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r | | DN 20 XN-Physios Unite 16 Pages 15 NUMERCIALS 15.1 The freezing point of mercury is -39°C, convert It into °F and the comfort level temperature of 20° into Kelvin. Data: Solution: a) T °C = -39°C (a) For °F T¥Fe? Using b) = T°C = 20°C T°F-32_ TC TK=? os. T°F= T°F= 27°C +32 T°F=1.8(-39) +32 IT °F = -38.2 °F (b) For TK TK =T °C +273 TK =20 + 273 15.2 The boiling point of liquid nitrogen is -321°F. Change it into equivalent Kelvin temperature. T °F = - 321°F TK-273_ T°F-32 TK=? ;.~COS TK = 3 °F. 32)+273 TK = 0.555 (- 321 - 32) + 273 TK =- 195.915 + 273 rk = 77.00 15.3 Calculate the volume occupied by a gram-mole of a gas at 0°C and a pressure of 1.0 atmosphere n=] PV =nRT T= O°C +273 = 273K (1.01 x 10°)(V) = (1(8.314)(273) P=1.01 x 10°N/m* Vv = £28.314)(273) Ve? 1.01 x10° V = 0.0224m° V = 0.0224m’ x 1000 Lim* Fe aa.< vine —— aa eae ee f air ata 184 An air storage tank whose volume is 112 liters contain 3kg o Pressure of atmospheres. How much air would have to be forced into the tank to increase e pressure to 21 atmospheres, assuming no change in temperature? Xil-Physics Unit# 16 Pages % Data: VieV2e 12L ee ee m; =3 kg For m: P; = 18 atm PY, _ PY, P; = 21 atm m mM, Am=? (18)(112) | (21)(112) | om _ (21)(3) | 18 | m= 3.5kg | putin 1 . | Am =3.5-3 155 A balloon contains 0.04m* of air at a pressure of 120KPa. Calculate the pressure required to reduce its volume to 0.025 m’* at constant temperature. V; = 0.04 m? PiVi = P2V2 P,= 120 KPa=120x 10° Pa (120 x 10*)(0.04) = P2(0.025) V2 = 0.025 m? p, = 120x 10° x 0.04 P,=? >" 0.025 15.6 The molar mass of nitrogen gas N; is 28 mol. For 100g of nitrogen, calculate. (a) The number of moles. -_(b) The volume occupied at room temperature (20°C) and pressure of 1.01x10° Pa. Data: b) M= 28 g/mol 8) M= 0.028 kg / mol 100 = 100g = ——=09, m Ig 1000 0. Ikg T=20°C +273 = 293K P=1.01 x 10°Pa a)n=? b)V=? Solution: m a) n= M eal 0.028 n = 3.57 mol b) PV= nRT y= aRT P Ve (3.57)(8.314)(293) 1,01x10° ee \V = 0.086 m° Xu-Physice Unité 15 Pagey 15.9 Calculate the root mean square speed of hydrogen molecule at 500K (mass og = 1.67 x 10° "kg and K = 1.38 x'10°J/molecule.-K) Data: T=500K mp = 1.67 x 107’ kg K = 1.38 x 10 Jmol K mms = ? m = 2mp = 2 x 1.67x 10?’ kg m = 3.34 x 107’ kg mv => xr z 3 3KT mn. <1 Wie Solution: Vom = «RE m vi. = .{30:38%10™ x 500) ae 3.3410 sn DADA a 10°C and at 40°C. (b) What is the Kinetic energy per mole of an ideal gas at these temperatures? — 15.10 (a) Determine the average value of the kinetic energy of the particles of an ideal 28 at a Of Broo, i) T= 10°C = 10+ 273 = 283K ii) T = 40°C = 40 +'273 = 313K K = 1.38 x 10 J/molK R= 8.314 J/molK a)( KE)avg =? b) (K.E) per mole =? Solution: KK w 8) KB)p= 2 Wh (KB )arg = SRT (KE avg = 338 x 10) (283) R K=— N, 3 R K.E)y = — —T (K.B) 2 N, (KB) Na= SRT (K.E)av per mole = SRT b) (K.E)per mol = SRT * |QK.E) per mol = 3527.25/mol a) (K.E) avg = — & d[wly (KE) y= 3 (1.38x 10? )(313) (EDs 647x107 b) (K.E) per mol = SRT (K.E) per mol ‘. 