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,