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Page 1: Solutions to Chemical and Engineering Thermodynamics, 3euser.engineering.uiowa.edu/~cbe_103/Problems/hwk6.pdf · Solutions to Chemical and Engineering Thermodynamics, 3e 4 4.1 Using

Solutions to Chemical and Engineering Thermodynamics, 3e

4

4.1 Using the Mollier diagram

µ=∂∂

FH IK = FH IK =−( )°

× − ×

= × ° = °−

T

P

T

PH H

∆∆

510 490

1241 10 7 929 10

4 463 10 4 463

7 6

6

C

Pa

C Pa C MPa

. .

. .

c h

κSS S

T

P

T

P=

∂∂

FH IK ≈ FH IK =−( )°

× − ×

= × ° = °−

∆∆

510 490

1069 10 9 515 10

1702 10 1702

7 6

5

C

Pa

C Pa C MPa

. .

. .

c h

∂ ∂∂ ∂

= ∂∂

× ∂∂

= ∂∂

× ∂∂

=∂∂

×∂∂

= = ( )

H S

H S

H T

S T

S P

H P

H T

H P

S P

S T

T H

P H

P S

T S

T

P

S

a fa f

a fa f

a fa f

a fa f

a fa f

a fa f

a fa f

,

,

,

,

,

,

,

,

,

,

,

,.

µκ

0262 unitless

4.2 (a) Start from eqn. 4.4-27

H T P H T P RT Z TdP

dTP dV

VV

V

, ,( ) − ( ) = −( ) + FH IK −LNM

OQP=∞

zIG 1

PRT

V b

a T

V bV b=

−−

( )+ −2 22

; ∂∂

FH IK =−

−+ −

P

T

R

V b

da dt

V bV bV2 22

so

H T P H T P

RT Z TR

V b

da dt

V bV b

RT

V b

a T

V bV bdV

RT Z a Tda

dT

dV

V bV b

V

V

V

V

, ,( ) − ( )

= −( ) +−

−+ −

RSTUVW −

−+

( )+ −

LNM

OQP

= −( ) + +FH IK + −

=∞

=∞

zz

IG

12 2

12

2 2 2 2

2 2

From integral tables we have

dx

a x b x c b a c

a x b b a c

a x b b a c′ + ′ + ′=

′ − ′ ′

′ + ′ − ′ − ′ ′

′ + ′ + ′ − ′ ′z 2 2

2

2

1

4

2 4

2 4ln for 4 02′ ′ − ′ <a c b

In our case ′ =a 1 , ′ =b b2 , c b= − 2 ; so 4 4 1 2 82 2 2 2′ ′ − ′ = ⋅ ⋅ − − ( ) = −a c b b b bc h and

′( ) − ′ ′ = =b a c b b2 24 8 2 2 .

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H T P H T P

RT Za Tda dT

b

V b b

V b b

V b b

V b b

RT Za Tda dT

b

V b

V b

V V

, ,

ln ln

ln

( ) − ( )

= −( ) + − + −+ +

− + −+ +

LNMM

OQPP

= −( ) +− + −

+ +

=∞

IG

12 2

2 2 2 2

2 2 2 2

2 2 2 2

2 2 2 2

12 2

1 2

1 2

a f

a f d id i

or finally

H T P H T P RT ZTda dT a

b

Z B

Z B, , ln( ) − ( ) = −( ) +

− + +

+ −

LNMM

OQPP

IG 12 2

1 2

1 2

a f d id i

(b) This part is similar except that we start from eqn. (4.4-28)

S T P S T P R ZdP

dT

R

Vdv

R ZR

V b

da dT

V bV b

R

VdV

R Z RV b

V

da dT

b

V b

V b

R Z Bda dT

b

Z B

Z B

VV

V

V

V

V

V

V

V

, , ln

ln

ln ln ln

ln ln

( ) − ( ) = + FH IK −LNM

OQP

= +−

−+ −

−LNM

OQP

= + − ++ +

+ −

= −( ) ++ +

+ −

FHG

IKJ

=∞

=∞

=∞ =∞

zz

IG

2 22

2 2

1 2

1 2

2 2

1 2

1 2

d id i

d id i

4.3 Start with eqn. (4.2-21): dU C dT TP

TP dV

V

= + FHG

IKJ −

LNMM

OQPPV

∂∂

. Thus ∂∂

U

TC

V

FHG

IKJ = V ;

∂∂

∂∂

∂∂

U

TC T

P

TP

V

TP V P

FHG

IKJ = + F

HGIKJ −

LNMM

OQPPFHG

IKJV and

∂∂

∂∂

∂∂

∂∂

U

T

U

TT

P

TP

V

TP V V P

FHG

IKJ − F

HGIKJ = F

HGIKJ −

LNMM

OQPPFHG

IKJ .

