felix termodinamica quimica ch05

8/13/2019 felix termodinamica quimica ch05

1/83

Chapter 5 SolutionsEngineering and Chemical Thermodynamics

Wyatt Tenhaeff

Milo Koretsky

Department of Chemical Engineering

Oregon State University

[email protected]


2/83

2

5.1

(a)

Following the example given by Equation 5.5a in the text

dPPudT

Tudu

TP

+

=

(b)

dss

udT

T

udu

Ts

+

=

(c)

dss

udh

h

udu

hs

+

=


3/83

3

5.2.

The internal energy can be written as follows

dvv

udT

T

udu

Tv

+

=

Substituting Equations 5.38 and 5.40

vv

cT

u=

and

=

P

T

PT

v

u

vT

into the above expression yields

dvP

T

PTdTcdu

v

v

+=

From the ideal gas law, we have

v

R

T

P

v

=

Therefore,

dvP

v

RTdTcdu v

+=

which upon noting thatv

RTP= for an ideal gas, becomes

dTcdu v=


4/83

4

5.3

The heat capacity at constant pressure can be defined mathematically as follows

( )

PvPPP

T

vP

T

u

T

Pvu

T

hc

+

=

+=

=

For an ideal gas:

P

R

T

v

P

=

Therefore,

RT

uc

vP +

=

One mathematical definition of du is

dPP

udT

T

udu

TP

+

=

We can now rewritevT

u

:

vvTvPv

cT

P

P

u

T

T

T

u

T

u

=

+

=

For an ideal gas:

0=

TP

u

so

Pv T

u

c

=

Substituting this result into our expression for Pc gives

Rcc vP +=


5/83

5

5.4 In terms of P, v,and T, the cyclic equation is

TPv P

v

v

T

T

P

=1

For the ideal gas law:

RTPv=

so the derivatives become:

v

R

T

P

v

=

R

P

v

T

P

=

P

v

P

RT

P

v

T

=

=

2

Therefore,

1=

=

P

v

R

P

v

R

P

v

v

T

T

P

TPv

The ideal gas law follows the cyclic rule.


6/83

6

5.5

For a pure species two independent, intensive properties constrains the state of the system. If wespecify these variables, all other properties are fixed. Thus, if we hold Tand P constanth cannotchange, i.e.,

0,

=

PTv

h


7/83

7

5.6

Expansion of the enthalpy term in the numerator results in

ss T

PvsT

T

h

+=

ss T

Pv

T

h

=

Using a Maxwell relation

Ps v

sv

T

h

=

PPs v

T

T

sv

T

h

=

We can show that

T

c

T

s P

P

=

(use thermodynamic web)

+

=

32

221

v

ab

v

a

bv

RT

Rv

T

P

(differentiate van der Waals EOS)

Therefore,

=

+

=

v

b

vRT

a

bv

vc

v

ab

v

a

bv

RT

RT

vc

T

hP

P

s

1222

32


8/83

8

5.7

:TP

h

vPsT

PPvsT

Ph

TTT

+

=

+=

( )2''1 PCPBP

R

T

v

P

s

PT

++=

=

( ) vvvPCPBP

RT

P

h

T

+=+++=

2''1

0=

TP

h

:sP

h

vP

PvsT

P

h

ss

=

+=

( )2''1 PCPBP

RT

P

h

s

++=

:PT

h

PP

cT

h=

(Definition of cP)

:

sT

h

sss T

Pv

T

PvsT

T

h

=

+=

( )2''1 PCPBRP

T

c

v

T

T

c

s

P

T

s

T

P P

P

P

TPs ++=

=

=


9/83

9

( ) ( )2

2

2 ''1

''1

''1

1

PCPB

PCPBc

PCPBRT

Pvc

T

hPP

s ++

++=

++=

Ps

cT

h=


10/83

10

5.8

(a)

A sketch of the process is provided below

mwell

insulated

s = 0T1

P1 T2

P2

The diagram shows an infinitesimal amount of mass being placed on top of the piston of apiston-cylinder assembly. The increase in mass causes the gas in the piston to be compressed.Because the mass increases infinitesimally and the piston is well insulated, the compression isreversible and adiabatic. For a reversible, adiabatic process the change in entropy is zero.Therefore, the compression changes the internal energy of the gas at constant entropy as thepressure increases.

(b)To determine the sign of the relation, consider an energy balance on the piston. Neglectingpotential and kinetic energy changes, we obtain

WQU +=

Since the process is adiabatic, the energy balance reduces to

WU=

As the pressure increases on the piston, the piston compresses. Positive work is done on the

system; hence, the change in internal energy is positive. We have justified the statement

0>

sP

u


11/83

11

5.9

(a)

By definition:

PT

v

v

=

1

and

TP

v

v

=

1

Dividing, we get:

TP

T

P

v

P

T

v

P

v

T

v

=

=

where derivative inversion was used. Applying the cyclic rule:

vTP P

T

v

P

T

v

=

1

Hence,

vT

P

=

(b)If we write T= T(v,P), we get:

dPP

Tdv

v

TdT

vP

+

=

(1)

From Equations 5.33 and 5.36

dPT

vdT

T

cdv

T

PdT

T

cds

P

P

v

v

=

+=

We can solve for dTto get:


12/83

12

dPT

v

cc

Tdv

T

P

cc

TdT

PvPvvP

+

=

(2)

For Equations 1 and 2 to be equal, each term on the left hand side must be equal. Hence,

vvPP T

P

cc

T

v

T

=

or

PPv

vPT

vT

T

v

T

PTcc

=

=

where the result from part a was used. Applying the definition of the thermal expansioncoefficient:

2Tv

T

v

T

PTcc

Pv

vP =

=


13/83

13

5.10

We need data for acetone, benzene, and copper. A table of values for the molar volume, thermalexpansion coefficient and isothermal compressibility are taken from Table 4.4:

Species

mol

m

10

36

v 1-3

K10 1-10

Pa10

Acetone 73.33 1.49 12.7

Benzene 86.89 1.24 9.4

Copper 7.11 0.0486 0.091

We can calculate the difference in heat capacity use the result from Problem 5.9b:

2vTcc vP =

or

( ) [ ]( )

[ ]

=

=

Kmol

J6.37

Pa107.12

K1049.1K293mol

m1033.73

1-10

21-33

6

vp cc

Species

Kmol

Jvp cc

Kmol

Jpc % difference

Acetone 37.6 125.6 30%

Benzene 41.6 135.6 31%

Copper 0.5 22.6 2%

We can compare values to that of the heat capacity given in Appendix A2.2. While we oftenassume that cPand cvare equal for condensed phases, this may not be the case.


