Ideal gases

2(a)(i) pV =NkT or pV =nRT and N =nN C1 A 2.3×105×3.5×10 −3 N = 1.38×10 −23×294 = 2.0 × 1023 A1 2(a)(ii) 1 C1 pV = Nmc2

9(a) constant voltage M1 that produces/dissipates same power as (the mean power of) the alternating voltage A1 9(b)(i) (maximum) rate of cutting of (magnetic) flux doubles B1 (peak and hence) r.m.s. induced e.m.f. doubles B1 9(b)(ii) sketch: (sinusoidal) wave of period 10 ms B1 peak E shown as ± 34V B2 (1 mark out of 2 awarded if peak E shown as ± 17V or ± 24V) 9(c) current in the coil results in forces that oppose its rotation B1 or current in the resistor dissipates the energy of rotation coil stops rotating B1 © UCLES 2021 Page 13 of 15

3(a)(i) no loss of kinetic energy B1 3(a)(ii) • molecules have negligible volume (compared with gas/container) B2 • no forces between molecules (except during collisions) • molecules are in random motion • collisions are instantaneous Any two points, 1 mark each 3(b)(i) 2mu A1 3(b)(ii) 2L / u A1 3(b)(iii) force = change in momentum / time = 2mu / (2L / u) A1 = mu2 / L 3(b)(iv) pressure = force / area = (mu2 / L) / L2 A1 = mu2 / L3 3(c) pV = NkT C1 NkT = ⅓Nm<c2> leading to ½m<c2> = (3/2)kT and ½m<c2> = E A1 K 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 © UCLES 2021 Page 9 of 19

9(a) constant voltage M1 that produces/dissipates same power as (the mean power of) the alternating voltage A1 9(b)(i) (maximum) rate of cutting of (magnetic) flux doubles B1 (peak and hence) r.m.s. induced e.m.f. doubles B1 9(b)(ii) sketch: (sinusoidal) wave of period 10 ms B1 peak E shown as ± 34V B2 (1 mark out of 2 awarded if peak E shown as ± 17V or ± 24V) 9(c) current in the coil results in forces that oppose its rotation B1 or current in the resistor dissipates the energy of rotation coil stops rotating B1 © UCLES 2021 Page 13 of 15

9(a) piezo-electric crystal B1 (ultrasound) wave causes shape change / vibrations (of crystal) B1 shape change / vibrations causes e.m.f. (which is detected) B1 9(b)(i) 93V A1 9(b)(ii) 2.7  107rads–1 A1 9(c)(i) kgm–2s–1 B1 9(c)(ii)  = Z / c = 1.7  106 / 1600 A1 = 1100 kg m–3 9(c)(iii) intensity reflection coefficient ≈ 1 or Z and Z are very different B1 1 2 almost no / no ultrasound transmitted (into air filled cavity) B1 © UCLES 2023 Page 17 of 18

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 E = ½m<c2> leading to E = (3/2) kT A1 K K 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 © UCLES 2023 Page 9 of 16

7(a)(i) P = I 2R A1 0 7(a)(ii) P = 4I 2R A1 0 7(b) sketch: square wave of period T, with P always non-zero B1 horizontal lines, from 0 to 0.5T and from 1.0T to 1.5T, all at the same level that the scale indicates to be I 2R B1 0 horizontal lines, from 0.5T to 1.0T and from 1.5T to 2.0T, at a level that is four times higher than the lower lines B1 7(c)(i) <P> = (5/2)I 2R A1 0 7(c)(ii) <P> = I 2R C1 r.m.s. I 2R = (5/2)I 2R A1 r.m.s. 0 I = √(5/2) I r.m.s. 0 © UCLES 2023 Page 13 of 16

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) E = m B1 P 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 c = 2200 m s–1 A1 r.m.s. 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 © UCLES 2023 Page 7 of 15

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 E = ½m<c2> leading to E = (3/2) kT A1 K K 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 © UCLES 2023 Page 9 of 16

7(a)(i) P = I 2R A1 0 7(a)(ii) P = 4I 2R A1 0 7(b) sketch: square wave of period T, with P always non-zero B1 horizontal lines, from 0 to 0.5T and from 1.0T to 1.5T, all at the same level that the scale indicates to be I 2R B1 0 horizontal lines, from 0.5T to 1.0T and from 1.5T to 2.0T, at a level that is four times higher than the lower lines B1 7(c)(i) <P> = (5/2)I 2R A1 0 7(c)(ii) <P> = I 2R C1 r.m.s. I 2R = (5/2)I 2R A1 r.m.s. 0 I = √(5/2) I r.m.s. 0 © UCLES 2023 Page 13 of 16

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) M1 ( = 507 K) 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) A1 = 1.78  1024 2(b)(iii)  (molecular) kinetic energy is proportional to temperature B2 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 r.m.s. speed (of molecules) in X is half r.m.s. speed (of molecules) in Y B1 © Cambridge University Press & Assessment 2024 Page 7 of 16

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 © Cambridge University Press & Assessment 2024 Page 8 of 14

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 E = m〈c2〉 leading to E = 3BT / 2A K K 2 4(c) line with positive gradient passing through the origin B1 smooth curve with decreasing positive gradient B1 © Cambridge University Press & Assessment 2025 Page 12 of 18

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 © Cambridge University Press & Assessment 2025 Page 9 of 17