講義上半學(xué)期eng sgwhk

Definition

of

entropy(熵的定義)：For

any

closed

system,

it

exists

at

equilibrium

asingle

valued

state

function

called

entropy,

denoted

S,which

is

an

extensive ty.

When

the

system

ischanged

from

equilibrium

A

to

equilibrium

B,

the

entropyincrement

S

equals

the

algebraic

sum

of

the

quotient

ofheat

over

temperature

in

any

reversible

processes,

i.e.,Q

S

S

B

S

A

i(QR)i

is

the

infinitesimalTity

of

energyd

asheat

to

the

system

reversibly

at

a

temperature

Ti.1Entropy

augmenting

principium

(熵增加原理)：For

aclosed

system

from

e

l A

to

equilibrium

B,forever The

entrop does

n

t an

adiabatiS=S

B

-S

A

02the

entropy

of

system

does

not

automatically

decrease.reversible

process,

and

augments

in

an

adiabatic

irreversibleprocess,

i.e.,>

adiabatic

irreversible=

adiabatic

reversibleDeduction：The

entropy

of

an

insulated

system

does

notautomatically

decrease

forever;

the

entropy

does

not

changeinreversible

processes,

and

augments

in

irreversible

processes，insulated(S) =

S

B

-

S

A

0>irreversible=

reversibleAny

system

at

equilibrium

has

a

state

function,

the

entropy,which

is

extensive

ty.

The

entropy

increment

of

systemundergone

any

process

consists

two

contributions

,i.e.,dS=deS

+diSdiSdeSdeS

Entropy

flow(熵流).

It

is

the

entropy

flowentering

into

the

system

when

the

system

exchangesenergy

and

substance

with the

surroundings

through

theboundary

,

it

has

not

defined

positive

or

negative

symboldiS

Entropy

production(熵產(chǎn)生).

It

isthe

entropyproduction

due

to

irreversible

processes

inside

the

system(such

as

diffusion,

heat

exchange,

chemical

reactions,

etc.)3Entropy

production

principium

(熵產(chǎn)生原理)：The

entropy

production

inside

a

system

could

notnever

be

negat.It

is

zero

in

reversible

processes,and

is

always

above

zero

in

irreversible

processes,

i.e.,4diS

0

{}>

irreversible=

reversibleThe

equilibrium

equation

of

entropy

for

any

systemdS/dt

=

(1/T)

Qi/dt

+

Sj

dnj/dt

+

diS/dtCalculation

of

the

change

in

entropy

of

thesurroundings(環(huán)境熵變的計算)We

assume

that

each

heat

source

is

very

huge,

andwith

constant

volume

and at

fixed

uniform

temperature，itis

that

the

change

of

the

heat

source

is

always

reversible,thusδQ(-δQ

)dSHeat

source

= or

ΔSHeat

source

=

-

sys

sys

Tsur

Tsurr5heat

rceen

inS

S

S

QsysIf

Tsurisconstant，surTHere，

QHeat

source＝-QsysAccording

to

the

Entropy

augmenting

principium,

theentropy

of

an

insulated

(or

isolated)

system

will

neverdecrease c

l

e

not i

e

iblprocesses,

and

will

increase

in

irreversible

processes.

Whenwe

use

entropy

to

determine

the

process,

we

often

combinethe

system

and

the

surroundings

as

an

isulated

system,

thuswe

have6Insulated(S) =

S+

Ssyssur0 {

>

irreversible

}=reversibleIII.

Calculation

of

the

entropy

change

in

non--8th

lecture-isothermal

processes(非等溫過程熵變的計算)(i)

Constant

pQp＝

dH

＝nCp,mdTT

T

nCp,mdTdS

QRFixed

composition

(no

chemical

reactions,

no

phasechange),

closed

system，constant

p，W′＝0So,T

2T

1nCp,

mdTS

δ

Q

p

Tp,mIf

C is

constant,

thenT1Tln

T

2p,mS

nC7Evidently，if

T↑，then

S↑.(ii)

Constant

VQV＝

dU

＝nCv,mdTT

T

nCV

,mdTdS

QVT

2

nCV

,mdTδ

QVFixed

composition

(no

chemical

reactions,

no

phasechange),closed

system，constant

V，W′＝0SoT1TSv,mIf

C

is

constant,

thenTln

T

2T

1V

,mS

nC8We

have

also，if

T↑，then

S↑.Example

2.

