Thermodynamics

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 nN = N (C1) A 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 B1 or mean square speed (directly) proportional to kinetic energy (of molecules) or kinetic energy (of molecules) (directly) proportional to (thermodynamic) temperature 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 © UCLES 2021 Page 9 of 18

2(a) pV = nRT C1 pV = nRT and N = nN C1 A or pV = NkT 3.1 × 10–3 × 8.5 × 105 = (N × 290 × 8.31) / (6.02 × 1023) A1 so N = 6.6 × 1023 or 3.1 × 10–3 × 8.5 × 105 = N × 1.38 × 10–23 × 290 so N = 6.6 × 1023 2(b)(i) (3.1 × 10–3 × 8.5 × 105) / 290 = (6.3 × 10–3 × 2.7 × 105) / T A1 so T = 190 K or 6.3 × 10–3 × 2.7 × 105 = 6.6 × 1023 × 1.38 × 10–23 × T so T = 190 K 2(b)(ii) ΔU = 3/2 × k × ΔT × N C1 = 3/2 × 1.38 × 10–23 × (190 – 290) × 6.6 × 1023 C1 = –1400 J A1 2(c) ΔU = q + w M1 q = 0 so ΔU = w A1 © UCLES 2021 Page 9 of 19

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 nN = N (C1) A 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 B1 or mean square speed (directly) proportional to kinetic energy (of molecules) or kinetic energy (of molecules) (directly) proportional to (thermodynamic) temperature 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 © UCLES 2021 Page 9 of 18

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> M1 or pV = NkT and pV = ⅓Nm<c2> leading to ½m<c2> = (3/2)kT and ½m<c2> = E A1 K 3(b)(iii) internal energy = ΣE (of molecules) + ΣE (of molecules) B1 K P or no forces between molecules 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 © UCLES 2021 Page 9 of 15

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> M1 or pV = NkT and pV = ⅓Nm<c2> leading to ½m<c2> = (3/2)kT and ½m<c2> = E A1 K 3(b)(iii) internal energy = ΣE (of molecules) + ΣE (of molecules) B1 K P or no forces between molecules 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 © UCLES 2021 Page 9 of 15

2(a) 0 B1 2(b) pV = nRT C1 (n =) 1.5 × 105 × 4.2 × 10–3 / 8.31 × 540 = 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 © UCLES 2022 Page 7 of 17 2(d)(ii) volume increases and work is done by the gas B1 temperature decreases and internal energy decreases B1 Question Answer Marks

3(a)(i) a gas that obeys pV  T M1 where p = pressure, V = volume, T = thermodynamic temperature A1 3(a)(ii) T = (273 + 17) K C1 n = pV / RT A1 = (1.2  105  0.24) / [8.31  (273 + 17)] = 12 mol 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 A1 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 ) © UCLES 2022 Page 9 of 16

3(a) (thermal) energy per unit mass B1 energy to change state between liquid and gas at constant temperature B1 3(b)(i) q = mL = 0.37  2.3  106 A1 = 8.5  105 J 3(b)(ii) pV = nRT and T = 373 K C1 n = 370 / 18 C1 V = [(370 / 18)  8.31  373] / (1.0  105) = 0.64 m3 A1 3(b)(iii) w = pV C1 = 1.0  105  0.64 A1 = 6.4  104 J 3(b)(iv) (water does work against atmosphere so) work done on water is negative B1 increase in internal energy = (8.5 – 0.64)  105 = 7.9  105 J A1 3(c) valid reasoning of how work done by water is affected M1 correct use of first law to draw conclusion about effect on specific latent heat that is consistent with work done A1 © UCLES 2022 Page 9 of 16

3(a)(i) a gas that obeys pV  T M1 where p = pressure, V = volume, T = thermodynamic temperature A1 3(a)(ii) T = (273 + 17) K C1 n = pV / RT A1 = (1.2  105  0.24) / [8.31  (273 + 17)] = 12 mol 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 A1 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 ) © UCLES 2022 Page 9 of 16

2(a) (thermal) energy per unit mass (to cause temperature change) B1 (thermal) energy per unit change in temperature B1 2(b)(i) work done correct (0) B1 increase in internal energy correct (+E) B1 2(b)(ii) work done correct (–W) and increase in internal energy same as (b)(i) B1 thermal energy correct so that it adds to work done to give increase in internal energy B1 2(c) more thermal energy needed so specific heat capacity is greater B1 © UCLES 2022 Page 7 of 16

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106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 © UCLES 2023 Page 7 of 18 2(b)(iii) pV = NkT e.g. C1 1.1010554010−6 N = 1.3810−23300 = 1.435  1022 E = (3 / 2) kTN k 1.10105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 © UCLES 2023 Page 8 of 18

