1

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 .

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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.

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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 ,,,,

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• 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

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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

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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

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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]

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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]

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(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

.

.

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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?