3(8.314)313) ) per mol = 3901.45/mol] (K.E) per mol = 3(8.314)283) ap ewe meee paewwen iii) Isothermal Process iv) Adiabatic Process 16.10 NUMERICALS are of 300 K. If the gas 1. A gas undergoes isothermal expansion at @ constant oom e by the 225: absorbs 500 J of heat during the process, calculate any aoUwnvaee = 2 ——- — T: =i Using first law of thermodynamic sQ=AU+SW AQ = 5003 q Work =? -| at Constant temperature AU =0 : 1. AQ= AW aes 3 and the final . If the initial volame is 0.02 m A mn compresses @ g2S adiabatically eal m’ and the initial pressure is 200 KPa determine the final pressure Assume the gas behaves ideally. Data Solution V; =0.02 m? P;Vy=P2V2 V, =0.01 m? PM op, P, = 200 KPa = 200 x 10°Pa Vz” ; m _ (200x10°)(0.02) iam, 7, (0.02) P, = 400000Pa P,= 400 x 10°Pa 6. A gas expands from 0. the work done in both a re Data: ay V, = 0.03 m” Irreversible work = PA’ V2 =.0.06 m? = (100 x 10°X(0.06 - 0.03) P= 100KPa = 100 x 10°Pa aoe 180s 10°(0.03) Workdone is reversible=? ~ Work = 3000) Workdone is irreversible = ? Reversible Zero water at 20°C. If the final 7.- A 50 g piece of copper at : , of the system is 30°C, calculate heat capacity of water = 4.18J/g° C). : . Ma = SOB = 5 = 005K Heat loss by Copper = heat gain by water - aa - Mg Ca(Ti -Ts)a = MheCw(Ts~ Ta) 1 0.05Cq(100-30) = 0.2 x 4180 (30-20) tw = 200g = 22 = o.02Kg 0.05Cq, x 70 = 0.2 x 4180 x10 T, =20°C , : eae 1 _ 0.2%4180x10 Ts = 30°C Cu = 9.05%70 Ce = 4.18 JiKg°C = 4.18 x 1000 = 4180 J/Kg°C C=? “ Cu = 2388.57 VK gC 4 sti waich bist ves ed th Vikan etinajete | (Specific heat capacity of lad = 0.128)/¢°O) ture of 1g of lead from 25° C10 100° C* Solution: | T, =25°C AQ= T> = 100°C Q=mCAT = me (T;- Ti) AQ= 1 x 128 x (100-25) m =1Kg a= C = 0.128/g°C = 0.128 x 1000 = 128 J/Kg°C 4Q-7 representation of Camot's cycle. oir 1. A Carnot engine takes 200J of heat from ® rded, how much work does the some act engine takes 2001 er ion emoee bent ie OS® engine do, and what is the efficiency? —— Data: Solution: T. Q; = 2000) ») n=(1- 2} T, = 500K : T, = 350K . . (1-355) Q=? M 500 Work =? aa na b) ao 1% W= 00 30 x 2000 Work = To9 c) Qi = Q - work Q:= 2000 - 600 water at 0°C. Compute its cbanlt! iilogram of ice at O°C is melted and converted £0 2. One entropy: { Data: : Solution: é m=1 T=0°C AS=— T =273K = 273.15K For Q - Hy=334000 J/kg Q=mH; S=? Q = (1334000) Pp = 334000 334000 AS= 773.15 Ah preg engine performs work of 0.4166 watts in one hour and re| sink, What is the efficiency of engine? jects 4500J of heat to the| re EIN, P = 0.4166 watt T=1 hr= 3600 Sec Q: = 4500) n=? Solution | n-("St + @ | A For work Work = 741663600) Work = 1499.76 J fo. — | Q; = 4500 + 1499.76 Q = 5999-76) Put ini | | 1499.76 — 4 76 = 25% : 100 eratures 850K and 300K. the engine perforn oe [ Data: as00 oO : Work = a= ‘ t in - = x 5999 7 7. A Carnot engine operates between the temp’ 1200J of work in each cycle, which takes 0.25 sec (a) What is the efficiency of this (b) What is the average power 0 (c) How much energy is extracted as heat from the (d) How much energy is deliv | Data: _ work a) T; t Tr = Work = ‘ fp ” _ work engine? of this engine? high temperature eae i ered as heat to the low temperature reserv b) Power = —— =850K = 300K 1200 = 6.2500 =? c) 7 = —— 100 Power =? Q eae 64.7 = 2.100 Q@ =? Q . = 120000 Solution: 64.7 A = 1854.75 r= (1-H ho = 1650753 ' d) Q: = Qi - work Q: = 1854.75 - 1200 Q = 654,75 J} oO Eh ei =k cease SESSION 2024-2025 ‘\ul-Physics Unit# 17 Paget 15 8 A Carnot engine absorbs 52kJ as heat and exhaust 36kJ as heat in cach cycle. ’ Calculate: (2) The engine efficiency "(b) The work done per cycle in kilojoules. Q, = 52KJ =520003 n-[1-2 ~ Q:= 236KJ = 36000J Q, ae n = (1-2 Work =? $2000 7 = 30.769 b) work =Q:-Q: work = 52000 — 36000 - work = 16000) = 16x 10° a= Ff SESSION 2024-2025 
