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Unformatted text preview: PHYS4125 Spring 2011 _ Exam 3 [Schroeder Ch. 4 — 6.3)
Name ilgﬂj‘dﬁ 3 4/15/11 1. (25 points) The operation of a gasoiine combustion engine is represented
by the Otto cycle' in the figure below. We assume that the gasoﬁne air
intake mixture is an ideal gas and use a compression ratio of 4. 1 (V4 —
4V1). Assume that 102 — 3131. V '3 1/3 1 LIKILH/a/ P9: 333,)
U329“ '’ gifta/ : H— 94')
OWN!) 3/9: y’i I VI Volume 41/1 3) Determine the pressure and temperature at each of the vertex points of the
PuV diagram in terms of p1, T1 and 7. For” 'Powxi $1? ‘Tl—
Fcr ?OW°\+ J“39”] —“—> we Know +LDC6' 1""? $33i€m UMQLE‘ra'CGS en chi‘ctLaceﬁt CWXLOJQSSSQA town ﬁe? £33m ‘1 510mm (“a we an :M 2:13” as,
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W E93? ﬂws W ”‘6 {Judged Lewd EVE, Maﬁa b) What is the maximum efficiency of this engine?
V ‘ LU _ 5T: _. C: 57:
may? 6 1'52" = 91,9... z ’5“ "’ ‘7
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CHE” QéiqlacL+Z‘ LEW 2.) (25 points) In this problem you will consider how to measure the
change in internal energy of a system held at constant particle number by
measuring the temperature, pressure, and several easily measured
quantities of the system. (a)Write down the thennodynarnic identity for dU remembering that dN=0
for this problem. Then expand both d8 and dVin terms of the partial derivatives with respect to P and T. [Hint Remember that any
thermodynamic quantity can be written as the sum of. partial derivatives of
three other thermodynamic quantities Since N is constant in this problem 6— 6
x x dz ]. Use these expressions to
8—32 ('32 , substitute for ds and dVin the therrjnogynamic identity. 3 Q a!
ClUwf‘“ iss RivMW “1 “0‘3 <43: ,ﬁfll EU}? CW : "rag rev 2 will do, such as 5155: (137+ Z  93 '3
M W ‘ 37”? MJJ ”75%)er gbkgl‘i‘l'alll/g 9mg
M, 93 C, as r—W_r__9_,r_ ”my ciT
AU, ﬁlmjcll +3? WI JT la? a; E I aT’ﬁlJ
,3 as ——  .. as Q: .
Or JO 7 Cng”? ”‘3'?— FP] "hill“ 945 aPlr?] b.) By starting with the thermodynamic identity for 616, and taking second
order partial derivatives, calculate a Maxwell relation relating S, P, V. and T. 6S at), so (la“role + Pew ___ saws? : “slaw was a» CWﬁ/ﬁ/‘H
“L6 anl v: “lg—Hi)? Use this relation to substitute for in your expression for d U. 65 ET C.) By remembering the definitions of CP=T (the heat capacity at P 1 constant ressure) B—— a—V
p ’ V a'r (the thermal expansion coefficient), and
P —1 6V K2? 5; (the isothermal compressibility) substitute into the
T expression in part (a). You should have a fairly simple expression for CH]
in terms of {H and dP, and readily measurable quantities of the system. So. 011‘; :. CF J7“... [El/”PAT gwls’V'Td'P a, KVFdTﬁ 3. (20 points) In the textbook and in class we derived the electrical
potential difference, 8 of a galvanic cell in terms of the change in the Gibbs free energy, AG, and the number of electrons exchanged in the
reaction.
(a) Write down an expression of the electrical potential difference in terms of AG. (hint, it may be helpful to consider the work done on the electron as it is , _ ‘
traverses the circuit.) Sim g6 : U HTS 1?l/ ‘33 dG : $1.7“ng “and +9?” + V4?
ms gm: maﬁa»; AG— :QQ £135) .. m + vase (ww— .