(a) Ideal gas PV RT=

TP

TP

U

T

U

TV P V

∂∂

∂∂

∂∂

FHG

IKJ − = ⇒ F

HGIKJ = F

HGIKJ0

(b) van der Waals gas

PRT

V b

a

V=

−− 2 ;

∂∂

P

T

R

V bV

FHG

IKJ =

−; ⇒

FHG

IKJ −

LNMM

OQPP =T

P

TP

a

VV

∂∂ 2

Also: dPRdT

V b

RT

V bdV

a

VdV=

−−

−( )+

2 3

2

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4.6 d SC

TdT

V

TdP

P

= − FHGIKJ

P ∂∂

[eqn. (4.2-20)]

For the ideal gas d SC

TdT

R

PdPIG P= −

*

. Thus, at constant temperature

d S SR

P

V

TdP

P

− = − FHGIKJ

LNM

OQP

IGd i ∂∂

and

S T P S T P S T P S T PR

P

V

TdP

PP

P

, , , ,( ) − ( ) − =( ) − =( ) = − FHG

IKJ

RSTUVW=

zIG IG0 00

∂∂

However, S T P S T P, ,=( ) − =( ) =0 0 0IG , since all fluids are ideal at P = 0 . Also

PV Z T P RTr r= ,a f . Thus

∂∂

∂∂

V

T PRZ T P RT

Z T P

TPr r

r r

P

FHG

IKJ = + F

HGIKJ

RSTUVW

1,

,a f a f

and

R

P

V

T

R

PZ T P

RT

P

Z

T

S T P S T P RZ

P

T

P

Z

TdP

RZ T P

P

T

P

Z

TdP

Pr r

P

PT P

T P

r r

r

r

r r Pr

T P

T P

rr r

r r

−FHG

IKJ = − − −

FHG

IKJ

⇒ ( ) − ( ) = − − + FHG

IKJ

LNM

OQP

= −−

+FHG

IKJ

LNMM

OQPP

=

=

zz

∂∂

∂∂

∂∂

∂∂

,

, ,

,

,

,

,

,

a fk p

a f

1

1

1

0

0

IG

4.7 (a) Ideal gas

PV NRT

N

=

=( )

+( ) × × ⋅=−

50 bar 100 m

27315 150 8314 10142 12

3

3K bar m kmol K kmol

c h. .

.

Energy balance, closed nonflow system

∆U Q PdV Q W= − = +z .

However, for ideal gas ∆U = 0 since T is constant (isothermal). Thus

W Q PdVNRTV

dV NRTVV

NRTPP

Q

= − = − = − = − = −

= − × × +( ) × FH IK= × == −

z z ln ln

. . ln

. .

.

2

1

1

2

3

142 1 27315 15050

300

895 9 10 895 9

895 9

kmol 8.314 J mol K

kJ MJ

MJ

Also, by Ideal Gas Law at fixed T and N

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Solutions to Chemical and Engineering Thermodynamics, 3e

PV PV V VP

P1 1 2 2 2 11

2

100 m50

3001667= ⇒ = = × =3 3 m.

(b) Corresponding states

T P ZH H

TS Sr r

C

IGIG

initial state

final state 1.391300

73.76cal

mol K

cal

mol K

− −

+ = =

=

150 27315

304 21391

50

73760679 094 07 0 4

4067 0765 45 2 4

.

..

.. . . .

. . . .

Number of moles of gas = = = =NPV

ZRT

142 1

094151 2

.