4(a)  particles are in (continuous) random motion B2  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 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 B1 (completely correct answer: change U Q w X to Y –U –U 0 Y to Z [ +U ] U + pV –pV Z to X 0 –W [ +W ] ) © UCLES 2023 Page 10 of 16

3(a) change in internal energy = work done + energy transfer by heating C1 increase in internal energy = work done on system + energy transferred to the system by heating A1 3(b)(i) AB change in internal energy: decrease B1 AB work done on gas: positive B1 BC change in internal energy: increase B1 BC work done on gas: zero B1 3(b)(ii) more work done by gas in CD than is done on gas in AB B1 or (no work done on gas in BC and DA so) (overall) gas does work (overall) change in internal energy is zero B1 (must be an overall) input of thermal energy B1 © UCLES 2023 Page 8 of 15

4(a)  particles are in (continuous) random motion B2  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 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 B1 (completely correct answer: change U Q w X to Y –U –U 0 Y to Z [ +U ] U + pV –pV Z to X 0 –W [ +W ] ) © UCLES 2023 Page 10 of 16

2(a) (thermal) energy per unit mass (to change temperature) B1 (thermal) energy per unit change in temperature B1 2(b)(i) work done = pV A1 = (2.0  105)  (0.063 – 0.038) = 5000 J 2(b)(ii) gas is expanding (against external pressure) B1 gas does work / work is done by gas, so (work done on gas is) negative B1 2(b)(iii) U = q + W C1 7600 = q + (–5000) A1 q = 12 600 J 2(b)(iv) specific heat capacity = q / mT C1 = 12600 / (0.35  56) = 640 J kg–1 K–1 A1 2(c) same gain in internal energy so same temperature rise B1 no change in volume so no work done B1 or no work done so less thermal energy needed (for same change in internal energy) less thermal energy needed (for same temperature change) so lower specific heat capacity B1 © UCLES 2023 Page 8 of 16

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 © UCLES 2023 Page 8 of 15

2(a) (thermal) energy per unit mass (to change temperature) B1 (thermal) energy per unit change in temperature B1 2(b)(i) work done = pV A1 = (2.0  105)  (0.063 – 0.038) = 5000 J 2(b)(ii) gas is expanding (against external pressure) B1 gas does work / work is done by gas, so (work done on gas is) negative B1 2(b)(iii) U = q + W C1 7600 = q + (–5000) A1 q = 12 600 J 2(b)(iv) specific heat capacity = q / mT C1 = 12600 / (0.35  56) = 640 J kg–1 K–1 A1 2(c) same gain in internal energy so same temperature rise B1 no change in volume so no work done B1 or no work done so less thermal energy needed (for same change in internal energy) less thermal energy needed (for same temperature change) so lower specific heat capacity B1 © UCLES 2023 Page 8 of 16

2(a) total kinetic energy associated with random motion of molecules M1 plus total potential energy (of molecules) but potential energy is zero A1 2(b)(i) W = pV C1 = 1.01  105  5.20  10–5 A1 = (+)5.25 J 2(b)(ii) V  T or V / T = constant C1 1.24 / (273 + 20) = (1.24 + 0.520) / T A1 T = 416 K 2(b)(iii) c = Q / mT C1 = 960 / (0.016  (416 – 293)) A1 = 490 J kg–1 K–1 2(c) no change in volume so no work is done (by the gas) B1 (same temperature change so) same change in internal energy B1 less thermal energy needs to be supplied so c is less B1 Question Answer Marks

3(a)(i) gas for which pV  T 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) A1 = 1.3  1025 3(b)(ii) E = (3/2) kT C1 K E = (3/2)  1.38  10–23  290 A1 K = 6.0  10–21 J 3(b)(iii) internal energy = total KE + PE of molecules B1 or PE = 0 so internal energy = total KE of molecules internal energy = 1.3  1025  6.0  10–21 A1 = 7.8  104 J 3(c) straight line with positive gradient B1 line passing through the origin B1 © Cambridge University Press & Assessment 2024 Page 8 of 15

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 © Cambridge University Press & Assessment 2024 Page 8 of 16

3(a)(i) gas for which pV  T 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) A1 = 1.3  1025 3(b)(ii) E = (3/2) kT C1 K E = (3/2)  1.38  10–23  290 A1 K = 6.0  10–21 J 3(b)(iii) internal energy = total KE + PE of molecules B1 or PE = 0 so internal energy = total KE of molecules internal energy = 1.3  1025  6.0  10–21 A1 = 7.8  104 J 3(c) straight line with positive gradient B1 line passing through the origin B1 © Cambridge University Press & Assessment 2024 Page 8 of 15

