physical chemistry i for biochemists chem340 lecture 18 (2...
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Physical Chemistry I for Biochemists
Chem340Chem340
Lecture 18 (2/23/11)
Yoshitaka Ishii
Ch5.8-5.11 & HW6Review of Ch. 5 for Quiz 2
Announcement
• Quiz 2 has a similar format with Quiz1. • Time is the same ~20 minsTime is the same. 20 mins.
• Answer for HW5 will be uploaded this afternoon. Study it well for Quiz 2
• Questions from HW6 to be covered in Quiz 2:
P5.2, 5.5, 5.6, 5.7, 5.8, 5.13, 5.14, 5.20, 5.22, 5.30 & Q1-Q2Q2,
• where questions in red will be studied today & green were finished in Lecture 17 .
2
w & q in various process for ideal gas Type of work
w q U T
Expansion for Pext = U= CV T & H= CP T
ext
const
isotherm -PextV -w 0 0adiabatic -PextV 0 -PextV U/CV
Reversible expansion/
V P
compression
isotherm -nRT ln(Vfin/Vini) -w 0 0
adiabatic CvT 0 CvT Tini{(Vfin/Vini)a -1)}
a=1-CP/CV=1-
S for ideal gas for constant Type of work
w q S T
Irreversible
P t = constPext = const
Isotherm -PextV -w nRln(Vf/Vi) 0Adiabatic -PextV 0 nRln(Vf/Vi)
+nCvln(Tf/Ti)
Tf = Ti +T
w/CV
ReversibleReversible
P =Pext
isotherm -nRT ln(Vf/Vi) -w nRln(Vf/Vi) 0
adiabatic CvT 0 0 Tini{(Vfin/Vini)a -1)}
a=1-CP/CV=1-
3
Calculated Change in Enthalpy (continued)
Case 5 (reversible) & 5’(irreversible) process: (Ti, Vi) (Tf, Vf)
By calculating S for (Ti, Vi) (Ti, Vf) (Tf, Vf)
Case 6 & 6’ (Ti, Pi) (Tf, Pf)
By calculating S for (Ti, Pi) (Ti, Pf) (Tf, Pf) (Pi/Pf = Vf/Vi)
)/ln()/ln( , ifmVif TTnCVVnRS
)/ln()/ln()/ln()/ln( ifmPififmPif TTnCPPnRTTnCVVnRS
[Q1 ]
[Q2 ] )()()()( ,, ifmPififmPif [ ]
P5.22) One mole of an ideal gas with CV,m = 3/2 R is transformed from an initial state T = 600. K and P = 1.00 bar to a final state T = 250. K and P = 4.50 bar. Calculate U, H, and S for this process.
600K)-(250K Cn ΔU mV,
600K)-(250KR) (Cn ΔH mV,
)/ln()/ln( ifmPif TTnCPPnRS
[Q1 ]
[Q2 ]
[Q3 ] )()( , ifmPif
4
5.8 Absolute Entropies and The Third Law of Thermodynamics
• The entropy of an element or a compound is experimentally determined from – Dqreversible =CpdT
Cp,m for O2
T gasTb Liquid
f
mfusionTf Solid
pmmm
dTCHdTC
T
H
T
dTCKSTS
'""
'
')()( ,
0
0
Molar Entropy for Gas
T
Tb
gasmp
b
monvaporizatiTb
Tf
Liquidmp
T
dTC
T
H
T
dTC
'"
'"
"
" ,,,
The entropy of a pure, perfectly crystalline
substance (element or compound) is zero at 0K.
Third Law of thermodynamics - What is Sm(0K)?
• P5.14) The standard entropy Sm at 298.15K of Pb(s) is 64.80 J K–1 mol–1. Assume that the heat capacity of Pb(s) is given by
Th l i i i 327 4°C d h h f f i
CP,m
Pb,s J mol1 K1
22.13 0.01172T
K1.00 105 T2
K 2
• The melting point is 327.4°C and the heat of fusion under these conditions is 4770. J mol–1. Assume that the heat capacity of Pb(l) is given by
a. Calculate the standard entropy Sm of Pb(l) at 500°C.
CP,m
Pb,l J K1 mol1
32.51 0.00301T
K
2CHC T LiquidTf Solid [Q2] [Q3]
b. Calculate H for transformation Pb(s, 25°C) Pb(l, 500°C).
""
''
)()(2
,,12
1
dTT
C
T
HdT
T
CTSTS
T
Tf
Liquidmp
f
mfusionTf
T
Solidpm
mm
2
,,21 "')(1
T
Tf
Liquidmpmfusion
Tf
T
Solidpm dTCHdTCTTH
[Q1] [Q2] [Q3]
5
T Dependence of SmCp,m /T > 0 and Svaporization, Sfusion > 0 Sm increases as T increases
The origin of Cp,m and Sm for Solids• Number of degrees of freedom for molecule made
of n atoms.
NT: Translational: 3
NR: Rotational: Atom 0, Linear 2, Non-Linear 3
NV: Vibrational: 3n – NT – NR
Cp,m & Sm higher for a molecule with more atoms.
