2.0 1.5 A C 1.0 3.0 4.0 5.0 6.0 7.0 8.0 V / 10ï m3 Fig. 2.1 The temperature of the gas at A is 290 K. The temperature at B is 870 K. Determine (i) the amount, in mol, of gas, amount = … mol [2] (ii) the temperature of the gas at C. temperature = … K [2] (c) Explain why the change from C to A involves external work and a change in internal energy. … … … [2]

8 marks

Mark scheme: 2 (a) (sum of) potential energy and kinetic energy of molecules/atoms/particles M1 mention of random motion/distribution A1 [2] (b) (i) pV = nRT either at A, 1.2 × 105 × 4.0 × 10−3 = n × 8.31 × 290 or at B, 3.6 × 105 × 4.0 × 10−3 = n × 8.31 × 870 C1 n = 0.20 mol A1 [2] (ii) 1.2 × 105 × 7.75 × 10–3 = 0.20 × 8.31 × T or T = (7.75 / 4.0) × 290 C1 T = 560 K A1 [2] (Allow tolerance from graph: 7.7–7.8 × 10–3 m3) (c) temperature changes/decreases so internal energy changes/decreases B1 volume changes (at constant pressure) so work is done B1 [2]

This question in 9702/41 May/June 2015

2 (a) State what is meant by internal energy. … … … [2] (b) The variation with volume V of the pressure p of an ideal gas as it undergoes a cycle ABCA of changes is shown in Fig. 2.1. 4.0 p / 105 Pa B 3.5 3.0 2.5 2.0 1.5 A C 1.0 3.0 4.0 5.0 6.0 7.0 8.0 V / 10ï m3 Fig. 2.1 The temperature of the gas at A is 290 K. The temperature at B is 870 K. Determine (i) the amount, in mol, of gas, amount = … mol [2] (ii) the temperature of the gas at C. temperature = … K [2] (c) Explain why the change from C to A involves external work and a change in internal energy. … … … [2]

8 marks

Mark scheme: 2 (a) (sum of) potential energy and kinetic energy of molecules/atoms/particles M1 mention of random motion/distribution A1 [2] (b) (i) pV = nRT either at A, 1.2 × 105 × 4.0 × 10−3 = n × 8.31 × 290 or at B, 3.6 × 105 × 4.0 × 10−3 = n × 8.31 × 870 C1 n = 0.20 mol A1 [2] (ii) 1.2 × 105 × 7.75 × 10–3 = 0.20 × 8.31 × T or T = (7.75 / 4.0) × 290 C1 T = 560 K A1 [2] (Allow tolerance from graph: 7.7–7.8 × 10–3 m3) (c) temperature changes/decreases so internal energy changes/decreases B1 volume changes (at constant pressure) so work is done B1 [2]

This question in 9702/43 May/June 2015

3 (a) State an expression, in terms of work done and heating, that is used to calculate the increase in internal energy of a system. … … … [2] (b) State and explain, in terms of your expression in (a), the change, if any, in the internal energy (i) of the water in an ice cube when the ice melts, at atmospheric pressure, to form a liquid without any change of temperature, … … … … [3] (ii) of the gas in a tyre when the tyre bursts so that the gas suddenly increases in volume. Assume that the gas is ideal. … … … … [3]

8 marks

Mark scheme: 3 (a) heat/thermal energy gained by system or energy transferred to system by heating B1 plus work done on the system or minus work done by the system B1 [2] (b) (i) either volume decreases so work done on the system or small volume change so work done on system negligible M1 (thermal) energy absorbed to break lattice structure M1 internal energy increases A1 [3] (ii) gas expands so work done by gas (against atmosphere) M1 no time for thermal energy to enter or leave the gas M1 internal energy decreases A1 [3]

This question in 9702/41 Oct/Nov 2015

3 (a) State an expression, in terms of work done and heating, that is used to calculate the increase in internal energy of a system. … … … [2] (b) State and explain, in terms of your expression in (a), the change, if any, in the internal energy (i) of the water in an ice cube when the ice melts, at atmospheric pressure, to form a liquid without any change of temperature, … … … … [3] (ii) of the gas in a tyre when the tyre bursts so that the gas suddenly increases in volume. Assume that the gas is ideal. … … … … [3]