1mol

silver

is

heated

from

273K

to

303K

through

twofollowing

isochoric

processesa. Quasistatic

heating

by

using

a

series

of

heaters of

infinitesimaldifference

in

temperature,

no

friction;b.

Direct

heating

using

a

303K

heat

sourceKnown

that

the

CV,m=24.48

J.K-1.mol-1

is

constant,

please

evaluate

theentropy

changesof

the

system

and

its

surroundings,and

predict

theproceeding

direction

of

the

process.準靜態(tài)加熱溫差無限小熱源273.0000000002K,……273.0000000001K302.9999999998K302.9999999999KTfV

,mln

Tf

2.55J.K

1Solution

(a)

S

sys

nCV

,mdT

nC1mol

Ag,

273K,p1mol

Ag,

303K,piTTTiS

?sursurS

2.55J.K

1surdS

nCV

,mdT9TS

S

S

0isolatedsys

surreversible1mol

Ag,

303K,pb.

Direct

heating

using

a

303K

heat

sourceSolution

(b)1mol

Ag,

273K,p等壓加熱303K物體（熱源）直接接觸iTfV

,mTln

TfTTinCV

,mdT

nC

2.55J.K

1sys

sysS

?

S

System

absorbs

heatQsys

nCV

,m

(Tf

Ti

)

734.4JsursurTS

Qsys

734.4

2.42J.K

1S

insulated10303

2.55

2.42

0.13J.K

1

0

S

S

sys

surIrreversibleIV.

Calculation

of

the

entropy

change

in

phasetransition

of

pure

substances①

Reversible

phase

change

at

equilibrium

T

and

pHeat

of

*

*,

*,A()

?→

A()

phase

Hm

Hm

(

A)

?

Hm

(

A)change*

*11hase

chan

emT

m

m

H

*

S*m

S

(

)

S

()

Molar

phase

changeentropy，J.K-1.mol-12

special

cases:1.

At

p,

T,

the

phase

c occur

respectivelyboiling

point

Tb, sublimation

point

Ts,

and

fusion

point

Tf2.

The

phase

change

can

occur

also

under

saturationvapor

pressure

pS

at

a

given

temperature12②

Calculation

of

entropy

change

in

irreversiblephase

change

at

non

equilibrium

T

and

pFor

irreversible hase

chan

e

rocess we

should

find

areversible

route

to

calculate

the

entropy

change.

ForexampleB(,T1,p1)13S1B(,Teq,peq)B(,T2,p2)S3B(,

Teq,peq)S=?Irreversiblephase

changeS2Reversiblephase

changeThen,

S＝S１＋S２＋S３Such

as，H2O（l,

90℃,

101325Pa）S1H2O（g,

90℃,101325Pa)S3H2O(g,

100℃,101325Pa)H2O(l,

100℃,

101325Pa)dTT

eqS

=?Irreversiblephase

changeS2Reversiblephase

changeTTnCp,m

(H2O,l)11S

TTTnC

(H

O,

g)

dT2p,m32T2eqS

nvap

H

m,SS＝S１＋S２＋S３14The

rinci les

to

find

a

reversible

routeEach

step

of

the

route

should

be

reversible；The

calculation

of

S

of

each

step

of

the

route

canbe

done

by

existing

formula；(iii）The

heat

data for

calculating

S

of

each

step

ofthe

route

can

be

found.15Example

3.

乙醇沸

351K的g

H

=

39.53

KJ.mol-1,

液態(tài)密l

m度=0.7600

g.dm-3,將351K

、p下的1mol液態(tài)乙醇在等溫下始終對抗外壓為0.5

p蒸發(fā)為同溫的0.5p的乙醇蒸汽,假設(shè)蒸汽為理想氣體,請計算體系乙醇的熵變及環(huán)境熵變,并判斷該過程是否為不可逆過程?設(shè)計可逆途徑計算體系熵變C2H5OH,

1mol,

l,351K,

pC2H5OH,

1mol,

g,351K,

0.5pS體fTg

HS

l

m122ppTQS

R

nRln

1C2H5OH,

1mol,

g,351K,

pU=0,R

R

1

2Q

=-W

=nRTln(p

/p

)2113

ln

118.39J.K0.539.531035118.314S

S

1

S

體計算體系與環(huán)境實際交換的熱量C2H5OH,

1mol,

l,351K,

pC2H5OH,

1mol,

g,351K,

0.5pQsysQ2

W2Q

g

H1

l

m2

1

0.5

p

V

V

C2H5OH,

1mol,

g,351K,

pIsothermalprocessQsys=Q1+Q2surg

H

l m

S

1

Qsys

116.78J.KsursurTTT0.5

p

nRT

nRT

p

psur

2

1

S

insulatedsys

S

S

118.39

116.78

1.61J.K

1

0sur17irreversibleExample

4.