.. kmol ( 142 1. =

PV

RT from above)

Final volume = =VZNRT

Pff

=× × × × +( )−0765 151 2 8 314 10 273 15 150

300

2. . . .=1356 m. 3

Energy balance on gas: ∆U Q W= +

Entropy balance on gas processes in gas are reversible: ∆SQ

TS S= + ⇒ =gen gen 0 . Therefore

∆SQ

T= or Q T S= ∆

S S S N S S N S S S S S S

NP

P

N N

Q T S

W U Q N U U Q N H H N P V PV Q

f i f i f f f i i i

f

i

f i f i f f i i

= − = − = − + − − −

= − × − − − ×( )RSTUVW

= − −RSTUVW = −

= = +( ) × × −( ) = −

= − = − − = − − − −

c h c h c h c hm rIG IG IG IG

J K

kmol MJ

2 4 4184 8 314 04 4184

8 368 8314300

502326

27315 150 151 2 2326 14885

. . . ln . .

. . ln .

. . . .

=−

+LNMMN T

H H

TCf f

C

IGd iH Hf

IGiIG−c h

0Since process

is isothermal.

−−

− +OQPP −T

H H

TZ RT Z RT QC

i i

Cf f i i

IGd i

=− − −( )( ) × −

× +( ) × −( )LNM

OQP +

= × − + + = − +=

151 2304 2 45 07 4184 8314

273 15 150 0 765 09414885

151 2 10 48365 615 7 14885 638 2 14885

850 3

3

.. . . . .

. . ..

. . . . . .

.

kmol MJ

J MJ MJ

MJ

(c) Peng-Robinson E.O.S.Using the program PR1 with T = 27315. , P = 1 bar as the reference state, we obtainT = °150 C , P = 50 barZ = 0 9202. ; V = × −06475 10 3. m mol3 ; H = 470248 . J mol ; S = −1757. J mol K .

T = °150 C , P = 300 bar

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Z = 0 7842. ; V = × −0 9197 10 4. m mol3 ; H = −6009. J mol ; S = −4124. J mol .

NV

V= =

×=−

100 m

06475 10154 443

3

3 m mol kmol

..

Q TN S= = +( ) × × − − −( )( ) = −∆ 273 15 150 154 44 4124 1757 1546 9. . . . . MJ

W U Q H PV H PV Q

N H PV H PV Q

f i

f i

= − = −( ) − −( ) −

= −( ) − −( ) −

=

− − × ×

× ⋅ −

+ × × ×

L

N

MMMM

O

Q

PPPP× + ×

=

154 44

6009 300 0 9197 10

10 4702 48

50 0 6475 10 10

10 1546 9 10

885 25

4

5

3 5

3 6.

. .

.

.

.

.

J bar m

MJ

3

{Note that N, Q and W are close to values obtained from corresponding states.}

4.8 ∂∂

∂∂

∂∂

∂∂

∂ ∂∂ ∂

T

P

T S

P S

T S

P T

P T

P S

S T P T

S P T PS

FHG

IKJ =

( )( )

=( )( )

⋅( )

= −( ) ( )( ) ( )

,

,

,

,

,

,

, ,

, ,a f =

−= =

∂ ∂∂ ∂

∂ ∂ αS P

S T

V T

C T

V T

CT

P

Pa fa f

a fP P

and

κκ

∂∂ ∂∂ ∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

S

T

S

T

V P

V V dP

V V dP

V S P S

V T P T

V S

V T

P T

P S

S V

T V

T P

S P

S

T

T

S

C

T

T

C

C

C

= =( ) ( )( ) ( )

=( )( )

⋅( )( )

=( )( )

⋅( )( )

= FHG

IKJ ⋅ FHG

IKJ = ⋅ =

1

1

a fa fa fa f

, ,

, ,

,

,

,

,

,

,

,

,V

P

V

P

4.9 (a)∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

H

V

H T

V T

H T

P T

P T

V T

H

P

P

VT T T

FHG

IKJ =

( )( )

=( )( )

⋅( )( )

= FHG

IKJ

FHG

IKJ

,

,

,

,

,

,

Since ∂∂

P

V T

FHG

IKJ ≠ 0 (except at the critical point)