14/83

14

5.11

We know from Equations 4.71 and 4.72

PT

v

v

=

1 and

TP

v

v

=

1

Maxwell relation:

vT T

P

v

s

=

Employing the cyclic rule gives

PTv T

v

v

P

T

P

=

which can be rewritten as

T

P

vT

v

P

v

T

v

v

T

P

v

s

=

=

1

1

Therefore,

=

Tvs

Maxwell Relation:

PT T

v

P

s

=

From Equation 4.71:

vTv

P

=

Therefore,

vP

s

T

=


15/83

15

5.12

(a)

An isochor on a Mollier diagram can be represented mathematically as

vsh

This can be rewritten:

vvv s

PT

s

PvsT

s

h

+=

+=

Employing the appropriate Maxwell relation and cyclic rule results in

Tvv vs

sTvT

sh

+=

We know

vv c

T

s

T=

and

vT T

P

v

s

=

For an ideal gas:

vR

TP

vs

vT

=

=

Therefore,

+=+=

vvv c

RT

v

R

c

TvT

s

h1

(b)

In Part (a), we found

vvv T

P

c

TvT

s

h

+=

For a van der Waals gas:


16/83

16

bv

R

T

P

v =

Therefore,

+=

bv

v

c

RTT

s

h

vv


17/83

17

5.13

(a)

The cyclic rule can be employed to give

TPs Ps

sT

PT

=

Substitution of Equations 5.19 and 5.31 yields

PPs T

v

c

T

P

T

=

For an ideal gas:

PR

Tv

P

=

Therefore,

PPs c

v

cP

RT

P

T==

1

(b)

Separation of variables provides

P

P

c

R

T

T

P

=

Integration provides

Pc

R

P

P

T

T

=

1

2

1

2 lnln


Pc

R

P

P

T

T

=

1

2

1

2

The ideal gas law is now employed


18/83

18

Pc

R

P

P

vP

vP

=

1

2

11

22

1

1

12

1

2 vPvP PP

c

R

c

R

=

where

kc

c

c

Rc

c

R

P

v

P

P

P

11 ==

=

If we raise both sides of the equation by a power of k, we find

kk vPvP1122

=

.constPvk =

(c)

In Part (a), we found

PPs T

v

c

T

P

T

=

Using the derivative inversion rule, we find for the van der Waals equation

( )

( )233

2 bvaRTv

bvRv

T

v

P

=

Therefore,

( )

( )23

3

2

1

bvaRTv

bvRTv

cP

T

Ps

=


19/83

19

5.14

The development of Equation 5.48 is analogous to the development of Equation E5.3D. Wewant to know how the heat capacity changes with pressure, so consider

T

P

P

c


PTTPT

P

P

h

TT

h

PP

c

=

=

Consider theTP

h

term:

vT

vTv

P

sT

P

PvsT

P

h

PTTT

+

=+

=

+=

Substitution of this expression back into the equation forT

P

P

c

results in

PPT

P vT

vT

TP

c

+

=

PPPT

PTv

TvT

Tv

TT

Pc

+

=

2

2

PT

P

T

vT

P

c

=

2

2

Therefore,

=

real

ideal

real

P

ideal

P

P

P P

c

c

P dPT

vTdc

2

2

and

=

real

ideal

P

P P

idealP

realP dP

T

vTcc

2

2


20/83


21/83

21

5.16

A schematic of the process follows:

Propane in

v1= 600 cm 3/mol

T1= 350 oC

Turbine

P2 = 1 atm

ws

Propane

out

We also know the ideal gas heat capacity from Table A.2.1:

263 10824.810785.28213.1 TTR

cP +=

Since this process is isentropic (s=0), we can construct a path such that the sum of s is zero.

(a) T, vas independent variables Choosing Tand v as the independent variables, (and changing T under ideal gas conditions), weget:

Temperature

volum

e

IdealGas

step1

step 2

v1,T1

s=0

v2,T2

s1

s2

or in mathematical terms:

0=

+

= dv

v

sdT

T

sds

Tv

However, From Equation 5.33:


22/83

22

ds=cv

TdT+

P

T

v

dv

To get s1

P= RT

vb

a

v2 so

P

T

v

= R

vb

and

s1= d s= P

T

v

dvv1

v2

= R

vbdv=R ln

v1

v2

v2b

v1b

or, using the ideal gas law, we can put s1 in terms of T2:

s1=R ln

RT2P2 bv1b

For step 2

+

==2

1

2

K15.623

263

210824.810785.28213.0

T

T

Tv dT

T

TTRdT

T

cs

Now add both steps

( ) ( )[ ]2226

232

1

2

2

21

K15.6232

10824.8K15.62310785.28

15.623ln213.0ln

0

+

+

=

=+=

TT

T

bv

bP

RT

sss

Substitute

K15.6231=T /molcm600 31=v

atm12=P

=

Kmol

atmcm06.82

3

R


23/83

23

and solve for T2:

[ ]K3.4482=T

(b) T, P as independent variables

Choosing T and P as the independent variables, (and changing T under ideal gas conditions), weget:

Temperature

Pressure

IdealGas

step1

step 2

P1,T1s=0

P2,T2

s1

s2

Mathematically, the entropy is defined as follows

0=

+

= dP

P

sdT

T

sds

TP

Using the appropriate relationships, the expression can be rewritten as

0=

= dP

T

vdT

T

cds

P

P

For the van der Waals equation

( )

( )

+

=

322

v

a

bv

RT

bv

R

T

v

P

Therefore,


24/83

24

( )

( )

02

2

1

2

132

=

+

= dP

v

a

bv

RT

bv

R

dTT

cs

P

P

T

T

P

We cant integrate the second term of the expression as it is, so we need to rewrite dP in terms

of the other variables. For the van der Waals equation at constant temperature:

( )dv

bv

RT

v

adP

=

23

2

Substituting this into the entropy expression, we get

( )010824.810785.28213.1

2

1

2

1

263

=

+=

dvbv

RdTT

TTs

v

v

T

T

Upon substituting

=

=

=

=

=

Kmol

atmcm06.82

mol

cm91b

atm)1atideallyacts(gas

cm600

K15.623

3

3

2

22

31

1

R

P

RTv

v

T

we obtain one equation for one unknown. Solving, we get

K3.4482=T


25/83

25

5.17

(a)

Attractive forces dominate. If we examine the expression forz, we see that at any absolutetemperature and pressure, .1


26/83

26

Taking the derivative gives:

5.05.0 +=

aRT

P

R

T

v

P

so

( dPaRTdTcdh P 5.05.0+=

For step 1

( )

=== mol

J2525.05.0

0

bar50

5.01

5.011 PaRTdPaRTh

For step 2

( )

=+=

mol

J7961875.01002.358.3

K500

K300

5.032 dTTTRh

For step 3:

( )

=== mol

J3235.05.0

bar50

0

5.02

5.023 PaRTdPaRTh

Finally summing up the three terms, we get,

=++=

mol

J7888321 hhhq

Alternative 2: real heat capacity

For a real gas

realP

ch=

From Equation 5.48:

=

real

ideak

P

P P

idealP

realP

dPT

vTcc

2

2

For the given EOS


27/83

27

+= 2/1

1aT

PRv

Therefore,

5.12

2

25.0 =

aRT

T

v

P

and

[ ]( ) 5.01/2bar50

bar0

5.0

2

2

K875.025.0 =

=

==

RTdPaRTdP

T

vT

real

ideak

real

ideak

P

P

P

P P

We can combine this result with the expression for realP

c and find the enthalpy change.

( )dTTTRh +=K500

K300

5.03 875.01002.358.3

==

mol

J7888hq

The answers is equivalent to that calculated in alternative 1


28/83

28

5.18

(a)

Calculate the temperature of the gas using the van der Waals equation. The van der Waalsequation is given by:

2v

a

bv

RTP

=

First, we need to find the molar volume and pressure of state 1.

( ) [ ]( )[ ]

====

mol

m00016.0

mol250

m4.0m1.0 3211

n

Al

n

Vv

[ ]( )

[ ] [ ] [ ]Pa1008.1Pa1001325.1m1.0s

m81.9kg10000

65

2

2

1 =+

=+= atmPA

mg

P

Substituting these equations into the van der Waals equation above gives

[ ]2

3

3

35

3

16

mol

m00016.0

mol

mJ.50

mol

m104

mol

m00016.0

Kmol

J314.8

Pa1008.1

=

T

K5.2971=T

Since the process is isothermal, the following path can be used to calculate internal energy:

v

T

v1,T1

v2,T2= T1

u

s

Thus, we can write the change in internal energy as:


29/83


30/83

30

(b)

From the definition of entropy:

surrsysuniv sss +=

First, lets solve for syss using the thermodynamic web.

dvv

sdT

T

sds

Tvsys

+

=

Since the process is isothermal,

dvv

sds

Tsys

=

=

2

1

v

v vsys dv

T

Ps

Again, for the van der Waals equation,

bv

R

T

P

v =

Substitution of this expression into the equation for entropy yields

=

2

1

v

v

sys dvbv

Rs

=

=

Kmol

J17.44

mol

m104

Kmol

J314.80244.0

00016.03

5

dv

v

ssys

K

J5.11042

= sysS

The change in entropy of the surroundings will be calculated as follows

surr

surrsurr

T

Qs =

where


31/83

31

QQsurr = ( Qis the heat transfer for the system)

Application of the first law provides

WUQ =

We know the change in internal energy from part a, so lets calculate Wusing

=2

1

v

v

PdvnW

Since the external pressure is constant,

[ ]( ) [ ]( )

=

molm00016.0

molm0244.0Pa1001325.1mol250

335W

[ ]J614030=W

Now calculate heat transfer.

[ ] ( ) [ ] [ ]J1039.1J614030J776100 6==Q

Therefore,

[ ][ ]

=

=

K

J4672

K5.297

J1039.1 6

surrS

and the entropy change of the universe is:

=

=

K

J5.6370

K

J4672

K

J5.11042univS


32/83

32

5.19

First, calculate the initial and final pressure of the system.

[ ] [ ]( )

[ ]

[ ]Pa1092.4m05.0

m/s81.9kg20000Pa1010 6

2

25 =+=iP

[ ] [ ]( )

[ ] [ ]Pa1089.6

m05.0

m/s81.9kg30000Pa1010 6

2

25 =+=fP

To find the final temperature, we can perform an energy balance. Since the system is well-insulated, all of the work done by adding the third block is converted into internal energy. Theenergy balance is

wu=

To find the work, we need the initial and final molar volumes, which we can obtain from thegiven EOS:

/molm1037.8 34=iv

( )[ ]/molm103.2

2511089.6

314.8 35-

6

+

+

=

f

ff

T

Tv

Now, calculate the work

( ) ( )( )

+

+

== 45-

6

6 1037.8103.225

11089.6

314.8Pa1089.6

f

fiff

T

TvvPw

We also need to find an expression for the change in internal energy with only one variable: Tf.To find the change in internal energy, we can create a hypothetical path shown below:

v

T

vi,Ti

vf,Tf

step1

step3

step 2

ideal gas

500 Tf

u= - Pf( vf-vi)


33/83


34/83

34

( )( ) ( )

( ) ( )( )( )

( )( )

+

+

=

++

+++

+

45-

6

6

62

2

222

2

1037.8103.225

11089.6

314.8Pa1089.6

//2511089.6

314.8ln

500025.0500686.11/

ln

f

f

lowff

f

f

f

ffi

lowi

i

i

T

T

bPRTT

T

aT

aRT

TTbv

bPRT

aT

aRT

Solving for Tf we get

K2.536=fT

The piston-cylinder assembly is well-insulated, so

sysuniv ss =

Since the gas in the cylinder is not ideal, we must construct a hypothetical path, such as oneshown below, to calculate the change in entropy during this process.

P

T

Pi,Ti

Pf,Tf

step1

step3

step 2

ideal gas

500 536

Plow

4.9

6.9

For steps 1 and 3

=

=

low

i

low

i

P

P P

P

P T

dPT

vdP

P

ss1


35/83

35

=

=

f

low

f

low

P

P P

P

P T

dPT

vdP

P

ss3

We can differentiate the given EOS as required:

( )

( )

( )

( )

+

+=

+

+=

i

low

i

iiP

P i

ii

P

P

Ta

TaRTdP

PTa

TaRTs

low

i

ln22

221

( )( )

( )( )

+

+=

+

+=

low

f

f

ffP

P f

ff

P

P

Ta

TaRTdP

PTa

TaRTs

f

low

ln22

223

For step 2

+==

=

f

i

f

i

f

i

T

T

T

T

PT

T P

dTT

dTT

cdT

T

ss 05.0

202

( )ifi

fTT

T

Ts +

= 05.0ln202

Sum all of the steps to obtain the change in entropy for the entire process

321 sssss sysuniv ++==

( )( ) ( ) ( )