在373K,p下,

水的

l

H

=

40627

J.mol-1,

=0.9583gmg.dm-3,將1mol

水由373K,

p下經(jīng)下列兩種過程變?yōu)橥瑴赝瑝合碌乃羝?a.等溫、等壓無摩擦準靜態(tài)過程b.等溫向真空 蒸發(fā)求水及環(huán)境熵變,并分別判斷兩過程的方向性Solution（a）g

*l

m108.90J.KTphase1sysS

n

H

1

40627

373surTS

Qsys

108.90J.K

1surS

S

S

0insulatedsys

surReversible18Solution（b）Free

expansionS

?sys1Ssys

108.90J.KWe=0，W’=0U

Q

We

W

'

Qsyssince

H

U

pV

sysglso

Q

H

pV

H

pV

p

V

H

p

Vg

H

nRTg

H

*sur19ST

T

l

mT

Qsys

H

nRT

nR1S

insulated

S

S

sys

sur

nR

8.314J.K

0irreversible-1=9874J.mol .已知s

mExample

5.在p下,苯的Tf

=

278.2K,l在T=268.2K

、p

下的

HCJ.K-1.mol-1,=122.6

J.K-1.mol-1,并將兩者當作常數(shù),請用s

ml

H

=

9916

J.mol-1.lp

,m=

126.8p

,mCs熵增加原理判斷268.2K、p下的1mol液態(tài)苯能否經(jīng)等溫、等壓過程變?yōu)楣虘B(tài)苯（或哪種苯更穩(wěn)定）？設(shè)計可逆途徑計算體系熵變1mol

C6H6,

s,268.2K,

ps1mol

C6H6,

l,

268.2K,

p

S

體系l278.2

nCp,mdTS1

268.2

T3278.2268.2

nCp,mdTTS

1mol

C6H6,

l,

278.2K,

p1mol

C6H6,

s,278.2K,

p20T

l

HS2

s

msys123S

S

S

S

35.49J.K

1surT

TT1sl

H

)

9874m268.2S

Qsys

36.82J.KS

insulated

1.33J.K

1

0

S

S

sys

surirreversible即在268.2K，p下，液態(tài)苯可變成固態(tài)苯，固態(tài)苯是穩(wěn)定相（或固態(tài)苯不能變成液態(tài)苯）21V.Calculation

of

entropy

change

for

mixingprocessofinertperfectgas(不同物質(zhì)惰性理想氣體混合過程的熵變求算)（a）Mixing

at

constant

T

and

V

(等溫等容混合)AnA,

pA,T,V+BnB,

pA,

BnA,

n

,PpApA+BpABB

p22BAp

nA

nB

RT

nART

nART

pV

V

VpABpA+BpABU=0

(pV)

=pV-(pAVA+pBVB)=V

(pA+pB)-

V

(pA+pB)=0H=

U

+

(pV)