∂∂

H

V T

FHG

IKJ = 0 if

∂∂

H

P T

FHG

IKJ = 0

(b)∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂ ∂ α∂∂

α

S

V

S P

V P

S P

T P

T P

V P

S

T

T

V

C

T VV

dT

dV

C TV

V V T

C

TV

S

V

P P P

P P P

FHG

IKJ =

( )( )

=( )( )

⋅( )( )

=FHG

IKJ ⋅

FHG

IKJ

= ⋅ ⋅ FHG

IKJ = = ⇒ F

HGIKJ

,

,

,

,

,

,

~P P P1

11

a fa f4.10 (a) We start by using the method of Jacobians to reduce the derivatives

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Solutions to Chemical and Engineering Thermodynamics, 3e

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂ ∂∂ ∂

∂∂

∂ ∂∂ ∂

∂ ∂∂

TV

T HV H

T HT P

T PT V

T VV H

H T

P T

P T V T

H V T V

H

P

P V

H T

H V

d H T

H

T

T

V

T

V

FHG

IKJ =

( )( )

=( )( )

⋅( )( )

⋅( )

= −( )( )

( ) ( )( ) ( )

= −FHGIKJ

= −

,,

,,

,,

,,

,

,

, ,

, ,

a f

a fa f

a fa f

Now from Table 4.1 we have that

∂∂

∂∂

H

PV T

V

TT P

FHG

IKJ = − F

HGIKJ and

∂∂

∂∂

∂∂

HT

C V TVT

PTV P V

FHG

IKJ = + − F

HGIKJ

LNM

OQPFHG

IKJP

alternatively, since H U PV= +

∂∂

∂∂

∂∂

H

T

U

T

PV

TC V

dP

dTV V V V

FHG

IKJ = F

HGIKJ +

( )FHG

IKJ = + FH IKV

Thus

∂∂

∂ ∂ ∂ ∂∂ ∂

∂ ∂ ∂ ∂

∂ ∂T

V

P V V T V T

C V P T

V P V T P T

C V P TH

T P

V

T V

V

FHG

IKJ =

− −+

=− +

+a f a f

a fa f a f

a fV V

Note: I have used ∂∂

∂∂

∂∂

P

V

V

T

P

TT P V

FHG

IKJ

FHG

IKJ = −FHG

IKJ .

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

T

V

T S

V S

T S

V T

V T

V S

S T

V T

T V

S V

S

V

T

S

T

C

P

T

S

T V V

FHG

IKJ =

( )( )

=( )( )

⋅( )( )

= −( )( )

⋅( )( )

=FHG

IKJFHG

IKJ = −

FHG

IKJ

,

,

,

,

,

,

,

,

,

,

V

(b) For the van der Waals fluid

∂∂

P

T

R

V bV

FHG

IKJ =

−,

∂∂

P

V

RT

V b

a

VT

FHG

IKJ = −

−( )+2 3

2

Thus

∂∂

T

V

RTV V b a V RT V b

C V R V bH

FHG

IKJ =

− − −( ) + + −( )

+ −

2 22n sV

after simplification we obtain

∂∂

T

V

a V b RTV b

C V b V R V b VH

FHG

IKJ =

− −( ) −

−( ) + −( )

2 2 2

2 2 3C

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Solutions to Chemical and Engineering Thermodynamics, 3e

and

∂∂

T

V

RT

C V bS

FHG

IKJ = −

−( )V

4.11 There are a number of ways to solve this problem. The method I use is a little unusual, but thesimplest that I know of. At the critical point all three roots of V are equal, and equal to V C .

Mathematically this can be expressed as V V C− =a f3 0 which, on expansion, becomes

V V V V V VC C C3 2 2 33 3 0− + − = (1)

compare this with

PRT

V b

a

V V b b V b

RT

V b

a

V bV b=

−−

+( ) + −( )=

−−

+ −2 22

which multiplying through by the denominators can be written as

V V bRT

Pb

bRT

P

a

PV b

RTb

P

ab

P3 2 2 3

2

32

0+ −FH IK + − − +FH IK + + −FHG

IKJ = (2)

Comparing the coefficients of V in Eqns. (1) and (2) gives TC , PC

V bRT

PVC

CC

2 3: − = − (3)