+++

+

++=

low

f

f

ffif

i

f

i

low

i

iiuniv

PP

TaTaRTTT

TT

PP

TaTaRTs ln205.0ln20ln2

22

Arbitrarily choose Plow(try 100 Pa), substitute numerical values, and evaluate:

( )

=

=

=

K

J766.0

Kmol

J388.0mol2

Kmol

J388.0

univ

univ

S

s


36/83

36

5.20

A schematic of the process is given by:

V = 1 L

T = 500 K

wellinsulated

Vacuum

V = 1 L

nCO=1 mole

State i

V = 2 L

T = ?

nCO=1 mole

State f

(a)

The following equation was developed in Chapter 5:

dvT

PTcc

v

v v

idealv

realv

ideal

+=

2

2

For the van der Waals EOS

02

2

=

T

P

Therefore,

idealv

realv cc =

From Appendix A.2:

( )

=

+=

Kmol

J0.22

500

31005001057.5376.3

24 RRcrealv

(b)As the diaphragm ruptures, the total internal energy of the system remains constant. Because thevolume available to the molecules increases, the average distance between molecules alsoincreases. Due to the increase in intermolecular distances, the potential energies increase. Sincethe total internal energy does not change, the kinetic energy must compensate by decreasing.Therefore, the temperature, which is a manifestation of molecular kinetic energy, decreases.


37/83

37

(c)

Because the heat capacity is ideal under these circumstances we can create a two-stephypothetical path to connect the initial and final states. One hypothetical path is shown below:

v

T

vi,Ti

vf,Tf

step2

step 1

500Tf

u2

s2

u1, s1

For the first section of the path, we have

==f

i

f

i

T

T

idealv

T

T

realv dTcdTcu1

( )

( ) ( ) 4.105074.2577375.191032.2

31001057.5376.2

231

K5002

41

++=

+=

fff

T

TTTu

dTT

TRu

f

i

For the second step, we can use the following equation

=

f

i

v

v T

dvv

uu2

If we apply Equation 5.40, we can rewrite the above equation as

=

f

i

v

v v

dvPT

PTu

2

For the van der Waals EOS,2v

a

bv

RTP

= ,


38/83


39/83

39

( )

+=002.0

001.0

K497

K500

dvbv

RdT

T

cs vsys

( ) ( )

==

+

+=

Kmol

J80.5

1095.3

31001057.5

376.2002.0

001.05

K497

K5003

4

sysuniv

sys

ss

v

dv

dTTT

Rs


40/83

40

5.21

A schematic of the process is given by:

V = 0.1 m3

T = 300 K

wellinsulated

Vacuum

nA=400 moles

State i

T = ?

State f

V = 0.1 m3

nA=400 moles

V = 0.2 m3

Energy balance:

0=u

Because the gas is not ideal under these conditions, we have to create a hypothetical path thatconnects the initial and final states through three steps. One hypothetical path is shown below:

v [m3/mol]

T [K]

vi,Ti

vf,Tf

step1

step3

step 2

ideal gas

300Tf

u

0.2/400

0.1/400


=

=

v

v Ti

dvv

u

u1

If we apply Equation 5.40, we can rewrite the above equation as

=

=

v

v vi

dvPT

PTu1


41/83

41

For the van der Waals EOS

22vT

a

bv

R

T

P

v

+

=

Therefore,

==

+

=

=

=

=

= mol

J1120

2

44 105.22

105.221

v

v i

v

vii

dvvT

advP

Tv

a

bv

RTu

Similarly for step 3:

==

+

=

=

=

=

=Kmol

J1680002

44

105

2

105

23f

v

v f

v

v T

dvvT

advP

Tv

a

bv

RTu

ff

For step 2, the molar volume is infinite, so we can use the ideal heat capacity given in theproblem statement to calculate the change in internal energy:

( )K3002

32 = fTRu

If we set sum of the changes in internal energy for each step, we obtain one equation for oneunknown:

( ) 0168000K300Kmol

J314.8

2

3

mol

J1120321 =

+

+

=++

ff

TTuuu

Solve for Tf:

K6.261=fT


42/83

42

5.22

A schematic of the process is shown below:

Ethane 3 MPa; 500K

Initially:vacuum

Tsurr= 293 K

(a)

Consider the tank as the system. Since kinetic and potential energy effects are negligible, theopen system, unsteady-state energy balance (Equation 2.47) is

++=

out

soutout

in

ininsys

WQhnhndt

dU&&&&

The process is adiabatic and no shaft work is done. Furthermore, there is one inlet stream and nooutlet stream. The energy balance reduces to

ininsys

hndt

dU&=

Integration must now be performed

=t

inin

U

U

dthndU0

2

1

&

ininin

t

inin hnnhndtnhunun )( 120

1122 === &

Since the tank is initially a vacuum, n1=0, and the relation reduces to:

inhu =2


43/83

43

As is typical for problems involving the thermodynamic web, this problem can be solved inseveral possible ways. To illustrate we present two alternatives below:

Alternative 1: path through ideal gas state

Substituting the definition of enthalpy:

ininin vPuu +=2

or

( ) ( ) ininin vPuTu = K500MPa,3atMPa,3at2 (1)

From the equation of state:

( ) ( ) ( )[ ]

=

=+=

mol

J800,3Pa103108.21K552

Kmol

J314.8'1 68PBRTvP inin (2)

The change in internal energy can be found from the following path:

Plow

T

step1

step3

step 2 ideal gas

500 K T2

u1

3 MPa

u3

u2

P

For steps 1 and 3, we need to determine how the internal energy changes with pressure atconstant temperature: From the fundamental property relation and the appropriate Maxwellrelation:

TPTTT P

vPT

vTP

vPP

sTP

u

=

=

From the equation of state

( ) RTBP

RTPB'P

P

RT

P

u '

T

=

+=

2

1


44/83

44

So for step 1:

[J/mol]3490

'

0

'

1 ===

= in

Pin PinT

PRTBRTdPBdPP

uu (3)

and for step 3:

TPRTBRTdPBdPP

uu

P P

T

7.02

2

0

'

2

0

'

3 ===

= (4)

For step 2

[ ]dTRcdTT

Pv

T

hdT

T

uu

T

P

T

PP

T

P

=

=

=

500500500

2

or

[ ]=

+=2

1 K500

263

2 10561.510225.19131.0

T

T

dTTTRu (5)

Substituting Equations 2, 3, 4, and 5 into 1 and solving for Tgives:

K5522=T

Alternative 2: real heat capacityStarting with:

inhu =2

The above equation is equivalent to

inhvPh = 222

222 vPhh in=

To calculate the enthalpy difference, we can use the real heat capacity

dPT

vTcc

P

P P

idealP

realP

ideal

=

2

2

For the truncated viral equation,


45/83

45

02

2

=

PT

v

Therefore,

idealP

realP cc =

Now, we can calculate the change in enthalpy and equate it to the flow work term.

