2022 RI Prelims H2 Phy Paper 3 Soln

Raffles Institution Year 5-6 Physics Department 1 2022 H2 Physics Preliminary Examination Solution Paper 3 – Section A 1 (a) The gas molecules exert no intermolecular forces on one another except during collisions. (b) (i) 1.      522.0 10 3.0 10 8.31 300 2.4067 2.4mol (shown) BB B B BB B B pV nR T pVn RT       2. Solution 1       52 52 total internal energy 33 22 333.0 10 2.0 10 2.0 10 3.0 1022 18000 J AB AA BB UU pV pV         Solution 2           total internal energy 33 22 331.8 8.31 400 2.4 8.31 30022 17949.6 17900 J AB AA BB UU nR T nR T     (ii) 1. Total internal energy is conserved as this container is a closed system.  total internal energy ' ' 17900 33 1790022 3 179002 AB AB AB UU nR T nR T RT n n        2117900 38 . 3 1 1 . 8 2 . 4 341.91 342 K T      

Raffles Institution Year 5-6 Physics Department 2 2. As gas A is at a higher pressure than gas B, it expands and does work against the pressure of gas B, hence work done W on gas A is negative. As gas A is at a higher temperature than gas B, heat is transferred from gas A to gas B, hence heat supplied Q to gas A is negative. Since UQW , the increase in internal energy U is negative which implies that the internal ener gy of gas A decreases. Since UT  (or Un 3 2 RT ), the temperature of gas A decreases. (iii) Vacuum does not contain any particles and has no pressure. Hence in expanding, gas A does no work and does not transfer heat to any other body. Hence by the first law of thermodynamics, there is no change in internal energy and no change in temperature. Comments (a) The direct reason why the microscopic potential energy of an ideal gas is zero is due to the absence of intermolecular forces between the molecules. This is the only acceptable answer. Candidates who wrote that the molecules ar e assumed to be very far apart and do not interact did not get any credit. Many candidates mentioned that in termolecular forces are attr active in nature, which is incorrect. It could be both attractive and repulsive. (b) (i) 1. This part was well done. However, many students did not explicitly state the numerical value of nB to a greater number of significant figure than the final answer of 2.4 mol, which is mandatory for questions that require candidates to prove a specific numerical value of a particular physical quantity. 2. Most candidates arrived at the correct answer by using 33 or22 AB A A B BUU p V p V  33 22 AB A A B BUU n R T n R T  . There are a handful who solved it using 33 22 AB A a v o A B a v o BU U n N kT n N kT  , but this is more tedious. The most common mistake here is probably due to candidates’ own carelessness such as not multiplying by the factor of 3 2 or not multiplying by the correct constants. (ii) 1. This part was poorly done. In addition, those who approached this question from the perspective that both pressure and temperature of gases A and B will eventually equalise in the final state of equilibrium, met with a dead end and could not get an answer. Yet many other candidates appr oached this part by applying pVpV nRT T nR  to the entire system of gas A and gas B. The problem with this is what will the final pressure p of the system be?

Raffles Institution Year 5-6 Physics Department 3 Many candidates erroneously ca lculated final pressure as 55 52.0 10 3.0 10 2.5 10 Pa2p   which has no basis. Candidates who did this correctly approached this problem from the perspective that since the container is insulated all round, there is no heat transfer to or from the system of gas A and B, hence the total internal energy of the system is conserved, and things become very easy! 2. This part exposes candidates’ lack of understanding and focus in both the requirement and the context of the question. Many candidates overlooked the fact t hat the question made a reference to gas A and not the entire system thus missing out on the context. That being the case, many candidates concluded that heat transferred to the system is zero (Q  0) when they should have consider ed heat transfer from gas A to gas B (as TA > TB) which is negative (QA < 0). A number of students failed to mention the important point that (change in) internal energy of ideal gas is propor tional to (change in) its thermodynamic temperature hence when internal energy of gas A decreases, its temperature decreases; not getting full credit as a result. (iii) This is another question in which candidates performed poorly. Many candidates did not realise that the piston has been removed and gas A is free to expand into the section where gas B was before and being replaced with a vacuum. This expansion does not require work to be done by gas A; it’s a free expansion due to the vacuum where there are no gas molecules hence work done by A is zero. Some candidates came to the conclusion that work done by gas A is zero because there is no change in volume of gas A (so pV = 0) which is not true, as gas A expands. Many other candidates approached this question using the ideal gas equation pVn R T and wrote that as gas A expands, its volume V increases and its pressure p decreases proportionately thus its temperature remains unchanged. By doing so, candidates are actually already assuming that the temperature remains constant and not proving it, hence not answering the question. Some candidates wrote that since the syste m is a closed one, the internal energy of the system is conserved hence its te mperature did not change. This was not given credit as it lacks explanation as internal energy is always conserved for this closed system. Candidates are expected to consider changes in both work and heat energy transferred to or from gas A and how these affect its internal energy and therefore should explicitly mention them in their explanation.

Raffles Institution Year 5-6 Physics Department 4 2 (a) (i) 0x is the decrease in height of the load and is the extension of the spring when the load is at the equilibrium position. decrease in G.P.E. 0mgx Solution 1 magnitude of the external force at the equilibrium position, Fm g increase in E.P.E. 00 11 22Fx mgx Solution 2 increase in E.P.E.  2 00 0 0 11 1 22 2kx kx x mgx  where k is the spring constant and 0kx mg at equilibrium comparing the expressions for G.P.E. and E.P.E, decrease in G.P.E. = 2  increase in E.P.E. (shown) (ii) The decrease in G.P.E. is greater than the increase in E.P.E. as there is negative work done by the external force in lowering the load slowly. (b) (i) at equilibrium, 0 0 mgkx mg k x  where k is the spring constant  2 2 0 0 0 1E.P.E. at lowest point 2 1 22 2 ke mg xx mgx      (ii) 1. G.P.E. mgh where h is the distance from the lowest point G.P.E. varies linearly with distance. at lowest point: G.P.E. 0 at highest point:  00G.P.E. 2 2mg x mgx at equilibrium position: 0G.P.E. mgx 2. 2 0 E.P.E. 2 mg ex where e is the extension of the spring and the distance from the highest point. E.P.E. and distance follows a quadratic relationship. from (b)(i), at lowest point: 0E.P.E. 2 mgx at highest point, spring is at its natural length: E.P.E. 0 at equilibrium position: 0 1E.P.E. 2 mgx

Raffles Institution Year 5-6 Physics Department 5 3.    2 22 2 22 2 2 2 2 00 0 11 1 1K.E. 22 2 2m x x m x x mx mx        K.E. and displacement follows a quadractic relationship. at lowest and highest points: K.E. 0 at equilibrium position: max. 22 2 2 00 0 0 0 11 1 1K.E. 22 2 2 km gmx m x x m g xmx      *total energy can be deduced from the energies at the lowest and highest points *maximum