6
5.9 Standard States in Entropy Calculation (P102
• For enthalpy, we defined Hf,A0 for the most stable
pure elements at 298.15K (T0) and 1 bar (P0) as 0.
• We define standard state of Entropy as S0
m = Sm(P0, T0) • What is the relationship between S0
m(P) and S0
m(1 bar)?
S (1 bar P) = −Rln(P/P ) Ent
ropy
Sm(1 bar P) = −Rln(P/P0)
Mol
ar E
S0m
Sm(P) = S0m(P0) −Rln(P/P0)
5.10 Entropy Change in Chemical Reaction
At 298.15 K and 1 barA + 2B 2C + D • Sreaction
0 = 2SC,m0 + SD,m
0 – SA,m0 – 2SB,m
0
I lIn general
A(T) + 2B(T) 2C(T) + D(T)For a constant pressure
X
mXXR STS 00,)(
T
p
T
dTCKSTS
298
0
298'
')()(
00
XmpXp CC ,
Q. How much is Cp0 for the above reaction?
00000 22BmpAmpDmpCmpp CCCCC ,,,,
7
• P5.20) Consider the formation of glucose from carbon dioxide and water, that is, the reaction of the following photosynthetic process: 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g). The following table of information will be useful in working this problem:p
• Calculate the entropy and enthalpy changes for this chemical system at (a) T = 298 K and (b) T = 330. K. Calculate also the entropy of the surrounding and the universe at both temperatures.
(a) SR0 = XSX
0
(b)
T
p
T
dTCKSTS
298
0
298'
')()(
00 pXXp CC
5.2 Heat Engines and the Second Law of Thermodyamics
Isotherm rev.: -PdV = -(nRT/V)dVa b: qab = -wab =nRThotln(Vb/Va)>0
d RT l (V /V )<0
Carnot Cycle
c d: qcd = -wcd = nRTcoldln(Vd/Vc)<0Adiabatic: q= 0da & bc: qcd = qbc=0
For Cycle:U = 0 wcycle + qab+qcd =0 w = -(q +q ) <0 (|q |>|q |) wcycle = -(qab+qcd) <0 (|qab|>|qcd|) Efficiency =|wcycle|/|qab| = |qab+qcd|/|qab| = 1 – Tcold/Thot < 1
8
P5.2 Consider the reversible Carnot cycle shown in Figure 5.2 with 1 mol of an ideal gas with CV = 3/2R as the working substance. The initial isothermal expansion occurs at the hot reservoir temperature of Thot = 600°C from an initial volume of 3.50 L (Va) to a volume of 10.0 L (Vb). The system then undergoes an adiabatic expansion until the temperature falls to Tcold = 150.°C. The system then undergoes an isothermal compression and a subsequent adiabatic compression until the initial state described by Ta = 600.°C and Va = 3.50 L is reached.
a. Calculate Vc and Vd. (Vc/Vb)1- =(Tc/Tb) Vc = Vb(Tc/Tb)(1/1-)
(Va/Vd)1- = (Ta/Td)
b Calculate w for each step in the cycle and for the total cycle
[Q1 ] x [Q2 ]
b. Calculate w for each step in the cycle and for the total cycle.
wab = -nRTaln(Vb/Va) wbc = Ubc = Cv(Tc-Tb)
c. Calculate and the amount of heat that is extracted from the hot reservoir to do 1.00 kJ of work in the surroundings.
=|w|/|qab|
[Q3 ] [Q4 ]
5.11 Refrigerator, Heat Pump and Real Engines
Carnot Cycle: wcycle < 0 & qcycle > 0 Reversed Carnot cycle: wcycle’ > 0 & qcycle’ < 0 Reversed Carnot cycle: wcycle 0 & qcycle 0
qab >0 qab’ < 0 Hot sink heated
qcd <0 qcd’ > 0Cold sink cooled
9
Carnot Cycle Refrigerator
qab >0 qab’ < 0qcd < 0 qcd’ > 0qcd 0 qcd 0
Refrigeration efficiency for a reversibleCarnot refrigerator
/ T /(T T )
Impossible Heat Pump
r = qcold/w = Tcold/(Thot – Tcold)
Refrigeration efficiency for a reversibleCarnot heat pump
hp = qhot/w = Thot/(Thot – Tcold)
Ex. Tcold = 0.9 Thot =9 (1 J of work 9 J of cooling)Tcold = 0.8 Thot = 4
Calculated Change in Enthalpy (continued)• Case3 (3’). Reversible change in T for a fixed V
Case4 (4’) Reversible change in T for a fixed P
)/ln(,,
ifmVmvreversible TTnC
T
dTnC
T
DqS
Case4 (4 ). Reversible change in T for a fixed P
Case 5 (reversible) & 5’(irreversible) process: (Ti, Vi) (Tf, Vf)
By calculating S for (T V ) (T V ) (T V )