=

=2

1K500

22

T

T

idealP vPdTc

[ ] ( )=

+==+2

1 K500

2222

263 '110561.510225.19131.1

T

T

PBRTvPdTTTR

Integrate and solve for T2:

K5522=T

(b)

In order to solve the problem, we will need to find the final pressure. To do so, first we need tocalculate the molar volume. Using the information from Part (a) and the truncated virialequation to do this

( )( )

( )[ ]Pa103108.21Pa103

K552Kmol

J314.8

'1 686

=+= PB

P

RTv

=

mol

m0014.0

3

v

This quantity will not change as the tank cools, so now we can calculate the final pressure.

( )( ) 28

3

2

108.21

K293Kmol

J314.8

mol

m0014.0

P

P

=

Solve for 2P :


46/83

46

Pa1066.1 62 =P

The entropy change of the universe can be expressed as follows:

surrsysuniv SSS +=

To solve for the change in entropy of the system start with the following relationship:

dPP

sdT

T

sds

TPsys

+

=

Alternative 1: path through ideal gas state

Using the proper relationships, the above equation can be rewritten as

dPT

vdT

T

cds

P

Psys

+=

We can then use the following solution path:

Plow

T

step1

step3

step 2 ideal gas

500 K T2

s1

3 MPa

s3

s2

P

1.66 MPa

Choosing a value of 1 Pa for Plow, for step 1:

( ) +=

=

Pa1

MPa66.1

'

Pa1

MPa66.1

1 1 dPPBP

R

dPT

v

sP

For step 3,

( ) +=

=

MPa3

Pa1

'

MPa3

Pa1

1 1 dPPBP

RdP

T

vs

P


47/83

47

For step 2:

+==

K293

K552

63

K293

K552

2 10561.510225.19131.1

dTTT

RdTT

cs P

Adding together steps 1, 2 and 3:

=

Kmol

J9.46syss

______________________________________________________________________________

Alternative 2: real heat capacity

Using the proper relationships, the above equation can be rewritten as

dPT

vdT

T

cds

P

realPsys

+=

For the truncated virial equation

+=

'

1B

PR

T

v

P

Now, substitute the proper values into the expression for entropy and integrate:

++

+=

Pa1066.1

Pa103

K293

K552

63

6

^

'1

10561.510225.19131.1

dPBP

RdTTT

Rssys

=

Kmol

J9.46syss

______________________________________________________________________________

In order to calculate the change in entropy of the surroundings, first perform an energy balance.

qu=

Rewrite the above equation as follows

( ) qPvh =


48/83

48

Since the real heat capacity is equal to ideal heat capacity and the molar volume does not change,we obtain the following equation

( ) =f

i

T

T

ifidealP qPPvdTc

[ ] ( )=

=

=

+

K293

K552

663

263 Pa103-Pa1066.1mol

m0014.010561.510225.19131.1

f

i

T

T

qdTTTR

=

mol

J15845q

Therefore,

=

mol

J15845surrq

and

=

Kmol

J08.54surrs

Before combining the two entropies to obtain the entropy change of the universe, find the

number of moles in the tank.

mol7.75

mol

m0014.0

m05.0

3

3

=

=n

Now, calculate the entropy change of the universe.

( )

+

=

Kmol

J9.46

Kmol

J08.54mol7.75univS

=

K

J544univS


49/83

49

5.23

First, focus on the numerator of the second term of the expression given in the problemstatement. We can rewrite the numerator as follows:

idealvT

idealvT

idealvTvT

idealvTvT

rrrrrrrr

rrrr

uuuuuu==

=,,,,,,

For an ideal gas, we know

0,,

==

idealvT

idealvT

rrrr

uu

Therefore,

idealvTvT

idealvTvT

rrrr

rrrr

uuuu=

=,,,,

Substitute this relationship into the expression given in the problem statement:

c

idealvTvT

c

idealvTvT

c

dep

vT

RT

uu

RT

uu

RT

urr

rrrr

rrrr

=

=

=

,,,,,

Now, we need to find an expression for idealvTvT

rrrruu

=

,,. Note that the temperature is constant.

Equation 5.41 reduces to the following at constant temperature:

dvPTPTdu

vT

=

The pressure can be written as

v

zRTP=

and substituted into the expression for the differential internal energy

dvTz

vRTdv

vzRT

vRT

Tz

vRTTdu

vvvT

=

+

=

2

Applying the Principle of Corresponding States


50/83

50

rvrr

r

c

Tdv

T

z

v

T

RT

du

r

r

=

2

If we integrate the above expression, we obtain

==

=

=

=

r

r

vrrr

r

v

v

r

vrr

rv

v c

idealvTvT

c

Tdv

T

z

v

T

RT

uu

RT

du 2,,

Therefore,

=

=

=

=

=

r

r

rrrr

rrrr

rr

v

v

r

vrr

r

c

idealvTvT

c

idealvTvT

c

dep

vTdv

T

z

v

T

RT

uu

RT

uu

RT

u 2,,,,,


51/83

51

5.24

We write enthalpy in terms of the independent variables Tand v:

dvv

hdT

T

hdh

Tv

+

=

using the fundamental property relation:

vdPTdsdh +=

At constant temperature, we get:

dvv

Pv

T

PTdh

TvT

+

=

For the Redlich-Kwong EOS

( )bvvT

a

bv

R

T

P

v ++

=

2/32

1

( ) ( ) ( )22/122/12 bvvT

a

bvvT

a

bv

RT

v

P

T ++

++

=

Therefore,

( ) ( ) ( ) dvbvT

a

bvvT

a

bv

RTv

bv

RT

dhT

++++= 22/12/12 2

3

To find the enthalpy departure function, we can integrate as follows

( ) ( ) ( )==

++

++

==

v

v

v

v

Tdep dv

bvT

a

bvvT

a

bv

RTv

bv

RTdhh

22/12/12 2

3

Since temperature is constant, we obtain

( )bvTa

bvv

bT

abv

RTbhdep

++

++

=

2/12/1ln

2

3

To calculate the entropy departure we need to be careful. From Equation 5.64, we have:

gasideal0,

gasideal,

gasideal0,,

gasideal,, == = PTPTPTPTPTPT ssssss


52/83

52

However, since we have a Pexplicit equation of state, we want to put this equation in terms of v.Lets look at converting each state. The first two states are straight -forward

vTPT ss ,, =

and

gasideal,

gasideal0, == = vTPT ss

For the third state, however, we must realize that the ideal gas volume vat the Tand Pof thesystem is different from the volume of the system, v. In order to see this we can compare theequation of state for an ideal gas at T and P