8 marks

Mark scheme: 3 (a) heat/thermal energy gained by system or energy transferred to system by heating B1 plus work done on the system or minus work done by the system B1 [2] (b) (i) either volume decreases so work done on the system or small volume change so work done on system negligible M1 (thermal) energy absorbed to break lattice structure M1 internal energy increases A1 [3] (ii) gas expands so work done by gas (against atmosphere) M1 no time for thermal energy to enter or leave the gas M1 internal energy decreases A1 [3]

This question in 9702/42 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

3 (a) State what is meant by the internal energy of a system. … … … [2] (b) Explain, by reference to work done and heating, whether the internal energy of the following increases, decreases or remains constant: (i) the gas in a toy balloon when the balloon bursts suddenly, … … … … [3] (ii) ice melting at constant temperature and at atmospheric pressure to form water that is more dense than the ice. … … … … [3] [Total: 8]

8 marks

Mark scheme: 3 (a) (sum of/total) potential energy and kinetic energy of (all) molecules/particles M1 reference to random (distribution) A1 [2] (b) (i) no heat enters (gas)/leaves (gas)/no heating (of gas) B1 work done by gas (against atmosphere as it expands) M1 internal energy decreases A1 [3] (ii) volume decreases so work done on ice/water B1 (allow work done negligible because ∆V small) heating of ice (to break rigid forces/bonds) M1 internal energy increases A1 [3]

This question in 9702/42 Oct/Nov 2016

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

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 the internal energy of a system. … … … … [2] (b) An ideal gas undergoes a cycle of changes as shown in Fig. 2.1. 3.00 2.80 Q 372 K pressure / 105 Pa 2.60 97.0 J 2.40 2.20 280 K P R 332 K 2.00 900 950 1000 1050 1100 1150 volume / cm3 Fig. 2.1 At point P, the gas has volume 950 cm3, pressure 2.10 × 105 Pa and temperature 280 K. The gas is heated at constant volume and 97.0 J of thermal energy is transferred to the gas. Its pressure and temperature change so that the gas is at point Q on Fig. 2.1. The gas then undergoes the change from point Q to point R and then from point R back to point P, as shown on Fig. 2.1. Some energy changes that take place during the cycle PQRP are shown in Fig. 2.2. change P → Q change Q → R change R → P thermal energy transferred to gas / J +97.0 0 … work done on gas / J … –42.5 +37.0 increase in internal energy of gas / J … … … Fig. 2.2 (i) State the total change in internal energy of the gas during the complete cycle PQRP. Explain your answer. … … … [2] (ii) On Fig. 2.2, complete the energy changes for the gas during 1. the change P → Q, 2. the change Q → R, 3. the change R → P. [5] [Total: 9]

9 marks

Mark scheme: 2(a) sum of potential and kinetic energies (of molecules/atoms/particles) B1 (energy of) molecules/atoms/particles in random motion B1 2(b)(i) final temperature = initial temperature B1 no change in internal energy B1 2(b)(ii) 1. work done on gas (P→Q): 0 A1 increase in internal energy (P→Q): (+)97.0 J A1 2. increase in internal energy (Q→R): –42.5 J A1 3. increase in internal energy (R→P): –54.5 J A1 thermal energy supplied (R→P): –91.5 J A1

This question in 9702/41 Oct/Nov 2018

2 (a) State what is meant by the internal energy of a system. … … … … [2] (b) An ideal gas undergoes a cycle of changes as shown in Fig. 2.1. 3.00 2.80 Q 372 K pressure / 105 Pa 2.60 97.0 J 2.40 2.20 280 K P R 332 K 2.00 900 950 1000 1050 1100 1150 volume / cm3 Fig. 2.1 At point P, the gas has volume 950 cm3, pressure 2.10 × 105 Pa and temperature 280 K. The gas is heated at constant volume and 97.0 J of thermal energy is transferred to the gas. Its pressure and temperature change so that the gas is at point Q on Fig. 2.1. The gas then undergoes the change from point Q to point R and then from point R back to point P, as shown on Fig. 2.1. Some energy changes that take place during the cycle PQRP are shown in Fig. 2.2. change P → Q change Q → R change R → P thermal energy transferred to gas / J +97.0 0 … work done on gas / J … –42.5 +37.0 increase in internal energy of gas / J … … … Fig. 2.2 (i) State the total change in internal energy of the gas during the complete cycle PQRP. Explain your answer. … … … [2] (ii) On Fig. 2.2, complete the energy changes for the gas during 1. the change P → Q, 2. the change Q → R, 3. the change R → P. [5] [Total: 9]