=0

V=V-(V+V)=-VWork

done

by

surroundingsto

systemWsur=-p(-Vsys)=-pVWork

done

by

system

tosurroundingsWsys=-pA(0-V)-pB(0-V)=-V(pA+

pB)=pVS=0W=Wsys+Wsur=-pV+

pV

=0Since

U=0，W=0，so

Q=0Theorem

of

mixing

at

constant

T

and

V

(等溫等容混合定理)U=0,

H=0,

(pV)=0,

S=0,

V=

-V23（b）Mixing

at

constant

T

and

p

(等溫等壓混合)+VAnA,T,pVBnB,T,pnA,

nB,

T,

p,

V+U

0,Q

WS1VA+VBnA,T,pAS2VA+VBnB,T,pBS3=0V2

n Rln

VA

n Rln

xS

QR

n Rln

VAdV

VBS2

nB

Rln

xBV1

1V2

nRTVW

pdV

VA

AAAVT

V1

AS

=

S1

+

S2

+

S3

=

-R(nAlnxA

+

nBlnxB)24mix

iii1The

entropy

change

of

mixing

R

substancesr

of

perfectgas

at

constant

T

and

p

S

Rn

ln

x

0Example

6.在一個箱子中,用隔板將0.8mol的N2氣和0.2mol的O2氣隔開,兩氣體的T,p

相同,撤掉隔板使兩氣體混合.設(shè)N2和O2都是理想氣體,求混合熵變,并論證此混合為不可逆過程.解Smix251

8.314

0.8ln

0.8

0.2

ln

0.2

4.160J.K因為混合前后p=0,

V=0,

所以W=-pV0因為T=0,

所以U=0,

Q=U-W=0，為絕熱過程所以S環(huán)=0S

S

S

4.160J.K

1

0孤

體

環(huán)（不可逆）（c）Gibbs

Paradox(佯謬)Mixing

of

perfect

gases

of

the

same

substances(同種類的理想氣體（粒子）混合)AnA,

T,

p+AnA,

T,

pA2nA,

T,

PS

=

-R(nAlnxA

+

nAlnxA)

=

0

？？In

fact

is

should

be26S

0VI.Calculation

of

entropy

change

in

mixingliquidsoftwodifferenttemperatures(不同溫度液體混合過程的熵變求算)(恒壓、絕熱情況下相同液體混合）T1

T2Liquid

1：from

T1

to

(T1+T2)/2212T1TT1p,mdT

T

T

nCp,m

ln

1 2S

nCdTT

TT1T2Liquid

2：from

T2

to

(T1+T2)/2222T2T

nCp,m

ln

1 2T2p,mS

nC1

2ln

T1

T2p,m2

TTS

2nC01

211

221

2121

22712since

T

T

4TT

2T

T1

20

221

1

2T

T1

2

T

2TT

T

2

T

2

2TT2

1

1

2

T

2

4TT

02

1

2T

2

2TT

T

2

4TTT

T

24TTBecauseVII.

T eral

formula

to

calculate

the

entropychange

of

perfect

gas(理想氣體熵變的一般公式)The

entropy

change

of

any

two

different

states

ofperfect

gas

of

mass

nA(T1,V1)

→

B(T2,V2)S

=

S(T2,V2)

-

S(T1,V1)11

221

2121

228For

example,

design

three

reversibleroutes

to

evalua the

entropy

changeIIsothermalV1

V2n,

T1,P,V2

IsochoricP=(V1/V2)P1

T1

T2n,

T1,P1,V1n,

T2,P2,V2,

1,P2,VV=(P1/P2)V1IIIsothermalP1

P2IsobaricT1

T2III11

221

2121

229IsobaricV1

1

V2n,

T,P1,V2IsochoricT’=(V2/V1)T12

1

2

1P1

P22121

2Calculation

formulas:1isochoricT2

nCV

,m

dTTTS1

Sisothermal

S

nR

ln

V2

V1T12isothermalisobaricT2

nCp,m

dTp2TS

S

S

nR

ln

p1

T

'T2

nCV

,m1nCp,misochoricTdTTTdT

T

'S3

Sisobaric

S11

221

2121

230Justification

of

the

equality

of

the

three

formula

2T2T1TnCV

,mdTV1VS1

nRln2TTV

,mTT1nCV

,m

dTTnC

dT

nRln

V2

2TV

,mT1nC

dTn(Cp,m

R)dTTTTV1

nRln

V2

2TTV

,mTT1T1p

,mTnC

dTnRdT

TTT

nC

dTV1V1

nRln

V2

TT

nC

dTTnC

dT2p,mV

,mT1TTT

nR(

ln

V2

lnT)

nC

dT2T

T1dT

TTV

,mT

nCT1V1p,mTT

S3111VT

VFor

ln

V2T1

0，we

can

choose

T

V2

TExample

7.