'v

RTP=

to a real gas at T and P

( )bvvTa

bv

RTP

+

=

The volume calculated by the ideal gas equation, v, is clearly different from the volume, v,calculated by the Redlich-Kwong equation. Hence:

+==gasideal,gasideal,

gasideal,gasideal,gasideal, '' vTvTvTvTPT

sssss

Thus,

( ) ( )

= ==

gasideal,

gasideal

,

gasideal,

gasideal,

gasideal,,

gasideal,, ' vTvTvTvTvTvTPTPT

ssssssss

Using a Maxwell relation:

vT T

P

dv

ds

=

Therefore,

dvT

Pds

vT

=


53/83

53

For the Redlich-Kwong EOS

( )bvvT

a

bv

R

T

P

v ++

=

2/32

1

so

( )( )

=

=

++

=

v

v

vTvT dv

bvvT

a

bv

Rss

2/3

gasideal,, 2

1

For an ideal gas

v

R

T

P

v

=

so

( ) =

=

=

v

v

vTvT dvv

Rss

gasideal,

gasideal,

Finally:

Pv

RTR

v

vR

v

dvRss

v

v

vTvTlnln

'gasideal

,gasideal

,

'

' ===

Integrating and adding together the three terms gives:

( )Pv

RTR

bv

v

bT

a

v

bvRsdep lnln

2ln

2/3

++

=


54/83

54

5.25

Calculate the reduced temperature and pressure:

[ ][ ]

344.0

bar48.220

K3.647

=

=

=

w

P

T

c

c

(Table A.1.2)

[ ][ ]

04.1K647.3

K15.673

36.1bar20.482

bar300

==

==

r

r

T

P

By double interpolation of data from Tables C.3 and C.4

921.2

)0(

, =

c

dep

PT

RTh rr 459.1

)1(

, =

c

dep

PT

RTh rr

From Tables C.5 and C.6:

292.2

)0(

,=

R

sdep

PTrr 405.1

)1(

,=

R

sdep

PTrr

Now we can calculate the departure functions

+

=

)1(

,

)0(

,

c

dep

PT

c

dep

PTc

dep

RT

hw

RT

hRTh rrrr

( ) ( )( )

=+

=

mol

J18421459.1344.0921.23.647

Kmol

J314.8

deph

+

=

)1(

,

)0(

,

R

sw

R

sRs

depPT

depPTdep rrrr

( )( )

=+

=

Kmol

J07.23405.1344.0292.2

Kmol

J314.8

deps


55/83

55

To use the steam tables for calculating the departure functions, we can use the followingrelationships.

idealPTPT

dephhh ,, =

idealPTPT

dep

sss ,, =

From the steam tables

=

kg

kJ0.2151,PTh and

=

Kkg

kJ4728.4,PTs

We need to calculate the ideal enthalpies and entropies using the steam tables reference state.

( ) +=K673.15

K273.16

, C.010 dTchh idealpvapidealPT

We can get

=

mol

kJ1.45vaph from the steam tables and heat capacity data from Table A.2.2.

Using this information, we obtain

++

+

=

K673.15

K273.162

53

,10121.0

1045.147.3Kmol

kJ008314.0

mol

kJ1.45 dT

TTh

idealPT

=molkJ14.59,

idealPTh

Now, calculate the ideal entropy.

( )

+=

1

2K673.15

K273.16

, lnC.010P

PRdT

T

css

idealpvapideal

PT

From the steam tables:

( )

=

Kmol

kJ165.0C01.0

vaps

Substitute values into the entropy expression:


56/83

56

++

+=

000613.0

30ln

10121.01045.1

47.3

Kmol

kJ008314.0165.0

K673.15

K273.163

53

, dTTT

sidealPT

=

Kmol

J107,

idealPT

s

Now, calculate the departure functions:

[ ]( )

=

=

mol

kJ4.20

mol

kJ14.59kg/mol0180148.0

kg

kJ0.2151

deph

[ ]( )

=

=

Kmol

kJ0264.0

Kmol

kJ107.0kg/mol0180148.0

Kkg

kJ4728.4

deps

Table of Results

Generalized

Tables Steam Tables

Percent Difference

(Based on steam tables)

mol

kJ

deph -18.62 -20.4 9.9

Kmol

kJ

deps -0.0231 -0.0264 12.5


57/83

57

5.26

State 1 is at 300 K and 30 bar. State 2 is at 400 K and 50 bar. The reduced temperature andpressures are

[ ]

[ ]982.0

K305.4

K300

616.0

bar74.48

bar30

,1

,1

==

==

r

r

T

P

[ ]

[ ]31.1

K305.4

K400

026.1

bar74.48

bar50

,2

,2

==

==

r

r

T

P

and

099.0=

By double interpolation of data in Tables C.3 and C.4

825.0

)0(

,,1,1 =

c

depPT

RT

hrr

799.0

)1(

,,1,1 =

c

depPT

RT

hrr

711.0

)0(

,,2,2 =

c

dep

PT

RT

hrr

196.0

)1(

,,2,2 =

c

dep

PT

RT

hrr

Therefore,

( ) 904.0799.0099.0825.0,1,1,

=+=

c

dep

PT

RT

hrr

( ) 730.0196.0099.0711.0,2,2,

=+=

c

depPT

RT

hrr

The ideal enthalpy change from 300 K to 400 K can be calculated using ideal cPdata from Table

A.2.1.