)/ln(,,
ifmPmPreversible TTnC
T
dTnC
T
DqS
By calculating S for (Ti, Vi) (Ti, Vf) (Tf, Vf)
Case 6 & 6’ (Ti, Pi) (Tf, Pf)
By calculating S for (Ti, Pi) (Ti, Pf) (Tf, Pf) (Pi/Pf = Vf/Vi)
)/ln()/ln( , ifmVif TTnCVVnRS
)/ln()/ln()/ln()/ln( ,, ifmPififmPif TTnCPPnRTTnCVVnRS
10
P5.6) One mole of N2 at 20.5°C and 6.00 bar undergoes a transformation to the state described by 145°C and 2.75 bar. Calculate S if
C T T 2
When Cp or Cv is not constant
•
CP,m
J mol1 K1 30.8111.87 103 T
K 2.3968105 T 2
K2
1.0176108
T 3
K3
Tmp,f
bar6 00
bar 2.75lnndT
T
Cn
p
pln nRΔS
f
R[Q1] [Q2]
115.418
65.293
32
Ti
K JK
Td
/n
bar6.00Tpi
KT
KT
dKT
cKT
ba
Equations to be memorized for ideal gas
(0) For constant Cv,m or Cp, m
• U = nCv,mT [1]• H = nCP mT = n(Cv m+ R)T [2]P,m ( v,m ) [ ](1) For a reversible isothermal process, U = H = 0 S = nRln(Vf/Vi) w = -q = -nRTln(Vf/Vi)
(2) For a reversible adiabatic process (for Cp,m, Cv,m const)
U f l f H & U l i [1] & [2]
1
pT
1
ff VT Useful for H & U calc. in [1] & [2]
f
i
i
f
p
p
T
T
i
f
i
f
V
V
T
0 )/ln()/ln( , ifmPif TTnCPPnRS
0 )/ln()/ln( , ifmVif TTnCVVnRS0S [Q1]
11
P5.13) One mole of an ideal gas with CV,m = 5/2 R undergoes the transformations described in the following list from an initial state described by Ti = 250. K and Pi = 1.00 bar. Calculate q, w, U, H, and S for each process.
a. The gas undergoes a reversible adiabatic expansion until the final press re is half its initial al efinal pressure is half its initial value.
• (a) qrev = 0. dS = Dqrev/T = 0
Obtain Tf
1
f
i
1
i
f
1
i
f
i
f
p
p
T
T
V
V
T
T
1
f
i
i
f
p
p
T
T
1
f
i
i
f
p
p
T
T
Use Xa=Yb X =Yb/a
[Q1]
w = U = nCv,mT; H = nCP,mT
Use X Y X Y
[Q2] [Q3]
• P5.30) Calculate Ssurroundings and Stotal for the processes described in parts (a) and (b) of Problem P5.13. Which of the processes is a spontaneous process? The state of the surroundings for each partprocess? The state of the surroundings for each part is as follows:
• a. 250. K, 0.500 bar (a) Reversible adiabaticexpansion
• b. 300. K, 0.500 bar (b) Adiabatic expansion at constant P = 0 500 barconstant Pext = 0.500 bar
• (a, b) qsurroundings = 0 Ssurroundings = 0
(a) Ssystem = 0[Q1]
12
(3) For an ideal gas in a reversible adiabatic process for T-dependent Cv(T) or Cp(T) (Memorize)
dTTC )(
dTTnCU mV )(, dTTnCH mP )(,
T
dTTnCPPnRS mP
if
)()/ln( ,
T
dTTnCVVnRS mV
if
)()/ln( ,
(4) For an ideal gas in a irreversible adiabatic process at a constant external pressure (Cv, Cp const)
wU
i
i
f
fexternalifmV, p
T
p
T pR nTT C n
TnCH mP , )/ln()/ln( , ifmPif TTnCPPnRS Q. What are U, H, S for Pext = 0? )/ln()/ln( , ifmVif TTnCVVnRS
Solve this for Tf
P5.6) One mole of N2 at 20.5°C and 6.00 bar undergoes a transformation to the state described by 145°C and 2.75 bar. Calculate S if
C T T 2
•
CP,m
J mol1 K1 30.8111.87 103 T
K 2.3968105 T 2
K2
1.0176108
T 3
K3
Tmp,f
bar6 00
bar 2.75lnndT
T
C n
p
plnnR ΔS
f
R
1
Ti
KJ K
Tdn
bar6.00Tpi
K
K T
KT
dKT
cKT
ba15418
65293
32
.
.
13
P5.13) One mole of an ideal gas with CV,m = 5/2 Rundergoes the transformations described in the following list from an initial state described by T = 250. K and P = 1.00 bar. Calculate q, w, U, H,and S for each process.and S for each process.
b. The gas undergoes an adiabatic expansion against a constant external pressure of 0.500 bar until the final pressure is half its initial value.
wU
i
i
f
fexternalifmV, p
T
p
T pR nTT C n
Solve this for Tf
if pp
TnCH mP ,
)/ln()/ln( , ifmPif TTnCPPnRS
Q. Which equation should we use for S?
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