Chapter 3

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

Chapter 3 Problems Problem 3.1: Express the volume expansivity β and isothermal compressibility κ as functions of density ρ and its partial derivatives. The isothermal compressibility coefficient () of water at 50 oC and 1 bar is

−6

44.18∗10

bar-1. To what pressure must water be compressed at 50 oC to change its density by 1%? Assume that  is independent of P. Given Data:

Volume expansivity=β=

1 ∂V V ∂T

( )

P

 Or β=

1 dV V dT

( ) →(1) P

Isothermal Compressibilty=κ=

−1 ∂ V V ∂P

( )

T

 Or κ=

−1 dV V dP

( ) → ( 2) T

Temperature=T =50 Pressure=P1=1 ¯¿

¿¯ κ=44.18∗10−6 ¿ −1

(a)

 1

We know that

C

Density of water =ρ1=1 P2=?

Solution:

0

kg m3

ρ2=( 1+1 )

kg m3

ρ2=1.01

kg m3

August 20, 2013

ρ=

1 V

PROBLEMS

V=

ZAID YAHYA

11-CH-74

1 ρ

 Put in (1) & (2)

β=ρ

( dTd 1ρ )

β= P

−ρ dρ 2 ρ dT

( )

β=

P

−1 dρ ρ dT

( ) Proved P

 Now, κ=−ρ

(

d 1 dP ρ

)

κ=

T

ρ dρ ρ2 dP

( )

T

κ=

1 dρ ρ dP

( ) Proved T

(b)



As κ=

1 dρ ρ dP

( )

κdP=

T

dρ ρ

 Integrating on both sides P2

ρ2

κ ∫ dP=∫ P1

ρ1

dρ ρ

P2

ρ2

1

1

κ|P|P =|ln ρ|ρ

κ ( P2 −P 1 )=( ln ρ 2−ln ρ1 )

κ ( P2 −P 1 )=ln

ρ2 ρ1

 Putting values ¯¿ ( P2−1 )= ln1.01 1 −6 44.18∗10 ¿

¯¿ −6 44.18∗10 P2−1=0.00995∗¿

P2=( 225.22+1 ) ¯¿

P2=226.22 ¯¿

Answer

Problem 3.2: Generally, volume expansivity β and isothermal compressibility κ depend on T and P. Prove that

( ∂∂ Pβ ) =−( ∂T∂κ ) T

2

P

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

Solution: 

We know that Volume expansivity=β=

 Since

β

1 ∂V V ∂T

( )

P

is very small V=

1 ∂V ∂ β ∂T

( ) → (1)

Isothermal Compressibilty=κ=

P

−1 ∂ V V ∂P

( )

T

 Since κ is very small V=

−1 ∂ V ∂κ ∂ P

1 ∂V ∂β ∂T

∂V ( ) =−1 ∂κ ( ∂ P)

( )

T

P

1 1 ∂ β ∂T

1 ∂V ( ) ∂V =−1 ∂κ (∂P)

T

P

T

( ∂T∂ κ ) =( ∂∂ Pβ )

−

P

T

( ∂∂ Pβ ) =−( ∂T∂κ ) Proved T

P

Problem 3.3: The Tait equation for liquids is written for an isotherm as: AP V =V 0 1− B+ P

(

)

Where V is specific or molar volume, Vo is the hypothetical molar or specific volume at P = 0 and A & B are positive constant. Find an expression for the isothermal compressibility consistent with this equation.

Solution: 

We Know That, Isothermal Compressibilty=κ=

3

−1 ∂ V V ∂P

( ) → (1) T

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 Given that

(

V =V 0 1−

AP B+ P

)

 Where  V0 = Hypothetical molar/specific volume at zero pressure, so it is constant  V = Molar/specific volume  Now, AP V =V o − V B+ P o V −V o =

−AP V B+ P o

V −V o −AP = Vo B+ P

 Differentiate w.r.t Pressure 1 ∂ −∂ AP ( V −V o ) = ∂ P B + P V o ∂P

(

−1 ∂V AB+ AP− AP = Vo ∂ P ( B+ P )2

( )

)

[

A ( B+ P )− AP ( 1 ) 1 ∂V −0 =− V o ∂P ( B+ P )2

(

)

]

−1 ∂V AB = V o ∂ P ( B+ P )2

( )

 Since, Temperature is constant  Therefore, −1 ∂V AB = V o ∂ P T ( B+P )2

( )

 Or, From (1)

κ=

AB Proved ( B+ P )2

Problem 3.4: For liquid water the isothermal compressibility is given by: c κ= V ( P+b ) Where c & b are functions of temperature only if 1 kg of water is compressed isothermally & reversibly from 1 bar to 500 bars at 60 oC, how much work is required? At 60 oC, b=2700 bars and c = 0.125 cm3 g-1 4

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

Given Data:

Isothermal compressibility=κ=

c V ( P+ b )

Pressure=P1=1 ¯¿

Mass of water=m=1 kg

P2=500 bars Temperature=T =60 c=0.125 cm3 / g

b=2700 bars



Work=W =?