RdTTTRhideal

TT39.71710561.510225.19131.1 2

K400

K300

63

21

=+=

The total entropy change is


58/83

58

dep

PTideal

TT

dep

PTrrrr

hhhh,2,221,1,1

,, ++=

( )[ ]CC TTRh 730.039.717904.0 +=

( ) ( )[ ]K4.305730.0K39.717K4.305904.0Kmol

J314.8 +

=h

=

mol

J2.6406h

Using the data in Table C.5 and C.6

( ) 676.0756.0099.0601.0,1,1,

=+=

R

sdep

PTrr

( ) 416.0224.0099.0394.0,2,2 , =+=

R

sdep

PT rr

Substituting heat capacity data into Equation 3.62, we get

+=

bar30

bar50ln

10561.510225.19131.1K400

K300

263

dTT

TTRs ideal

Rsideal 542.1=

Therefore,

( )416.0542.1676.0,2,2,1,1

,, +=++= Rssss depPT

idealdep

PTrrrr

=

Kmol

J98.14s


59/83


60/83


61/83

61

5.28

A reversible process requires the minimum amount of work. Since the process is reversible andadiabatic

0=s


0,2,2,1,1

,, =++= dep

PTidealdep

PTrrrr

ssss

Calculate reduced temperature and pressures using data from Table A.1.1

[ ][ ]

0217.0bar0.46

bar1,1 ==rP

[ ][ ]

217.0bar6.04

bar10,2 ==rP

57.1

K190.6

K003,1 ==rT

From Tables C.5 and C.6:

( ) 0046.00028.0008.000457.0,1,1,

=+=

R

sdep

PTrr

Substituting heat capacity data into Equation 3.62, we get

+=

bar1

bar01ln

10164.210081.9702.12

K003

263dT

T

TTRs

Tideal

Therefore,

+

++=

R

s

dTT

TTRs

dep

PTT

rr ,2,2

2 ,

K003

263

303.210824.810785.28213.1

0046.0

We can solve using a guess-and-check method

K4002=T : 10.2,2 =rT

=

Kmol

J98.4s

K3852=T : 02.2,2 =rT


62/83

62

=

Kmol

J42.1s

K3792=T : 99.1,2 =rT

= KmolJ018.0s

Therefore,

K3792T

An energy balance reveals that

swhhh == 12

We can calculate the enthalpy using departure functions. From Tables C.3 and C.4:

( ) 0966.0011.0008.00965.0,1,1,

=+=

c

dep

PT

RT

hrr

( ) 0613.0015.00089.00614.0,2,2,

=+=

c

dep

PT

RT

hrr

Ideal heat capacity data can be used to determine the ideal change in enthalpy

+=

dTTTRhidealK379

K003

263 10164.210081.9702.1

Therefore,

( )( )

++

=

dTTTh

K379

K003

263 10164.210081.9702.10613.00966.0K6.190Kmol

J314.8

==

mol

J2.3034hws&

and

W1.101mol

J2.3034

s

mol30/1 =

=SW&


63/83

63

5.29

Equation 4.71 states

PT

v

v

=

1

PT

vv

=

This can be substituted into Equation 5.75 to give

( )

PJT

c

Tv 1=


64/83

64

5.30

For an ideal gas

P

R

T

v

P

=

Therefore,

( )0=

=

=PP

JTc

vv

c

vP

RT

This result could also be reasoned from a physical argument.


65/83

65

5.31

The van der Waals equation is given by:

2v

a

bv

RTP

= (1)

The thermal expansion coefficient is given by:

PP v

T

vT

v

v

=

=

11 (2)

Solving Equation 1 for T:

+=

R

bv

v

aPT

2

Differentiating by applying the chain rule,

3

3

32

221

Rv

abavPv

v

a

R

bv

Rv

aP

v

T

P

+=

+=

(3)

Substitution into Equation 2 gives

abavPv

Rv

23

2

+=

Substituting Equation 1 for Pgives b in terms of R, T, v, a, and b:

( )( )23

2

2 bvaRTv

bvRv

=

The isothermal compressibility is given by:

TT v

P

vP

v

v

=

=

11

From the van der Waals equation:

( )( )

( )23

23

32

22

bvv

bvaRTv

v

a

bv

RT

v

P

T

+=+

=

so


66/83


67/83

67

( )( )

( )( )

( )

+

+

=real

ideal

v

v

ideal

P

JT

dvbv

bvaRTv

bva

RTvc

bvaRTv

bvavbRTv

2

23

23

23

2

32

2

2


68/83

68

5.32

We can solve this problem by using the form of the Joule-Thomson coefficient given in Equation5.75. The following approximation can be made

PP T

v

T

v

At 300 C,

( ) ( )C250350

C,1MPa250C,1MPa350

=

vv

T

v

P

C250350

kg

m23268.0

kg

m28247.0

33

=

PT

v

=

=

Kkg

m0005.0

Ckg

m0005.0

33

PT

v

A similar process was followed to find cP.

PP

PT

h

dT

hc

=

At 300 C,

( ) ( )C250350

C,1MPa250C,1MPa350

=

hh

T

h

P

C250350

kg

kJ6.2942

kg

kJ7.3157

=

PT

h

=

=

=

Kkg

kJ15.2

Ckg

kJ15.2

PP

PT

h

dT

hc

Now, JTcan be found.


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69

( )

=

=

Kkg

kJ15.2

kg

m25794.0

Kkg

m0005.0K15.573

33

P

PJT

c

vT

vT

=

kJ

Km0133.0

3

JT


70/83

70

5.33

At the inversion line, the Joule-Thomson coefficient is zero. From Equation 5.75:

0

2

2 =

=real

ideal

P

P P

ideal

P

P

JT

dPT

vTc

vT

vT

This is true when the numerator is zero, i.e.,

0=

v

T

vT

P

For the van der Waals equation, we have

2v

a

bv

RTP

=

Solving for T:

+=

R

bv

v

aPT

2

so

3

3

32221

RvabavPv

va

Rbv

RvaP

vT

P

+=

+=

Substituting for P:

( )( ) 3

23 2

Rvbv

bvaRTv

v

T

P

=

Hence,

( )( )23

23

2

20bvaRTv

bvavbRTvv

T

vT

P +==

Solving for T:

( )3

22

bRv

bvavT

= (1)


71/83

71

Substituting this value of Tback into the van der Waals equation gives

( ) ( )223

322

bv

bva

v

a

bv

bvavP

=

= (2)

We can solve Equations 1 and 2 by picking a value of v and solving for Tand P. For N2, thecritical temperature and pressure are given by Tc= 126.2 [K] and Pc= 33.84 [bar], respectively.Thus, we can find the van der Waals constants aand b:

=

2

32

mol

Jm0.137=

6427

c

c

P

RTa

=

mol

m103.88=

8

35-

c

c

P

RTb

Using these values in Equations (1) and (2), we get the following plot:

Joule-Thomson inversion line

0

200

400

600

800

1000

0 100 200 300 400

T [K]


72/83

72

5.34

We can solve this problem using departure functions, so first find the reduced temperatures andpressures.