V bbRT

P

a

PVC

C CC: − − + =3

232 2 (4)

V bRT

Pb

ab

PVC

C CC

0 3 2 3: + − = − (5)

From Eqn. (3)

P b

RT

P V

RTZC

C

C C

CC− = − = −1

33 or

P b

RTZC

CC= −1 3 (6)

For convenience, let y ZC= −1 3 or Zy

C =−1

3. Then

P b

RTyC

C

=

From Eqn. (4)

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−FHG

IKJ − F

HGIKJ + =

⇒ − + + =−( )

3 2 3

3 23 1

9

2

22

22

2

P b

RT

P b

RT

aP

RTZ

y yaP

RT

y

C

C

C

C

C

CC

C

C

a f

a for expanding and rearranging

aP

RTy yC

Ca f c h221

310 4 1= + + (7)

Finally from eqn. (5)

P b

RT

bP

RT

P b

RT

aP

RTZC

C

C

C

C

C

C

CC

FHG

IKJ + F

HGIKJ − F

HGIKJFHG

IKJ = −

3 2

23

a fy y y y y y3 2 2 31

310 4 1

1

271+ − ⋅ + + = − −( )c h

or

64 6 12 1 03 2y y y+ + − = (8)

This equation has the solution y = 0077796074.

⇒ =bRT

PC

C

0077796074. (from Eqn. (6))

aRT

PC

C

= 04572355292

.a f

(from Eqn. (7))

Also Zy

C =−

=1

30307401309. .

Note that we have equated a and b to TC and PC only at the critical point. Therefore these functions

could have other values away from the critical point. However, as we have equated functions of V ,

we have assumed a and b would only be functions of T. Therefore, to be completely general wecould have

aRT

P

T

T

bRT

P

T

T

C

C C

C

C C

=FHG

IKJ

=FHG

IKJ

0457235529

0077796074

2

.

.

a fα

β

with αT

TC

FHG

IKJ → 1 as

T

TC

→ 1 and βT

TC

FHG

IKJ → 1 as

T

TC

→ 1 .

In fact, Peng and Robinson (and others) have set β= 1 at all temperatures and adjusted α a s a

function of temperature to give the correct vapor pressure (see chapter 5).

4.12 (also available as a Mathcad worksheet)

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dN

dtN N N N

dUdt

N H N H QQN

H H

= + = ⇒ = −

= + + = ⇒ = −

& & & &

& & &&

1 2 2 1

1 1 2 21

2 1

0

0

Also, now using the program PR1 with T = 27315. , P = 1 bar reference state we obtain

T = °100 C T = °150 CP = 30 bar P = 20 barZ = 0 9032. Z = 0 9583.

V = × −09340 10 3. m mol3 V = × −01686 10 2. m mol3

H = 3609 72. J mol H = 679606 . J mol

S = −1584. J mol K S = −4 68 . J mol

&

& . . .Q

N= − =679606 3609 72 318634 J mol

4.13 (also available as a Mathcad worksheet)

Since process is adiabatic and reversible ∆S = 0 or S Si f= , i.e.,

S S T310 K, 14 bar 345 bar( ) = =( )?, . Using the program PR1 with the T = 27315. K and P = 1

bar reference state we obtain T = 310 K , P = 14 bar , Z = 0 9733. , V = × −01792 10 2. m mol3 ,

H = 109083. J mol and S = 1575. J mol K .

By trial and error (knowing P and S , guessing T) we obtain T = 34191. K , P = 345 bar ,

Z = 0 9717. , V = × −08007 10 4. m mol3 , H = 188609. J mol , S = 1575. J mol K

⇒ =Tf 34191. K .

System = contents of compressor

M.B.: dN

dtN N N N= = + ⇒ = −0 1 2 2 1& & & &

E.B.: dU

dtN H N H= = + +0 1 1 2 2& & &Q

0

adiabatic

+ −&W Ps

dV

dt

0

volume ofcompressor

constant

& & &W N H N HS = − +1 1 2 2 or &

& . . .W

NH HS = − = − =2 1 188609 109083 7952 6 J mol

4.14 (a) Pa

VV b RT

PV

RT

V

V b

a

RTV+

FHG

IKJ −( ) = ⇒ =

−−2

i) lim limPV

V

PV

RT

V

V b

a

RTV→→∞

→∞=

−−F

HGIKJ =

01

ii) B VPV

RTV

V

V b

a

RTVPV

V= −FH IK =

−− −RST

UVW→→∞

→∞lim lim

01 1

=− −( )

−( )−RST

UVW =−( )

−RSTUVW = −

→∞ →∞lim lim

V VV

V V b

V b

a

RTV

bV

V b

a

RTb

a

RT

Q.

