This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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. perloci. T, we. need. a. leanbk 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 = LoL. == 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 rootmean—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 iaieqL dqs law PV 2 "RT M
e (C1)“ = , /3Pv7 = 3xiou.0i3xw’r’e tenurns? same;
' “’4 =iIo 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 KTlﬂ = 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=l3k
T —————“‘S Assum'em an ‘tcleaL acts 2 0.0I7o M3 LOIS! IDsPg, +
(iii) How much work is done by the water vapor? (3 points)
w = PAV aft consi'au't prél’SSu re .
5 I a sum: Louie: has (L
w 2: LemmasPa # (0.0t?ou3  ‘0 “'0 M ’5‘“) denslbﬂ aj ~ 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 Frisf Lat.) A5 = Q~W
AE ‘" Z2S'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 ::. nRTJu = (17mole5)¥(Q3N¢leole.k)(Wklﬂiuz
(iron 15* law) " it.
Q : w :: 392 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 98 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 changerpf 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 = 2174 2' #DOK
PAVA Pave UV) AEM = a CV (TA «‘13) ( even "thoudL it: Its a. coustautpressare precess) AEM : l x gx 834 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 ...
View
Full
Document
This note was uploaded on 04/03/2012 for the course PHY 252 taught by Professor Treacy during the Spring '08 term at ASU.
 Spring '08
 Treacy

Click to edit the document details