0

C

Solution:

We know that W =−∫ PdV →(1)

κ=

c →(2) V ( P+b )

 Also κ=

−1 dV V dP

( ) → ( 3) T

 Comparing (2) & (3) −1 dV c = V dP V ( P+b )

−dV =

c dP P+b

 Put in (1) c dP W =−∫−P P+b P2

P2

W =c ∫ dP−b c ∫ P1

P1

P2

P W =c ∫ dP P+b P

1 dP P+b

1

P2

P+b−b W =c ∫ dP P+ b P

P2

P2

1

1

[

W =c ( P2 −P 1) −bc ln

 Putting values 5

P2

1

1

P+b b W =c ∫ dP−c ∫ dP P+b P+ b P P

1

W =c|P|P −bc|ln (P+b)|P

P2

W =c ( P2 −P 1) −bc [ ln ( P2 +b ) −ln ( P1 +b ) ] P2 +b P1 +b

]

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74 3

3

cm ∗¯¿ g 3 ¯ cm ∗¿ −57.216 ¿ g W =62.375 ¿

3

cm ( ¯ ¯¿ 0.125 cm ∗ln 500+ 2700 W =0.125 ∗ 500−1 )−2700 g g 1+2700

3

cm ∗¯¿ g W =5.16¿

3

cm ∗¯¿ ∗1 m3 g ∗101325 N 1003 cm 3 ∗J 1.01325 ¯¿ m2 Nm W =5.16 ¿

W =0.516

J Answer g

Problem 3.5: Calculate the reversible work done in compressing 1 ft 3 of mercury at a constant temperature of 32F from 1(atm) to 3,000(atm). The isothermal compressibility of mercury at 32F is: κ/(atm)-1 = 3.9 x 10-6 - 0.1 x10-9P(atm) Given Data:

Work done=W =?

Volume=V =1 ft 3

Pressure=P2=3000 atm

Temperature=T =32 F

Pressure=P1=1 atm

κ /atm−1=3.9∗10−6−0.1∗10−9 P (atm)

 Where  Term,

3.9*10-6

has

unit

of

atm-1

&

Solution: 

We know that, work done for a reversible process is

W =−∫ PdV →(1)  Also κ=

6

−1 dV V dP

( )

T

dV =−κVdP

0.1*10-9

has

units

of

atm-2

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 Put in (1), P2

W =−∫ P (−κVdp )

W =V ∫ κ P dP P1

P2

P2

−6 −9 W =V ∫ ( 3.9∗10 −0.1∗10 P ) P dP

3000

W =3.9∗10 V

∫

P dP−0.1∗10 V

P1

∫

P2 dP

1

3000

||

P2 W =3.9∗10 V 2 W=

P1

−9

3000 −9

1

−6

2

W =V ∫ 3.9∗10 P dP−V ∫ 0.1∗10 P dP

P1

−6

P2 −6

1

3000

||

P3 −0 .1∗10 V 3 −9

1

1.95∗10−6∗1 ft 3 ( 3.333∗10−11∗1 ft 3 ( 30002−12 ) atm2− ∗ 30003−13 ) atm3 2 atm atm

W = (17.55−0.8991 ) atm∗ft 3

W =16.65atm∗ft 3 Answer

Problem 3.6: Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which the temperature change from 0 oC to 20oC. Determine ΔVt, W, Q, and ΔUt. The properties for liquid carbon tetrachloride at 1 bar & 0 oC may be assumed independent of temperature: β = 1.2 x 10-3 K-1 Cp = 0.84 kJ kg-1 K-1, ρ = 1590 kg m-3 Given Data:

Mass=m=5 kg

Pressure=P=1 ¯ ¿

T 1 =273.15 K

7

Temperature=T 1=0

0

C

August 20, 2013

PROBLEMS

ZAID YAHYA

Temperature=T 2=20 T 2 =( 20+273.15 ) K

T 2 =293.15 K

−3

β=1.2∗10 K

0

C

C P=0.84

−1

11-CH-74

kJ kg∗K

ρ=1590

kg m3

t

Q=?