N1

.N2

.M.B.

E.B.

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Solutions to Chemical and Engineering Thermodynamics, 3e

iii) C VPVRT

BV

VbV b V b

V bb V

V bb

PV

V V= − −F

HGIKJ = − −( )

−RST

UVW =−

=→→∞

→∞ →∞lim lim lim

0

22

21

⇒ =C b2

(b) At the Boyle temperature: limP

VPV

RTB

→−FH IK = ⇒ =

01 0 0

0 = −ba

RTB

, Ta

RBb

= but aV RTc c=

9

8, b

V c=3

(Eqns. 4.6-3a)

TV RT

RVT TB

c c

cc c= = =

9 8

3

27

83375.

4.15 (a) From table: TB ~ 320 K , i.e., B T BBa f = ( ) =320 K 0

The inversion temperature is the temperature at which

∂∂

∂∂

∂∂

∂∂

∂∂

T

P CV T

V

T

TP T

RTP

BRP

dBdT

V TVT

RTP

BRTP

TdBdT

B TdBdT

H P

P P

P

FHG

IKJ = = − − F

HGIKJ

LNM

OQP

FHG

IKJ = +L

NMOQP = +

− FHG

IKJ = + − − = −

01

P

Thus, T inv is the temperature at which B TdB

dt− = 0 .

Plot up B vs. T, obtain dB dT either graphically, or numerically from the tabular data. I find

T inv ~ 600 K . Also, dB dT decreases with increasing temperature (i.e.,

dB dT ~ .456 cm mol K3 at 87.5 K and 0.027 cm mol K3 at 650 K. Presumably it is negative

at even higher temperature!)(b) Generally

µ ∂∂

∂∂

= FHG

IKJ = − − F

HGIKJ

RSTUVW = − −RST

UVWTP C

V TVT C

B TdBdTH P

1 1

P P

Using the data in the table it is easy to show that for T T< inv , B TdB

dT− < ⇒ >0 0µ , while for

T T> inv , B TdB

dT− > ⇒ <0 0µ .

(c) Since Fig. 2.4-3 for nitrogen is an H-P plot is easiest to proceed as follows

∂∂ ∂ ∂ ∂ ∂

∂ ∂∂ ∂

T

P H T P H

H P

H TH P T

T

P

FHG

IKJ =

−=

−1

a f a fa fa f

Since ∂ ∂H T CPa f = P is > 0 and less than ∞ [Except at a phase transition—see Chap. 5 and

Problem 5.1—however, µ has no meaning in the two-phase region], if dT dP Ha f is to be zero,

then d H dP Ta f must equal zero. That is, an inversion point occurs when isotherms are parallel

to lines of constant H (vertical line). This occurs at low pressures (ideal gas region) and at high

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Solutions to Chemical and Engineering Thermodynamics, 3e

Pa

VV b RT

V V

T T

T

PRT

V b

a

V

ii

i

i f

f i

f

f

f

f f

+FHGG

IKJJ − =

⇒ = × = ×

− =− ⋅

× × ×= −

⇒ = − = = − °

=−

− = ×× − ×

−×

− −

− −

d id i

a f c h

d i c h

2

5 4

5

2 4 5

6 678 10 1336 10

02283

2 2725 6678 1062 73

29315 62 73 230 42 42 73

8314 230 42

1336 10 4 269 10

02283

1336

. ; . .

.

. ..

. . . .

. .

. .

.

