Tiêu đề 12 First Law of Thermodynamics Tutorial Soln.pdf ✅

St. Andrew’s Junior College H2 Physics 12-1 CONCEPT WORKSHEET Answer the following questions: (a) Which state variables remain constant during the following process. (i) A-B pressure (ii) B-C volume (iii) C-A maybe temperature but need one more point to confirm (b) Complete the table, filling in either “+”, “-“ or “0” to indicate gain, loss or no change for each process. (c) Work done on gas can be calculated from the area under the P-V graph. Are you able to determine the magnitudes of the work done for processes A to B and C to A accurately? If you cannot, state why. A to B: 2 Po (2 Vo - Vo) = 2 PoVo C to A: cannot determine as the graph is not plotted on a graph paper so cannot determine its area accurately. Path of cycle Heat supplied/J Work done on gas/J Increase in Internal Energy/J A to B + (by deduction from 1st Law) - (gas expands) + (PV increases, so T increases) B to C - (by deduction from 1st Law) 0 (gas volume remains constant) - (PV decreases so T decreases) C to A - (by deduction from 1st Law) + (gas is compressed) 0 (T remains constant, U remains constant) A to B to C to A + (by deduction from 1st Law) - (net work is negative) 0 (gas goes back to initial state) p 2Po Po Vo 2Vo V A B C

St. Andrew’s Junior College H2 Physics 12-3 1(d) In stage C to D, negative work is done on the gas (gas is expanded) AND no heat is supplied or removed from the gas (expansion is rapid). The change in internal energy is negative, meaning the internal energy DECREASES. This is an ideal gas, so the internal energy is equal to the microscopic kinetic energy (microscopic potential energy is zero) so the kinetic energy of the molecules will decrease and thus the root mean square speed of the molecules will decrease. 1(e) In stage C to D the gas is expanded, walls of the container do NEGATIVE WORK on the gas molecules during collision. This means energy is removed from the gas molecules, so the kinetic energy of the molecules decreases. 1(f) 23 23 3 2 3 (0.030 6.02 10 ) (1.38 10 )(1960) 2 732 J K N kT −    =       =        = 2(a) Water that has evaporated obtained its energy from water that remains on the clothing. 2(b) Let proportion of ice (by mass) be p. Water that has evaporated is therefore 1-p Energy required to evaporate water = energy taken from water to form ice (1 − p)2500 = p(333) (2500 + 333)p = 2500 p = 2500 2830 = 0.883 Assumption: no other form of energy is transferred between wet clothing and surroundings. 3(a) The temperature/time gradient just before melting = (63-53)/160 = 0.0625 K/s Power supplied before melting = mc ( dT dt) = m(1.6 × 103 )(0.0625) Melting of solid takes (400-160) s = 240 s Energy required = Power x time = mL m(1.6 × 103 )(0.0625)240 = mL L = 24000 Jkg −1 3(b) The same power is used to cause the temperature of the liquid nitrogen to increase m(1.6 × 103 )(0.0625) = mc dT dt = mc 73 − 63 600 − 400 c = 2000 JK −1kg −1