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110%,, :: £5— 00 E ‘haljb PM Feudkm Wmh ,
(1)) Use this relation to calculate potential difference created at the terminals 12
of a galvanic cell making use of the following reaction. Refer to the table below for some helpful data.
Ag(solid) + 1/2 C12(gas) = AgCl(solid) Ag(solid) +Cl'(aq) = AgCl(soh‘d) +e
1&3]; (gas) +9 2 C1‘(aq) Formula AH (k1) AG (u) l s (J/mole K) as (J/mole K) 33.95
25.5 4. (30 points) With modern lithographic and molecular growth techniques
model quantum mechanical systems are easily Cteated. For example, a one
dimensional quantum well for an electron can be fabricated in a
GaAs/AlGaAs heterostructure. Assume that an electron is placed into such
a structure which can be approximated as a quantum well with infinitely high walls 10 run apart. a) Draw a simple picture of the ground state and the first 2 excited State
wavefunctions for this potential (a one dimensional particle in a box
potential with hard walls) . Write down expressions for the quantized
wavelengths, wavenumbers (k), momentum ( hk ), and energies for this potential. _ p ‘ ’7
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\/ “r5 MiGMFR cLt +1354” fem—h M :_ Vt .
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15+ email; 3% . L ____3 P: “in
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W E: 1 am : T n
L ~—:> L b) Using the expression from (a) for the energies calculate the classical
(Boltzmann theory) partition function keeping only the first three terms of thesum. ﬁf 1;an '1
Z : 2 65: 5 :2 “ﬁg—’1; H
5 1? c an— (34‘ 4‘} we hﬁig‘kfmg
.aﬁa :2 .1 ﬂ? __?Eﬁn3
Z' ‘ éiﬁlﬁa ted/EM +6 ”C”
a w a as a at then a an an).
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Pram” 2: C dEd'CLLMF +6 c.) From the partition function found in (b) calculate the average energy of
an electron in such a potential well, again keeping only the first 3 terms of __;j % 1  ~43 he”
{ﬂy 9 dB ,LIIMCXOJCDCF] Bonus: (10 points) Calculate the average energy in meV for a 10 nm
quantum well at a temperature of 10 K and comment on the result. “211173. : Slﬁ‘ifwougaﬁ .t 81‘?§’2({o"“nj 3‘ EFL/WV
9mg)? game “ #3%Br i broom/’6’ a Perl£33 Matias g?Lgmxaa+4ymcm694—M%Wd9"[email protected]%
Cor/W Mtg 3M( QME‘? E:t&[email protected] M I ‘ ‘ , we E l 47. 68“ E I§[3~?%ML13 ”[1‘3’953394 $133,066,?ch +9 1.1%,“) J
E: épfﬂosmo‘ayv) : 37;, My almost all 'Pﬂmloljfly tig M W PHYS4125 Spring 2011
Exam 3 gSchroeder Ch. 2 — 3! Formuias and Constants
First and Second Law Equipartition Theorem: U them: = N % kBT
Vf
13t Law of Thermodynamics: U 2 Q + W Compressive Work: W I J‘ P (V) dV
VD
Adiabatic Processes: PW: constant; VT?2 = constant; T = (f+2) / f
6 6 U
Heat Capacity: C = QIAT or C =7— 9 = —
6T MWH 5T Mvm
( n E + n2) !
Entropy: S EkB 1n ( .Q ) Entropy of mixing: S :kB 111 m
nin!