∆ U =?

Solution: 

As V=

1 ρ

V 1=

1 ρ1

1 m3 V 1= 1590 kg

 Also,  we know that Volume expansivity=β=

1 dV V dT

( )

βdT =

P

1 dV V

 Integrating on both sides, T2

V2

β ∫ dT =∫ T1

V1

dV V

T2

V2

1

1

β|T |T =|lnV |V

β ( T 2−T 1 )=( lnV 2−ln V 1 )

β ( T 2−T 1 )=ln

V2 V1

 Putting values V ∗1590 kg 1.2∗10 ∗( 293.15−273.15 ) K =ln 2 K m3 −3

V 2=0.000644

e

0.024

=

V 2∗1590 kg m3

m3 kg

 Now, ∆ V =V 2−V 1

(

∆ V = 0.000644−

 Now, for total volume,

8

1 m3 1590 kg

)

∆ V =15.28∗10−6

3

m kg

t

∆ V =?

1.024 ∗m3 1590 V 2= kg

W =?

August 20, 2013

PROBLEMS

∆ V t =∆ V ∗m

ZAID YAHYA

∆ V t =( 15.28∗10−6∗5 )

m3 ∗kg kg

11-CH-74

∆ V t =7.638∗10−5 m3 Answer

 Now,  We know that for a reversible process,

Work done=W =−P ∆ V t W =−1 ¯¿ 7.638∗10−5

m3∗101325 N ∗J 1.01325 ¯¿ m2 ∗1 kJ Nm 1000 J

W =−7.638∗10−3 kJ Answer

 Now,  For a reversible process at constant pressure,we have Q=∆ H

Q=m C P ∆ T

Q=5 kg∗0.84

kJ ∗( 293.15−273.15 ) K kg∗K

Q=84 kJ Answer

 Now,  According to first law of thermodynamics, ∆ U t =Q+W

∆ U t =( 84−7.368∗10−3 ) kJ

∆ U t =83.99 kJ Answer

Problem 3.7: A substance for which k is a constant undergoes an isothermal, mechanically reversible process from initial state (P1, V1) to (P2, V2), where V is a molar volume. a) Starting with the definition of k, show that the path of the process is described by V = A ( T ) exp (−κP) b) Determine an exact expression which gives the isothermal work done on 1 mol of this constant-k substance.

Solution: (a) 

We know that

Isothermal compressibilty =κ=  Integrating on both sides,

9

−1 dV V dP

( )

T

dV =−κdP V

August 20, 2013

∫

PROBLEMS

ZAID YAHYA

dV =−κ ∫ dP V

lnV =−κP+lnA ( T )  Where ln A (T) is constant of integration & A depends on T only

lnV −lnAT =−κP

ln

V =−κP A (T )

 Taking anti log on both sides,

V =e−κP AT

V = A ( T ) e−κP

 Or

V = A ( T ) exp (−κP ) Proved

(b)

Work done=W=? 

For a mechanically reversible process, we have,

dW =−PdV → ( 1 )  Using,

d ( PV ) =PdV +VdP

−PdV =VdP −d ( PV )

 Put in (1)

dW =VdP−d ( PV ) → ( 2 ) 

We know that

Isothermal compressibilty =κ=  Put in (2)

dW =

−dV −d ( PV ) κ

 Integrating on both sides,

10

−1 dV V dP

( )

T

−dV =VdP κ

11-CH-74

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

−1

∫ dW = κ ∫ dV −∫ d ( PV ) W=

−1 Δ V −Δ ( PV ) κ

 Since volume changes from V1 to V2 & pressure changes from P1 to P2 ,  Therefore,

W=

−1 ( V 2−V 1 ) −( P2 V 2−P1 V 1 ) κ

W=

( V 1−V 2 ) κ

+ P1 V 1−P2 V 2 Proved

Problem 3.8: One mole of an ideal gas with C V = 5/2 R, CP = 7/2 R expands from P 1 = 8 bars & T1= 600 K to P2 = 1 bar by each of the following path: a) Constant volume b) Constant temperature c) Adiabatically Assuming mechanical reversibility, calculate W, Q, ∆ U, and ∆ H for each of the three processes. Sketch each path in a single PV diagram. Given Data:

5 CV = R 2

7 C P= R 2

P1=8 ¯¿

T 1 =600 K

Solution: (a) 

According to first law of thermodynamics,

∆ U =Q+W →(1)  For a constant volume process,

W =0

∆ U =CV ∆T

 Put in (1)

Q=∆U =C V ∆ T

11

Q=∆U =C V ( T 2 −T 1 ) →(2)

P2=1 ¯¿

W =?