請分別用上述S三種表達式,求算2mol

CO理想氣體從300K,

25dm3變到600K,

100dm3的熵變.已知CO的3

1

1CV,m=Cp,m-R，

Cp,m

{

27.61

5.02

10

(

T

/

K

)}J

.K

.mol1isochoricTT2

nCV

,m

dTV1

TS1

Sisothermal

S

nR

ln

V2

60030012513

52.81J

.KTdT27.61

R

5.0310

T

S

nRln

100

n12isothermalisobaricTT2

nCp,m

dTp2TS

S

S

nR

ln

p1

6003002213

52.81J

.KTdT27.61

5.0310

T

pS

nRln

p1

n132isochoricTT

'

nCp,mT2nCV

,mdTTT3S

Sisobaric

SdT

T'VIII.Calculation

of

entropy

change

for

transferringdirectlyQfromhighT2

hotsourcetolowT1

hotsource(從高溫?zé)嵩碩2直接傳遞熱量Q到低溫?zé)嵩碩1的熵變求算)Q

QsyssysQsys

Qsyssys2iiTS

T

T

dT

T

2dT T

dT

22

2

QR

i

QsysQsys21

Qsys

Qsys

T

T

dT T

2dT

T

dT

1

1s

s1Q

Qs

s2

T

TT2T2-dT…..

T12

1T

T

TT

T

Q

T

sys

1

2

1

2

Since

the

systemreleases

heatsysQ

QS

Q

1sysS

Q

1

1

T

T

2 1

sur1T33Sinsulated

=

Ssys

+

Ssur

=|Q|/T2

0（spontaneous，irreversible）IX.Fundamental

equation

of

closed

system(封閉體系的基本公式)The

1st

law

gives

dU

=

Q

+

Wexp

+

W’，when

there

is

no

other

work34W’=0，

dU

=

Q+

Wexpin

a

reversible

processQ

=QR=

TdS，and

Wexp=-pdV，We

obtain

the

fundamental

equation:

dU

=TdS

-

pdVAny

processes

without

other

workFor

any

processesIn

reversible

processesdU=

Q+Wexp

=TdS

-

pdVQ

=

TdS，

Wexp

=

-pdVIn

irreversible

processes

Q

TdS，

Wexp

>

-pdVSince

the

circular

integration

dU

0so

theTdS

pdVi.e.,

in

a

circular

process,

the

work

done

by

system35

pdV

equals

TdSBased

on

this

point

we

may

use

the

T-S

chart

torepresent

work.T-Schart

of

Carnotc cle

of

erfeTT12I2IIIV過程I:

恒溫可逆膨脹(熵增加)IQ

Q2T12

2T

TSI

S2

S1

或QI=T2(S2-S1)=Q234III恒溫可逆膨脹中,體系從高溫?zé)嵩碩2取得熱Q2=T2S過程II:

絕熱可逆膨脹(熵不變)SII=0,

QII=0過程III:

恒溫可逆壓縮(熵減少)III1Q

QSIII

S1

S2

T1

T1S1

S2

S循環(huán)過程完成后,有U

dU

0;

S

dS

0W

Q

Qi

T2

S2

S1

T1

S2

S1

i熱機效率體系從低溫?zé)嵩碩1取熱Q1=-T1S過程IV:絕熱可逆壓縮(熵不變)SVI=0,

QVI=0循環(huán)過程包圍的面積

W

Q2

吸熱過程I線段下的面積Charts

of

Carnotc cle

of

erfeexpressed

usingdifferent

variablesp

TT1121224434UST33V1

2p321S2VH3

4414373SX.

The Entropy

change

in

chemical

reactions(化學(xué)反應(yīng)的熵變)The

entropy

change

in

chemical

reaction

can

be

alsoobtained

by

designing

a

reversible

route,

and

to

calculate

thesum

of

quotient

of

heat

over

temperature

in

this

route.However,

since

the

establishment

of

the

third

law

ofthermodynamics,

the

data

of

“absolute”

entropy(calorimetric

entropy)

of

each

substance

can

be

found

in

listof

thermodynamic

data.It

lists

often

the

data

at

1

standard

pressure

p

and

at298.15K.