.

m mol m mol

Pa m mol

J mol K m mol K

K C

3 3

6 2

3

10

8286 10 8286 bar

4 2

6

= × =

c hPf . . Pa

(d) Here we will use the program PR1. Using the 273.15 K and 1 bar reference state we find that atthe initial conditions Z = 11005. , Vi = × −05365 104. m mol3 , Hi = −328185. J mol and

Si = −5912. J mol K. Therefore

U H P Vi i i i= − = − − × × × = −−328185 500 05365 10 10 5964354 5. . . J mol

Now since U Uf i= = −596435. J mol and

V Vf i= = × −2 1073 104. m mol3 .

We must, by trial-and-error, find the temperature and pressure of the state having theseproperties. I find the following as the solution Tf = 230 9. K ; P~ 99 bar (for which

V = 01075. m mol3 and U = −5968 4. J mol ). To summarize, we have the following answers

for the different parts of the problem:

Pf T f

Ideal gas 250 bar 293.15 KCorresponding states 101.1 bar 237 Kvan der Waals 82.86 bar 230.42 KPeng-Robinson 99 bar 230.9 K

Once again, the ideal gas solution is seriously in error.

4.20 Mass balance (system = both tanks): N N Ni f f1 1 2= +

energy balance (system = both tanks): N U N U N Ui i f f f f1 1 1 1 2 2= +

entropy balance (system = portion of initial contents of tank 1, also in there finally): S Si f1 1=

Also, P P Pf f f1 2= = ; N

V

Vi

i11

1

= ; NV

Vf

f11

1

= and NV

Vf

f22

2

=

(a) Ideal gas solution: obtain P

T

P

T

P

T

i

i

f

f

f

f1

1

1

1

2

2

= + from mass balance and

P P P P Pi f f f f1 1 2

72 250 bar 25 10= + = ⇒ = = ×. Pa from energy balance

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Solutions to Chemical and Engineering Thermodynamics, 3e

T TP

PTf i

f

i

R Cf

1 11

11

8 314 35 565

20 273 151

2249 3

239

=FHG

IKJ ⇒ = +( )FH

IK =

= − °

P

K

C from entropy balance

. .

.

. .

and

1 2 1355 9 82 7

2 1 12

T T TT

f i ff= − ⇒ = = °. . K C

Also

N

N

P V

RT

RT

P V

f

i

f

f

i

i1

1

1 1

1

1

1 1

7

7

25 10

249 3

293 15

500 100588= ⋅ =

×⋅

×=

.

.

.

..

and

N

N

N

N

f

i

f

i2

1

1

1

1 0412= −FHG

IKJ = .

(b) Corresponding States Solution:

Initial conditions Tr = =29315

190 71538

.

.. ; Pr = ×

×=5 10

4 64 101077

7

6.. ; ⇒ =Z 122. ;

H H

TC

IG

J mol K−

= 180 . ; S SIG J mol K− = 96 . .

Mass balance:

P

Z TP

Z T Z T

i

i if

f f f f1

1 1 1 1 2 2

751 1 50 10

122 293 151398 10= +

RSTUVW

×= ×

.

. .. (1)

Entropy balance:

S S S S S S S Sf i f f i i

1 1 1 1 1 1 1 10− = = − + − − −IG IG, IG, IGd i d i d i

or

S S CT

RPf

f f

1 11

729315 50 109 6− + −

×= −IG

Pd i ln.

ln.

. (2)

Energy balance:

N U N U N Ui i f f f f1 1 1 1 2 2= + but N N N N U U N U Ui f f f f i f f i

1 1 2 1 1 1 2 2 2 0= + ⇒ − + − =d i d i

or

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Solutions to Chemical and Engineering Thermodynamics, 3e

P V

Z RTH P V H P V

P V

Z RTH P V H P V

Z TH H H H H H Z RT Z RT

Z TH H H H H H

f

f ff f f i i i

f

f ff f f i i i

f ff f f i i i f f i i

f ff f f i i i

1 1

1 11 1 1 1 1 1

2 2

2 22 2 2 1 1 1

1 11 1 1 1 1 1 1 1 1 1

2 22 2 2 1 1 1

0

1

1

− − − + − − − =

− + − − − − +

+ − + − − −

d i d in s d i d in s

d i d i d in s

d i d i

, IG , IG , IG , IG

, IG , IG , IG , IGd in s− + =Z RT Z RTf f i i2 2 1 1 0

Substituting in the known values gives

135565 29315 8314 6 406 5

135565 29315 8 314 6 406 5 0

1 11 1 1 1 1

2 22 2 2 2 2

Z TH H T Z T

Z TH H T Z T

f ff f f f f

f ff f f f f

− + − − +

+ − + − − + =

, IG

, IG

d i c hn s

d i c hn s

. . . , .