Q=?

∆ U =?

∆ H=?

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 For T2 , We know that for an ideal gas

T1 T 2 = P1 P2

T2=

T1 ∗P P1 2

8 ¯¿∗1 ¯¿ 600 K T2= ¿

T 2 =75 K

 Put in (2),

5 Q=∆U = R ( 75−600 ) K 2

Q=∆U =

−5 J ∗8.314 ∗525 K 2 mol∗K

Q=∆U =−10.912



Q=∆U =−10912

kJ Answer mol

 Also For a mechanically reversible process we have,

∆ H=C P ∆ T

7 ∆ H= R ( T 2−T 1 ) 2

7 J ∆ H= ∗8.314 ∗( 75−600 ) K 2 mol∗K ∆ H=−15.277

∆ H=−15277

kJ Answer mol

(b) 

For a constant temperature process,

∆ U =0

∆ H=0

 We know that at constant temperature, work done is

W =R T 1 ln W =8.314 

J 1 ∗600 K∗ln mol∗K 8

0=Q+ W

Q=−W

 Or

Q=10.373

kJ Answer mol

(c)

12

We know that for an adiabatic process

P2 P1

W =−10373

Now, according to first law of thermodynamics,

∆ U =Q+W



J mol

J mol

W =−10.373

kJ Answer mol

J mol

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

Q=0  Now, according to first law of thermodynamics,

∆ U =Q+W

∆ U =W →(1)

∆ U =CV ∆T  Put in (1)

W =∆ U=C V ∆ T

W =∆ U=C V ( T 2−T 1 ) →(2)

 For T2 , We know that for an adiabatic process

T1 P

( 1−γ ) γ 1

=T 2 P

( 1−γ ) γ 2

T 2 =T 1

P1 P2

( )

( 1−γ ) γ

T 2 =600 K

8 1

()

( 1−1.4 ) 1.4

T 2 =331.23 K

 Put in (2)

5 W =∆ U= R∗( 331.23−600 ) K 2

W =∆ U=

−5 J ∗8.314 ∗268.77 K 2 mol∗K

W =∆ U=5.5864 

W =∆ U=−5586.4

J mol

J Answer mol

For a mechanically reversible adiabatic process we have

7 ∆ H= R ( T 2−T 1 ) 2

∆ H=C P ∆ T ∆ H=−7.821

7 J ∆ H= ∗8.314 ∗( 331.23−600 ) K 2 mol∗K

J Answer mol

Problem 3.9: An ideal gas initially at 600k and 10 bar undergoes a four-step mechanically reversible cycle in a closed system. In step 12, pressure decreases isothermally to 3 bars; in step 23, pressure decreases at constant volume to 2 bars; in step 34, volume decreases at constant pressure; and in step 41, the gas returns adiabatically to its initial state. Take CP = (7/2) R and CV = (5/2) R. a) Sketch the cycle on a PV diagram. b) Determine (where unknown) both T and P for states 1, 2, 3, and 4. c) Calculate Q, W, ∆ U, and ∆ H for each step of the cycle. Given Data: 13

August 20, 2013

PROBLEMS

ZAID YAHYA

Initial Temperature=T 1=600 K

Initial Pressure=P1=10 ¯¿

11-CH-74

7 C P= R 2

5 CV = R 2

Solution: (b)

Step 12, an Isothermal process,  Since  For an isothermal process, temperature is constant  Therefore,

T 2 =T 1=600 K

P2=3 ¯¿

 We know that, for an ideal gas

P2 V 2 =R T 2

V 2=

RT 2 P2

2 ¯ ¿∗1.01325 ¯ ¿m ∗N∗m 101325 N mol∗K∗3 J 8.314∗J∗600 K V 2= ¿

3

V 2=0.0166

m mol

Step 23, an Isochoric process,  Since  For an isochoric process, Volume is constant  Therefore,

m3 V 3=V 2=0.0166 mol

P3=2 ¯¿

 We know that, for an ideal gas

P3 V 3 =RT 3

T3=

P3 V 3 R

2 ¯¿ 0.0166 m 3∗mol∗K ∗J mol∗8.314 J ∗101325 N N∗m T3= 1.01325 ¯¿ m2

Step 34, an Isobaric process,  Since  For an isobaric process, pressure is constant 14