Using

the

temperature

dependence

of

heat

capacitywe

can

calculate

the

entropy

change

in

chemical

reaction

at38other

temperature.:

the

differenceThe

standard

entropy

of

reaction,

r

Sbetween

the

molar

entropies

of

the

pure,

separated

productsand

the

pure,

separated

reactants,

all

substances

being

intheir

standard

states

at

the

specified

temperature.Fora

reaction0

νBBr

S

νB

Sm

(B)θ

θr

m

mProducts

Reactants

S

ν

S

ν

Sv

stoichiometric

coefficient39Example

8：Evaluate

thechange

in

standard

molarentropy

of

the

chemical

reaction

below

C2

H5OH

(g,298K

,

p

)

→

C2

H

4

(g,298K

,

p

)

H

2O(g,298K

,

p

)S

(C

H

OH

,

g,298K

)

282.59J

.K

?1.mol

?1Known：

m

2

6S

(C

H

,

g,298K

)

219.45J

.K

1.mol

140m

2

4S

(H O,

g,298K

)

188.74J

.K

1.mol

1m

2Solution：S

S

(C

H

,

g,298K

)

S

(H O,

g,298K

)m

m

2

4

m

2

S

(

C

H

OH

,

g,298K

)

125.60J.K

1

.mol

1m

2

6XI.

Calculation

of

the

entropy

change

of

simplepVT

systems(簡單pVT體系過程熵變的計算)ty

of

single

phase

pu bstance,

oraty

of ponent

system

of

fixedFor

a

givengivencomposit,

t eeds

only

two

independent

state

variablesto

determine

the

state

of

the

system,

so

we

can

use

twoformulaations:S

S

T

pS

T

,V41(1)

From

S=S(T,p)

S

S

T

pdS

T

dT

p

dp(1)dU=TdS-pdVFrom

the

basic

formulaof

closed

systemFrom

H=U+pVdH=dU+d(pV)=TdS-pdV+Vdp+pdV+dpdV=TdS+VdpSo,dS

dH

V

dpT

TA

given ty

of

single

phasepure

substance,

or

of

-ponent

system

of

fixedcomposition

(no

phase

change,

nochemical

reaction),

no

other

work.since

H=H

T

H

H

Introduce

it

intothe

above

equation

T

pdH

T

dT

p

dp42dS

dT

dp

dpT

T

T

p

T1

H

V1

H

T

pCp1

H

V

dpdS

TdT

T T

p

T

p

T

dS

S

dT

S

dp

p

(I)In

comparison

with

S

or

S

1

H

p

T

Relation

of

entropywith

T

(II)

T

CpT

pT

T

p

p

T

p

T

T

S

1

H

V

Relation

of

entropywith

p

(III)43(III)

can

bewritten

as

S

T

p

V

T

T

p

H

Differentiating

bothside

at

constant

p

T

p

T

p

S

V

T

S

T

p

T

pp

H

T

T

p

at

constant

TppTTT

p

S

T

T

p

Differentiate

(II)

H

We

obtain

then

S

V

T

p

S

T

T

p

It

may

change

order

H

T

p

T

p

of

differentiationfor

state

functionsp

p

T

T44

S

CpTRelation

of

S

with

T

p

T

Relation

of

S

with

pp

Tp

p

TT

T

S

1

H

V

V

S

S

Cp

1

H

T

pdS

T

dT

p

dpobtainfinally

TdS

dT

V

dpT

T

pT

T

T

p

T

pdS

Cp

dT

V

dp

nCp,m

dT

V

dp45(2)

From

S=S(T,V)

dS

S

dT

S

dVV

T

V

TdS

dU

p

dVT

TRelation

of

S

with

T

S

CVIn

the

same

waywe

can

justify

:T

V

T

p

Relation

of

S

with

p

V

p

T

T

S

1

U

T

p

T

V

So,CV1

U

dS

dT

pdVT T

p

TTT

T

T

dS

CV

dT

p

dV

nCV

,m

dT

p

dV

V

V46（3)

p,V,T

changes

of

perfect

gasFrom

PV=nRT

P

nR

V

nR

T

V

p

V

p

T

dS

nCp,m

dT

nRdpT

pdS

nCV

,mdT

nRdVT

VWeS

ST,pS

ST,V

47dS

nCV

,mdT

nRdVFromT

VTaking

logarithm

of

both

side

of

pV＝nRT，and

operate

the

differentiationp,mV,mand

introduce

C

－C

＝R，it

givesdp

dV

dTp

V

TVpp,mV

,mV

,m

C

V

dp

dV

nCdVdS

nCdS

nCV

,mdp

nCp,m

dVp

VRearrangingS

Sp,V48dS

nCV

,mdT

nRdVIfCp,m

,

CV,m

are

constantIntegrating