. . . , . (3)

Eqns. (1-3) now must be solved. One possible procedure isi) Guess Pf

ii) Use Eqn. (2) to find T f1

iii) Use Eqn. (1) to find T f2

iv) Use Eqn. (3), together with T f1 and T f

2 to see if guessed Pf is correct. If not, go back to

step i.After many iterations, I found the following solution Pf = 9787. bar; T f

1 2216 K= . ;

T f2 259 4= . K ; N Nf i

1 1 0 645= . ; N Nf i2 1 0 355= . .

(c) Peng-Robinson equation of stateHere we use the equations

N N Ni f f1 1 2= + (4)

N U N U N Ui i f f f f1 1 1 1 2 2= + with U H PV= − (5)

S Si f1 = (6)

P P Pf f f1 2= =

and N V Vi i1 1 1= ; N V Vf f

1 1 1= ; N V V V Vf f f2 2 2 1 2= = since V V1 2= (value of V1 cancels out

of problem, so any convenient value may be used). Procedure I used to solve problem was as

follows. From PR1 we know V Ni i1 1⇒c h and Si

1 given initial conditions. Then

1. Guess value of T f1 , find P Pf f

1 = that satisfies S Sf i1 1=

2. Use T f1 , Pf and V f

1 to get N f1 ; then N N Nf i f

2 1 1= − so V f2 is known.

3. From Pf and V f2 find (trial-and-error with PR1) T f

2

4. See if eqn. (5) energy balance is satisfied; if not go back to step 1. After a number of

iterations I find Pf =103 6 bar. ; T f1 222 3= . K ; T f

2 2555= . K ; N Nf i1 1 0 619= . ;

N Nf i2 1 0381= . .

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Solutions to Chemical and Engineering Thermodynamics, 3e

Summaryideal gas(part a)

Correspondingstates (part b)

P-R E.O.S.(part c)

Pf 250 bar 97.87 103.6

T f1

249.3 K 221.6 K 222.3 K

T f2

355.9 K 259.4 K 255.5 K

N Nf i1 1

0.588 0.645 0.619

N Nf i2 1

0.412 0.355 0.381

Clearly, the ideal gas assumption is seriously in error!

4.21 System = contents of compressor. This is a steady-state, open constant volume system.

mass balance: dN

dtN N= = +0 1 2& &

energy balance: dU

dtN H N H Q W Ps= = + + + −0 1 1 2 2& & & & dV

dt

0

⇒ = − + +0 1 1 2& & &N H H Q Wsa f

entropy balance: dS

dtN S N S

Q

T= = + + +0 1 1 2 2

& &&

S gen

0

.= − +&

&N S S

Q

T1 1 2a f

Thus,

& &Q TN S S= − −1 1 2a f&

&Q

NQ T S S

12 1= = −a f

and

& &

&W Q

NQ H HS +

= + = −1

2 1W

(a) Corresponding states solution

Q T S S T S S S S S S

T S S S S RP

P

= − = − + − − −

= − − − −RSTUVW

2 1 2 2 2 1 1 1

2 2 1 12

1

a f d i d i d in s

d i d i

IG IG IG IG

IG IG ln

Now Tr = =37315

405 6092

.

.. ; Pr , .

~ . 1 =1

112 80 009 ; Pr , .

. 2 = =50

112 80 443 . Thus

Q RT T S S S SPT

PT

r

r

r

r

= − + − − −RST

UVW= − × × + − −( ) = −

==

==

ln

. . ln . . .

, ,..

..

50

1

8314 37315 50 37315 523 0 14,088 1

2 2 0 4440 92

1 1 0 0090 92

2 1

IG IG

J mol

d i d i

and

W + = − = − + − − −

=

Q H H H H H H H H

T

2 1 2 2 20

1 1 1( ) ( ) ( )IG IG

sinceconstant

IG IG