9 marks

Mark scheme: 2(a) sum of potential and kinetic energies (of molecules/atoms/particles) B1 (energy of) molecules/atoms/particles in random motion B1 2(b)(i) final temperature = initial temperature B1 no change in internal energy B1 2(b)(ii) 1. work done on gas (P→Q): 0 A1 increase in internal energy (P→Q): (+)97.0 J A1 2. increase in internal energy (Q→R): –42.5 J A1 3. increase in internal energy (R→P): –54.5 J A1 thermal energy supplied (R→P): –91.5 J A1

This question in 9702/43 Oct/Nov 2018

2 (a) State what is meant by the internal energy of a system. … … … … [2] (b) By reference to intermolecular forces, explain why the change in internal energy of an ideal gas is equal to the change in total kinetic energy of its molecules. … … … [2] (c) State and explain the change, if any, in the internal energy of a solid metal ball as it falls under gravity in a vacuum. … … … … [3] [Total: 7]

7 marks

Mark scheme: 2(a) total potential energy and kinetic energy (of molecules/atoms) M1 reference to random motion of molecules/atoms A1 2(b) (in ideal gas,) no intermolecular forces B1 no potential energy (so change in kinetic energy is change in internal energy) B1 2(c) (random) potential energy of molecules does not change M1 (random) kinetic energy of molecules does not change M1 so internal energy does not change A1 or decrease in total potential energy = gain in total kinetic energy (M1) no external energy supplied (M1) so internal energy does not change (A1) or no compression (of ball) so no work done on the ball (M1) no resistive forces so no heating of the ball (M1) so internal energy does not change (A1) Question Answer Marks 2(c) or no change of state so potential energy (of molecules) unchanged (M1) no temperature rise so kinetic energy (of molecules) unchanged (M1) so internal energy does not change (A1)

This question in 9702/41 May/June 2020

2 (a) State what is meant by the internal energy of a system. … … … … [2] (b) By reference to intermolecular forces, explain why the change in internal energy of an ideal gas is equal to the change in total kinetic energy of its molecules. … … … [2] (c) State and explain the change, if any, in the internal energy of a solid metal ball as it falls under gravity in a vacuum. … … … … [3] [Total: 7]

7 marks

Mark scheme: 2(a) total potential energy and kinetic energy (of molecules/atoms) M1 reference to random motion of molecules/atoms A1 2(b) (in ideal gas,) no intermolecular forces B1 no potential energy (so change in kinetic energy is change in internal energy) B1 2(c) (random) potential energy of molecules does not change M1 (random) kinetic energy of molecules does not change M1 so internal energy does not change A1 or decrease in total potential energy = gain in total kinetic energy (M1) no external energy supplied (M1) so internal energy does not change (A1) or no compression (of ball) so no work done on the ball (M1) no resistive forces so no heating of the ball (M1) so internal energy does not change (A1) Question Answer Marks 2(c) or no change of state so potential energy (of molecules) unchanged (M1) no temperature rise so kinetic energy (of molecules) unchanged (M1) so internal energy does not change (A1)

This question in 9702/43 May/June 2020

2 (a) State what is meant by the internal energy of a system. … … … [2] (b) The atoms of an ideal gas occupy a container of volume 2.30 × 10–3 m3 at pressure 2.60 × 105 Pa and temperature 180 K, as illustrated in Fig. 2.1. 2.30 × 10–3 m3 3.80 × 10–3 m3 2.60 × 105 Pa 2.60 × 105 Pa 180 K T 980 J Fig. 2.1 The gas is heated at constant pressure so that its volume becomes 3.80 × 10–3 m3 at a temperature T. For the fixed mass of gas, calculate: (i) the amount of substance, in mol amount = … mol [2] (ii) the temperature T, in K. T = … K [2] (c) During the change in (b), the thermal energy supplied to the gas is 980 J. (i) Determine the work done on the gas during this change. Explain your working. work done = … J [3] (ii) Determine the change ΔU in internal energy of the gas. ΔU = … J [1] [Total: 10]