3(a)(i) number of particles per unit amount of substance B1 3(a)(ii) N = R / k B1 A 3(b)(i) X pressure and Y pressure both = NkT / V B1 X amount = N / N and Y amount = 2N / N B1 A A X mean-square speed = 3kT / m and Y mean-square speed = 3kT / 2m B1 X internal energy = 3NkT / 2 and Y internal energy = 3NkT B1 3(b)(ii) line passing through the origin and not returning to either axis B1 curve with positive decreasing gradient B1 © Cambridge University Press & Assessment 2024 Page 8 of 15

3(a) (thermal) energy per unit mass (to cause change of state) B1 (thermal) energy to change state at constant temperature B1 3(b)(i) W = pV C1 = 1.0  105  0.017 = 1700 J = 1.7 kJ A1 3(b)(ii) U = Q + W C1 Q = 17.6 + 1.7 A1 = 19.3 kJ 3(b)(iii) mass = 710  7.2  10–5 C1 ( = 0.051 kg) L = 19.3 / 0.051 A1 = 380 kJ kg–1 3(c) fusion involves (much) smaller volume change (than vaporisation) B1 smaller change in intermolecular spacing so smaller change in internal energy B1 negligible work done (by substance during fusion) so L is less (than L ) B1 F V © Cambridge University Press & Assessment 2024 Page 7 of 14

3(a)(i) number of particles per unit amount of substance B1 3(a)(ii) N = R / k B1 A 3(b)(i) X pressure and Y pressure both = NkT / V B1 X amount = N / N and Y amount = 2N / N B1 A A X mean-square speed = 3kT / m and Y mean-square speed = 3kT / 2m B1 X internal energy = 3NkT / 2 and Y internal energy = 3NkT B1 3(b)(ii) line passing through the origin and not returning to either axis B1 curve with positive decreasing gradient B1 © Cambridge University Press & Assessment 2024 Page 8 of 15

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 ) © Cambridge University Press & Assessment 2025 Page 12 of 19

3(a) (thermal) energy per unit mass (to cause temperature change) B1 (thermal) energy per unit change in temperature B1 3(b)(i) density = mass / volume C1 mass = 2.700 × 103 × 3.612 × 10–3 A1 = 9.752 kg 3(b)(ii) volume = 3.612 × 10–3 × (2.700 / 2.620) = 3.722 × 10–3 m3 A1 or volume = 9.752 / (2.620 × 103) = 3.722 × 10–3 m3 © Cambridge University Press & Assessment 2025 Page 10 of 18 3(b)(iii) W = pV C1 = 1.01 × 105 × (3.722 – 3.612) × 10–3 A1 = 11.1 J 3(b)(iv) volume (of block) increases B1 work is done against the atmosphere so work done (on block) is negative B1 3(b)(v) thermal energy = (4.38 × 106) + 11.1 B1 specific heat capacity = (4.38 × 106) / (9.75 × 500) C1 = 898 J kg–1 °C–1 A1 3(c) work done is negligible compared with (change in) internal energy so (answer in (b)(v) would be) unchanged B1 © Cambridge University Press & Assessment 2025 Page 11 of 18

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 ) © Cambridge University Press & Assessment 2025 Page 12 of 19

2(a) work done on / by system B1 thermal energy supplied to / removed from system B1 2(b)(i) no thermal energy transferred to / from system (due to lack of time) B1 work is done on the gas to compress it / to decrease its volume B1 internal energy increases so temperature increases B1 2(b)(ii) (during vaporisation) molecular separation increases B1 (heating causes) potential energy of molecules to increase B1 kinetic energy of molecules unchanged so temperature unchanged B1 © Cambridge University Press & Assessment 2025 Page 9 of 17

4(a) change in internal energy = work done + energy transfer by heating C1 increase in internal energy = work done on system + energy transferred to the system by heating A1 4(b)(i) U = (3 / 2) pV A1 4(b)(ii) pV = NkT and k identified as Boltzmann constant B1 U = (3 / 2) NkT A1 4(c)(i) W = (+)8XY A1 4(c)(ii) W = –20XY A1 4(d) work done during stages BC and DA = 0 B1 change in internal energy (over complete cycle) = 0 C1 thermal energy supplied = 20XY – 8XY A1 = (+)12XY © Cambridge University Press & Assessment 2025 Page 11 of 17

2(a) work done on / by system B1 thermal energy supplied to / removed from system B1 2(b)(i) no thermal energy transferred to / from system (due to lack of time) B1 work is done on the gas to compress it / to decrease its volume B1 internal energy increases so temperature increases B1 2(b)(ii) (during vaporisation) molecular separation increases B1 (heating causes) potential energy of molecules to increase B1 kinetic energy of molecules unchanged so temperature unchanged B1 © Cambridge University Press & Assessment 2025 Page 9 of 17