1 2 N 3le
1 V TT Ideal Gas Law: PV = nRT = ngT Ideal Gas: .QN mmw—gﬁ—ww—( 2mU)3N’2
N! h (3N/2)! 3/2
Sackur—Tetrode Equation: S :NkB 1n 3: 4 TTmU E
N 314312 2
Transport
6151" C
Heat Conduction: 22—krA— k ="1" "I"; V
A: dx ’ 2 V
1 V
Mean Free Path: 1 w 2 ~— ; RMS velocity: 12m = Germ)“
4 TU‘ N
—— I F du
Scattering time: A I '1”— Viscosity: —x{= n I
Wm A & . . . dn Diffnsmn, Ftck's Law: Jx= — D———
dx
W
N f ,1 _n N f
Binomial Distribution: PN(n)=— p rm 3 and Q (N, n):—
n!(IJn)! n!(IJn)!
+ N— 1 !
Einstein Solid: 9. (N , q): £§H__)
g .’(N m 1 )! Stiﬂing's Approximation: 111(N f) +13 Nln (N)~ N 1 _ —ﬁ 21, 2
Gaussian Distribution: PN( ﬂ)— 6 [(n ) 20 1 where 0: Npr or more generaﬂy
VZNU 2
0:\/— U for systems with only quadratic degrees of freedom and ﬂ: pN fN Interactions
. . a U
Deﬁnition of Temperature: T: —
a S N, V
C dT C
Entropy and heat: dS 3 ﬁg: V ,' A S 2f —V (3T for quasistatic processes.
T T T
Third Law ofThermodynamics: CV—‘> 0 as T >0,' S (T: 0)=0
a S a U
Deﬁnition of Pressure: P: T __...... —  —
a V U,N a V 5,»:
Thermodynamic Identity: d U : T 615— P dV + “ dN
a S a U
Deﬁnition of Chemical Potential: p = ~ T — = ——
6N U,V a N S,V Engines and Refrigerators
beneﬁt_ W __ Q}: —Qc < Tit—Tc
— — — x.
cost Q h Q h T h Deﬁnition of Coefﬁcient of Performance: beneﬁt : Q Q: Tc 2 s
cost W Qh—QC T—T h c Deﬁnition of Efﬁciency: 6 E COPE
J odeThompson 'llirottiing Process: Hf: H i Free Energy
Enthalpy: HE U +PV
Helmholtz Free Energy: F E U — TS
Gibbs Free Energy: G E U  TS + PV
Thermodynamic Identities: dH = TdS + VdP + p dN , tip 2 *SdT *— Pd V + u dN ,
dG=—SdT + VdPip dN
“l —— 1
For ﬁxed V, N: (ism; w dF For fixed P, N: dSmI= — dG
T T Chemical Potential: GZNLI ,‘u (T , P)=y°(T)+kBTln (JP/Pa)
dP L Clausius Clapeyron Relation for a Phase Boundary: '—' ""— dT TAV N2
van der Waals Equation of State: P+%5 (V  Nb}: NkBT Boltzmggg Statistics Boltzmann Factor: 8(‘E‘lkBT): 3‘3 El 1 —3 E! Probability function: P( 3): Z— 6 _ ’3 E:
Partition Function: Z ‘2 9 Average Values: X =2 X(S)P(S):i2 X(S)€_BE’ E =—l§—Z—
“”9 s z s “”9 Z . a B
Equipartition (Again): if 13(91): €42 then Eave=112kBT
Reference Data: 13 =1.3807 X1073 I/K = 8.617 x 10‘5 eVIK N1 = 6.022 111023
R=83151lmolK h=6.626x 10’34Js=4.136x 10‘15 eVs
c = 2.998 X 108 m/s G = 6.673 x 10'11 Nmza'kg2
e = 1.602 x 1019 c m. = 9.109;: 10—31115;
mp = 1.673 x 10'27 kg 63 = 9.2741x10'2“ J/T . Unit Conversions:
lann = 1.013 bar: 1.013 x105 Pa: 14.7 161162 = 760 mm Hg
(Tin°C)=(Ti31K)—273.15 (Tin‘mwis (Ti11“C)+32
1"R=5f9K 1ca1=4.186J
1Bt11=1054J 16V: 1.6021110ng ' 1u=1.661x10‘27kg ...
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