10 marks

Mark scheme: 2(a) sum of potential energy and kinetic energy (of particles) B1 (total) energy of random motion of particles B1 2(b)(i) pV = nRT C1 2.60 × 105 × 2.30 × 10–3 = n × 8.31 × 180 n = 0.400 mol A1 2(b)(ii) (2.30 × 10–3) / 180 = (3.80 × 10–3) / T or 2.60 × 105 × 3.80 × 10–3 = 0.400 × 8.31 × T C1 T = 297 K A1 2(c)(i) ΔW = pΔV = 2.60 × 105 × (2.30 – 3.80) × 10–3 C1 = (–)390 J A1 negative because work is done by gas or negative because work is done against atmospheric pressure or negative because volume of gas increases B1 2(c)(ii) ΔU = (980 – 390) = 590 J A1

This question in 9702/42 Oct/Nov 2020

2 A fixed mass of an ideal gas has a volume V and a pressure p. The gas undergoes a cycle of changes, X to Y to Z to X, as shown in Fig. 2.1. Z p Y X 0 0 V Fig. 2.1 Table 2.1 shows data for p, V and temperature T for the gas at points X, Y and Z. Table 2.1 p / 105 Pa V / 10–3 m3 T / K X 1.5 4.2 540 Y 230 Z 5.1 782 (a) State the change in internal energy ΔU for one complete cycle, XYZX. ΔU = … J [1] (b) Calculate the amount n of gas. n = … mol [2] (c) Complete Table 2.1. Use the space below for any working. [2] (d) (i) The first law of thermodynamics for a system may be represented by the equation ΔU = q + W. State, with reference to the system, what is meant by: ΔU : … q : … W : … [3] (ii) Explain how the first law of thermodynamics applies to the change Z to X. … … … … [2] [Total: 10]

10 marks

Mark scheme: 2(a) 0 B1 2(b) pV = nRT (n =) 1.5 × 105 × 4.2 × 10–3 / 8.31 × 540 C1 = 0.14 mol A1 2(c) missing pressure 1.5 (× 105) B1 both missing volumes 1.8 (× 10–3) B1 2(d)(i) (ΔU:) increase in internal energy (of the system) B1 (q:) thermal energy supplied to the system B1 (W:) work done on system B1 Question Answer Marks 2(d)(ii) volume increases and work is done by the gas B1 temperature decreases and internal energy decreases B1

This question in 9702/42 Feb/March 2022

3 (a) State what is meant by the internal energy of a system. … … … [2] (b) Use the first law of thermodynamics to explain what happens to the internal energy: (i) of a spring when it is stretched at constant temperature within its elastic limit … … … … … [3] (ii) of a sample of water when it evaporates from a rain puddle on a hot day. … … … … … [3] [Total: 8]

8 marks

Mark scheme: 3(a) sum of potential energy and kinetic energy (of particles) B1 (total) energy of random motion of particles B1 3(b)(i) no thermal energy transferred B1 work is done on the spring (increasing the potential energy of particles) M1 so internal energy increases A1 3(b)(ii) thermal energy transferred to water B1 work is done by water (expanding against atmosphere as it vaporises) B1 more thermal energy transferred than work done so internal energy increases B1

This question in 9702/42 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

3 (a) State what is meant by the internal energy of a system. … … … [2] (b) With reference to molecular kinetic and potential energies, describe and explain how the internal energy of the system changes when: (i) a gas is heated at constant volume so that its temperature increases … … … … … [3] (ii) a wire is stretched within its elastic limit at constant temperature. … … … … … [3] [Total: 8]

8 marks

Mark scheme: 3(a) sum of potential energy and kinetic energy B1 (total) energy of random motion of particles B1 3(b)(i) no change in separation so no change in (molecular) potential energy B1 temperature increases so kinetic energy (of molecules) increases B1 kinetic energy increases and potential energy unchanged, so internal energy increases B1 3(b)(ii) temperature constant so no change in (molecular) kinetic energy B1 separation increases so potential energy (of molecules) increases B1 potential energy increases and kinetic energy unchanged, so internal energy increases 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) (i) State what is meant by the internal energy of a system. … … … [2] (ii) Explain why the internal energy of an ideal gas is directly proportional to the thermodynamic temperature of the gas. … … … … [2] (b) A sample of an ideal gas at thermodynamic temperature T has internal energy U. The gas is compressed so that its temperature increases to 3T. During this compression, work W is done on the gas. The gas is then cooled at constant volume so that its temperature decreases to 2T. Complete Table 4.1 to show, in terms of some or all of W, T and U, 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 two processes. Table 4.1 thermal energy increase in internal work done on gas supplied to gas energy of gas compression +W cooling [4] [Total: 8]