[ ]

[ ]99.0

bar0.365

bar50,1 ==rP

[ ]

[ ]2.0

bar0.365

bar10,2 ==rP

967.0K282.4

K15.273,1 ==rT

Since the ethylene is in two-phase equilibrium when it leaves the throttling device, thetemperature is constrained. From the vapor-liquid dome in Figure 5.5:

6.214

76.0

2

,2

=

T

T r

The process is isenthalpic, so the following expression holds

0,2,221,1,1

,, =++=

dep

PTideal

TT

dep

PTrrrr

hhhh

Therefore,

idealTT

dep

PT

dep

PT hhh

rrrr 21,1,1,2,2,,

=

From Table A.2.1:

[ ] +=K6.214

K15.273

263 10392.410394.14424.121

dTTTRhidealTT

From Tables C.3 and C.4 ( )085.0= :

( ) 976.351.3085.0678.3,1,1,

=+=

c

depPT

RT

hrr

Now we can solve for the enthalpy departure at state 2.


73/83

73

+=

K6.214

K15.273

263, 10392.410394.14424.1K4.282

1976.3

,2,2dTTT

RT

h

c

dep

PT rr

01.3

,2,2 ,

=

c

dep

PT

RT

hrr

We can calculate the quality of the water using the following relation

( )c

vapdepPT

c

liqdepPT

c

depPT

RT

h

xRT

h

xRT

hrrrrrr

,,

,,, ,2,2,2,2,2,2

1

+

=

where x represents the quality. From Figures 5.5 and 5.6:

( ) 068.55.5085.06.4,, ,2,2 =+=

c

liqdepPT

RT

hrr

( ) 464.075.00859.04.0

,, ,2,2 =+=

c

vapdep

PT

RT

hrr

Thus,

447.0=x

55.3% of the inlet stream is liquefied.


74/83

74

5.35

Density is calculated from molar volume as follows:

v

MW=

Substitute the above into the expression for 2sound

V :

( )s

s

sound v

P

MW

v

MW

PV

=

=

/1

12

The following can be shown using differentials:

2

1

v

v

v

=

Therefore,

sssound

v

P

MW

vPV

=

=

22


75/83

75

5.36

From Problem 5.35:

sssound

v

P

MW

vPV

=

=

22

The thermodynamic web gives:

PTTvssPvs T

s

s

P

v

s

s

T

T

P

v

T

v

s

s

P

v

P

=

=

=

Pvv

PP

TTvs v

T

T

P

c

c

T

c

s

P

v

s

c

T

v

P

=

=

If we treat air as an ideal gas consisting of diatomic molecules only

5

7=

v

P

c

c

v

R

T

P

v

=

R

P

v

T

P

=

Therefore,

=

v

P

v

P

s 5

7

and

=

=

MW

RT

v

P

MW

vV

sound 5

7

5

72

[ ]m/s343=sound

V

The lightening bolt is 1360 m away.


76/83


77/83

77

So

=

=

3m

kPakg

001002.0009995.

34.25000

v

P

v

P

v

P

Ts

and

[ ]m/s1414

2 =

=

ssound v

PvV


78/83


79/83

79

TTz

Z

S

T

A

Z

=

zzT T

F

Z

A

T

=

( )0zzkT

F

Z

S

zT

=

=

Substituting the expressions for the partial derivatives into the expression for the entropydifferential, we obtain

( )dzzzkdTbT

andS 0

+=

(c)

First, start with an expression for the internal energy differential:

dzZ

UdT

T

UdU

Tz

+

=

From information given in the problem statement:

( )bTanT

U

z

+=

Using the expression for internal energy developed in Part (a) and information from Part (b)

( )( ) ( ) 000 =+=+

=

zzkTzzkTF

z

ST

Z

U

TT

Therefore,

( )[ ] ( )[ ]dTbTandzdTbTandU +=++= 0

(d)We showed in Part (c) that

0=

=

TU

z

UF


80/83

80

Using the expression forTz

S

developed in Part B, we obtain

( )0zzkT

z

STF

T

S =

=

(e)

First, perform an energy balance for the adiabatic process.

WdU =

Substitute expressions for internal energy and work.

( )[ ] ( )dzzzkTFdzdTbTan 0==+

Rearrangement gives

( )( )[ ]bTan

zzkT

dz

dT

+

= 0

The right-hand side of the above equation is always positive, so the temperature increases as therubber is stretched.


81/83

81

5.39

The second law states that for a process to be possible,

0 univs

To see if this condition is satisfied, we must add the entropy change of the system to the entropychange of the surroundings. For this isothermal process, the entropy change can be written

dvT

Pdv

T

PdT

T

cds

vv

v

=

+=

Applying the van der Waals equation:

dvbv

Rds

=

Integrating

==

Kmol

J5.11ln

1

2

v

vRssys

For the entropy change of the surroundings, we use the value of heat given in Example 5.2:

==

mol

J600surrqq

Hence the entropy change of the surroundings is:

=

==

Kmol

J6.1

373

600

surr

surrsurr

T

qs

and

=+=

Kmol

J9.9surrsysuniv sss

Since the entropy change of the universe is positive we say this process is possible and that it isirreversible.

Under these conditions propane exhibits attractive intermolecular forces (dispersion). The closerthey are together, on average, the lower the energy. That we need to put work into this systemsays that the work needed to separate the propane molecules is greater than the work we get outduring the irreversible expansion.


82/83

82

5.40 A schematic of the process is given by:

Gas A in

Pi = 100 barTi = 600 K

Turbine

Pf = 20 bar

Tf = 445 K

ws

Gas A

out

The energy balance for this process is provided below:

Swh=

Because the gas is not ideal under these conditions, we have to create a hypothetical path that

connects the initial and final states through three steps. One hypothetical path is shown below:

P[bar]

T [K]

Pi,Ti

Pf,Tf

step1

step3

step 2

ideal gas

600445

h

100

20

Plow


=

=

0

1

1

P

P T

dPP

hh

If we apply Equation 5.45 we can rewrite the above equation as

=

+

=

0

1

P

P Pi

i

dPvT

vTh

For the given EOS:


83/83

2

iP T

aP

P

R

T

v=

Therefore,

=

+=

++=

=

=

=

=mol

J2467

20

10100

0

10100

155

v

P i

P

P iii

dPbT

aPdPv

T

aPPh

Similarly for step 3

=

+=

=

=mol

J250

2

51020

0

3

fP

P f

dPbT

aPh

For step 2, the pressure is zero, so we can use the ideal heat capacity given in the problemstatement to calculate the enthalpy change.

( )

=+== mol

J627002.030

K445

K600

2 dTTdTch

f

i

T

T

P

Now sum each part to find the total change in enthalpy:

=++= molJ8487321 hhhh

=

mol

J8487sw

In other words, for every mole of gas that flows through the turbine, 8487 joules of work areproduced.

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