2021 HCI_H3_Physics_Prelim_2021_Solutions_Marker_sComments

9814 H3 Physics Prelim 2021 Suggested Solutions and Markers’ Comments 1 (a) 0 0 0 0 0 (3 ) CC AA CA C PV PV TT P V PV TT   TC = 3T0 A1 Comments: Quite well-done. Majority of candidates managed to get the correct expression for temperature at C. However, the approaches varied. Most used PV=nRT instead of the suggested method. It is expected that candidates compare temperatures at A and C, but there were a few compared temperatures at B and C, and gave the relationship without clear working - the mark was awarded, given the benefit of doubt. (b) Work done by gas during process A to B = 0 Work done by gas during process B to C Work done by gas during process C to A = - Po (VA - VC)  = - Po (3Vo - Vo) = - 2PoVo Net work done by gas = 3nRTo ln 3 - 2PoVo = 3nRTo ln 3 - 2nRTo since PoVo = nRTo = nRTo (3 ln 3 - 2) B1 B1 B1 A0 Comments: Quite well done. Most students could show the correct expression for net work done by gas. We preferred that candidates explicitly state that the net work done is the sum of work done by gas during each process AB, BC, CA before showing the math. More candidates had difficulty determining the work done by gas during process BC, as it required integration. There were signs that some candidates actually “reverse-engineered” the proof. (c) A to B: By First Law of Thermodynamics, as w = 0, U = q Since temperature increases, U and hence q are positive i.e. heat is added to the gas. B to C: as temperature is constant, U = 0, q = - w B1 3 3 ln 3 ln3 o o o o V B V V B V o PdV nRT dVV nRT V nRT      

2 © Hwa Chong Institution 9814 H3 Physics / Prelim 2021 Since gas is expanding, work done on the gas, w is negative. To keep U = 0, q must be positive i.e. heat must be added to the gas. C to A: Temperature decreases, U is negative, gas is compressed, hence w is positive. Hence q must be negative i.e. heat is lost. B1 B1 Comments: Quite well done. Most candidates were able to clearly explain whether heat is supplied during each process, analysing work done by gas using the change in volume, the change in internal energy using change in temperature or PV, and applying First Law of Thermodynamics. Candidates are reminded to explain/define all notations. Marks are deducted for incomplete/wrong explanations and undefined notations. 2 (a) By Newton’s 3rd law, the tension force (F) acting on both springs is the same. 1 1 2 2F k x k x The total extension/compression of the springs is 12e x x It can be found that 2 1 12 kxe kk  Since the system performs simple harmonic motion, F ma , where a is the acceleration of the oscillation.   2 2 11 2 122 1 1 2 1 2 2 2 2 k x m e m e T m k kkk