Exam2_Solution

# Exam2_Solution - Exam 2 Wednesday April 1“ 2009 PHY 252...

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Unformatted text preview: Exam 2. Wednesday, April 1“, 2009 PHY 252 SOL. UTIOIU Answer a_]l questions. Be careful to justify your answers — most f the redit will be for clear ex lanations of the h sics. ‘\ 1. The period of a simple pendulum of length L is Zia/Ll g , assuming \ L that the inertia of the thin rod can be ignored (ﬁg. on left). The length L\ \ L. L: i Name: of the rod changes with temperature, so a simple pendulum clock is inaccurate. Design a compensated clock pendulum out of two thin rods of lengths Li and L2 (see ﬁg. on right), made from brass and steel, such that the pendulum period is a constant 1.0 second, even when the temperature changes. What are the lengths l21 and L2 needed? Which rod should be brass, and which steel? (as, = 1.2 x10'5 K“ ,ab, = 2.0 ><10‘5 K") (10 points) ... 1." For- a. per-loci. T, we. need. a. lean-bk L -- ﬁgs 2 \$u9.80uj\$1:o,1+ezm ‘P UMEMT=15¢Q For our Compound pendulum L: 0.2482.“ + L. = Li"L2. For Qua A'T, tug want AL ='- ALI—ALI. = 0 ie. 4.1..“ - aszm’ :0 Since LI>L2., than we need al.<a£,_ , L. Is Sibel) L115 [crass—73F. L: Li—Lz = 2.. —-&2.. 0L1. => 2.. = ’3 = Oleﬁn. 20.6206m—-—-7\’- ’“ I- :22: mean —_-> L2 = Lo-L. == ages—0.24.32 =o.3?24M——-x C la m. 55) 2. An engine performs 27 J of work and dumps 370. J of heat into the environment. What is the efﬁciency of the engine? (6 points) —Ib 2 -I- . 5:.- W 2 W _ 7 ...0.068 “a 5.3%— ~54- Q in Qout‘l'w 370 +2? 3. 7.0 moles of argon gas are in a container of volume 17.24 liters at a pressure of 10. atm. Estimate the root-mean—square speed of a single argon atom in the gas. (Argon has an atomic mass of40.a.m.u.,and latm. a 1.013><105 Pa.) (9 points) <01)”; = \/ 313:, “Ml irom the iai-eqL dqs law PV 2 "RT M e (C1)“ = , /3Pv7 = 3xiou.0i3xw’r’e ten-urns? same;- ' “’4 =iI-o x. 0.0:on <¢;=i)"‘1 '- 433 ml; -——————-——————"’"* Inq. Exam 2. Wednesday, April 15', 2009 PHY 252 4. 10.0 g of liquid water at atmospheric pressure and 100. °C, are heated and converted to water vapor at atmospheric pressure and 100. °C. Assume that water vapor is an ideal gas. (1') How much heat is added to the water? (Lv = 2256 kJ/kg) (3 points) Q: MLV == lOXIo—a'KslﬂXZZsé K-Tlﬂ = 22.56 k3- (ii) What is the volume of the water vapor? (Molecular mass of water = 18 amu) (3 points) IO.03 01- water contains n = “lo/18.0 ~7- 0'555 “0'95 05’ “1.0- V = nn'r :__ £g3.3rz,xs=l-3k T -—————-“‘S Assum'em an ‘tcleaL acts 2 0.0I7o M3 LOIS! IDs-Pg, + (iii) How much work is done by the water vapor? (3 points) w = PAV aft consi'au't prél’S-Su re . -5 I a sum: Louie:- has (L w 2: Lemmas-Pa #- (0.0t?ou3 - ‘0 “'0 M ’5‘“) denslbﬂ a-j ~ lglcm‘ w 1‘. I '72 KIT 7% (iv) What is the change in internal energy of the water during this process? (5 points) From the Fri-sf Lat.) A5 = Q~W AE ‘-"- Z2-S'6KI— 1.723(3 AE :- 20.8 KG— 7‘" 5. 1 .7 moles of an ideal monatomic gas (C, = 3R I 2) are expanded isothermally to two times the initial volume at 127.0 °C. Find W, Q, AE and AS for this process. (11 points) lso‘tIqEr‘me process => 57:0 ‘7” A5 = 0 7‘4 .‘_ Q = w ::. nRTJ-u = (1-7mole5)¥(Q-3N¢-leole.k)(Wklﬂiuz (iron 15* law) " it. Q : w :: 3-92 RS 5- asOtuerMaL so AS 2 (1‘9. =- J. d0. = .2 : mud: J . T T . T v; I- L A g = (1.7 Males) (assailmrex) 21M). {— A5 2 9-8 Th1 Exam 2. Wednesday, April 1“, 2009 PHY 252 6. A 1.0 kg block of copper at 127.°C is brought into contact with another 1.0 kg block of copper at 27 .°C. Assume cCu = 386 J I Kg.K , and no heat is exchanged with the environment. (1') What is the ﬁnal temperature of the two blocks? (Show your working.) (4 points) C IOSecl sustem , “0 work is done -‘- Qto‘b ’- O . 1 Q‘ + Q1 = O _ .L MCCTs'Tt) + MCCTé’Tz) = 0 =7 T3 “ afﬁrm) 1;: £04.00 +300) : 350K (or 7'?‘°C.) '96 (ii) What is the total entropy changer-pf the system'jl} 1} (4 points) Ti- astot = As.+A52 = J is: +§ 019.: = m a}: +mcjel1 '1‘, T .l-" '1’ T T 2. T. 1 T1 AS = M C144. 13' 2 T5 = K It is?! «T'— . 3' £01: (1.. +Mc1u(ﬁ) Maln(_nn I 386 mam ?% IR 7. One mole of a monatomic gas, initially at 200K, is taken through a cycle ABCDA as drawn below. Find: 1') The work done during one complete cycle. (3 points) ii) TC (2 points) iii) TD (2 points) iv) The change in internal energy AEDA. (3 points) U) Work =3 area in : C2Pa“po) (2V9‘Vp) : Po v6 2V V V From IdeaLaqs law P.Vo=MQT4, EnRTA '—‘ “elm-"200 2'2'1‘62a‘ 7F ‘ ' lie. (I!) Pch. ___ Pal/A _.-—_-> Tc:- PcvaT = C2.%)C2Ua) ._ ‘636 T; n PA \/A A Po V0 Tﬂ- _ '— K * (lit) To = PBVDTA = p°(2V”)TA = 2-174 2' #DOK PAVA Pave UV) AEM = a CV (TA «‘13) ( even "thoudL it: Its a. coustautpressare precess) AEM : l x g-x 8-34 x (200 new) : —- 24.39.3- 8. A box contains 6 distinguishable gas atoms. The box has two regions, “left” and “right”. The macrostate shown in the ﬁgure is labeled “ 6/0 ” because there are 6 atoms in the left half and no atoms in the right half. (1') How many distinct left/right macrostates are there? (2 points) (ii) Make a table listing the number of microstates for each macrostate. (5 pts) ?— Hqcrosfa‘ée; *6 ...
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## This note was uploaded on 04/03/2012 for the course PHY 252 taught by Professor Treacy during the Spring '08 term at ASU.

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Exam2_Solution - Exam 2 Wednesday April 1“ 2009 PHY 252...

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