8 marks

Mark scheme: 4(a)(i) sum of potential energy and kinetic energy B1 (total) energy of random motion of particles B1 4(a)(ii) potential energy (of molecules) (in an ideal gas) is zero, so the internal energy of the gas is equal to the total kinetic energy B1 (of molecules) kinetic energy of molecules is proportional to (thermodynamic) temperature (so internal energy is proportional to B1 (thermodynamic) temperature)) 4(b) cooling work done = 0 B1 compression increase in internal energy = +2U B1 cooling change in internal energy = –U B1 both rows: thermal energy adds to work to give increase in internal energy in terms of U and/or W B1 (if fully correct, thermal energy for compression = 2U – W and thermal energy for cooling = –U: compression +W 2U – W +2U cooling 0 –U –U )

This question in 9702/41 May/June 2025

4 (a) (i) State what is meant by the internal energy of a system. … … … [2] (ii) Explain why the internal energy of an ideal gas is directly proportional to the thermodynamic temperature of the gas. … … … … [2] (b) A sample of an ideal gas at thermodynamic temperature T has internal energy U. The gas is compressed so that its temperature increases to 3T. During this compression, work W is done on the gas. The gas is then cooled at constant volume so that its temperature decreases to 2T. Complete Table 4.1 to show, in terms of some or all of W, T and U, 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 two processes. Table 4.1 thermal energy increase in internal work done on gas supplied to gas energy of gas compression +W cooling [4] [Total: 8]

8 marks

Mark scheme: 4(a)(i) sum of potential energy and kinetic energy B1 (total) energy of random motion of particles B1 4(a)(ii) potential energy (of molecules) (in an ideal gas) is zero, so the internal energy of the gas is equal to the total kinetic energy B1 (of molecules) kinetic energy of molecules is proportional to (thermodynamic) temperature (so internal energy is proportional to B1 (thermodynamic) temperature)) 4(b) cooling work done = 0 B1 compression increase in internal energy = +2U B1 cooling change in internal energy = –U B1 both rows: thermal energy adds to work to give increase in internal energy in terms of U and/or W B1 (if fully correct, thermal energy for compression = 2U – W and thermal energy for cooling = –U: compression +W 2U – W +2U cooling 0 –U –U )

This question in 9702/43 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

3 (a) With reference to molecular kinetic energy and molecular potential energy, explain what is meant by the internal energy of an ideal gas. … … … [2] (b) A sample of an ideal gas is initially in state A, at a pressure of 2.0 × 105 Pa and with a volume of 0.016 m3, as shown in Fig. 3.1. 6 pressure / 105 Pa 4 2 A 0 0 0.01 0.02 0.03 0.04 volume / m3 Fig. 3.1 In state A, the temperature of the gas is 400 K. The gas undergoes two successive changes X and Y. In change X, it is heated at constant volume to a pressure of 4.0 × 105 Pa. At the end of change X, the gas is in state B. In change Y, it is then allowed to expand at constant temperature back to its original pressure. At the end of change Y, the gas is in state C. (i) Determine the internal energy of the gas in state A. internal energy = … J [2] (ii) Determine the temperature of the gas in state B. temperature = … K [1] (iii) Determine the volume of the gas in state C. volume = … m3 [1] (iv) On Fig. 3.1, draw two lines, one to represent change X and one to represent change Y. Label your lines X and Y respectively. [3] [Total: 9]

9 marks

Mark scheme: 3(a) total kinetic energy associated with random motion of molecules B1 potential energy (of molecules) is zero B1 3(b)(i) pV = NkT and U = (3 / 2)NkT C1 U = (3 / 2)pV A1 = (3 / 2)  2.0  105  0.016 = 4800 J 3(b)(ii) temperature = 800 K A1 3(b)(iii) volume = 0.032 m3 A1 3(b)(iv) straight vertical line labelled X between A and (0.016, 4.0) B1 curve from B with continuously decreasing negative gradient, labelled Y B1 line labelled Y between (0.016, 4.0) and (0.032, 2.0) B1

This question in 9702/44 Oct/Nov 2025