T 3 =400 K

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 Therefore,

P4 =P 3=2 ¯¿  For T4 , we will use an adiabatic relation of temperature and pressure  As

T 4 P4 = T1 P1

( )

R CP

T 4=T 1

P4 P1

( )

R CP

2 10

( )

T 4=600 K∗

2∗R 7R

T 4=378.83 K

 We know that, for an ideal gas

P4 V 4 =RT 4

V 4=

2 ¯¿∗1.01325 ¯ ¿m ∗Nm 101325 N mol∗K∗2 J 8.314 J∗378.83 K V 4= ¿

RT4 P4

3

V 4 =0.0157

m mol

Step 41, an adiabatic process,  Since  Gas returns to its initial state adiabatically  Therefore,

T 1 =600 K

P1=10 ¯¿

 We know that, for an ideal gas

P1 V 1=R T 1

V 1=

RT 1 P1

¯¿∗1.01325 ¯¿ m2 ∗Nm 101325 N mol∗K∗10 J 8.314 J∗600 K V 1= ¿

(c)

Step 12, an Isothermal process,  Since  For an isothermal process, temperature is constant  Therefore

15

V 1=4.988∗10−3

3

m mol

August 20, 2013

∆ U 12 =0 

PROBLEMS

11-CH-74

∆ H 12=0

For an isothermal process, we have

Q=−R T 1 ln



ZAID YAHYA

P2 P1

Q=−8.314

J 3 ∗600 K∗ln mol∗K 10

Q=6006

J mol

Q=6.006∗103

J Answer mol

According to first law of thermodynamics

∆ U 12=Q12 +W 12

0=Q12+W 12

W 12=−Q12

W 12=−6.006∗103

J Answer mol

Step 23, an Isochoric process,  Since  For an isochoric process, Volume is constant  Therefore,

W 23=0  At constant volume we have

Q23=∆ U 23=CV ∆ T

Q23=∆ U 23=CV ( T 3−T 2 )

5 ∗8.314 J 2 Q23=∆ U 23= ∗(−200 ) K mol∗K

5 Q23=∆ U 23= R ( 400−600 ) K 2

Q23=∆ U 23=−4157

J mol

Q23=∆ U 23=−4.157∗10 3

J Answer mol

 We know that

∆ H 23=C P ∆ T

∆ H 23=C P ( T 3−T 2 )

∆ H 23=−5820

Step 34, an Isobaric process,  Since  For an isobaric process, pressure is constant  Therefore, at constant pressure we have, 16

7 ∆ H 23= R ( 400−600 ) K 2 J mol

∆ H 23=−5.82∗103

7 J ∆ H 23= ∗8.314 ∗(−200 ) K 2 mol∗K J Answer mol

August 20, 2013

PROBLEMS

Q34=∆ H 34 =C P ∆ T

ZAID YAHYA

7 Q34=∆ H 34 = R ( T 4−T 3 ) 2

11-CH-74

7 J Q34=∆ H 34 = ∗8.314 ∗( 378.83−400 ) K 2 mol∗K

Q34=∆ H 34 =−616

J Answer mol

 For an Isobaric process we have

W 34=−R ∆ T

W 34=−R ( T 4−T 3 )

W 34=−8.314

J ∗( 378.83−400 ) K mol∗K

W 34=176

J Answer mol

 We know that,

∆ U 34 =CV ∆ T

5 ∆ U 34 = R ( T 4 −T 3 ) 2

5 J ∆ U 34 = ∗8.314 ∗( 378.83−400 ) K 2 mol∗K

∆ U 34 =−440

J Answer mol

Step 41, an adiabatic process,  Since  For an adiabatic process there is no exchange of heat  Therefore,

Q41=0  We know that,

∆ U 41=C V ∆ T

5 ∆ U 41= R ( T 1−T 4 ) 2

5 J ∆ U 41= ∗8.314 ∗( 600−378.83 ) K 2 mol∗K

∆ U 41=4.597∗103

∆ U 41=4597

J mol

J Answer mol

 We know that

∆ H 41=C P ∆T

7 ∆ H 41= R ( T 1−T 4 ) 2

7 J ∆ H 41= ∗8.314 ∗( 600−378.83 ) K 2 mol∗K

∆ H 41=6.4358∗103 

17

According to first law of thermodynamics

J Answer mol

∆ H 41=6435.8

J mol

August 20, 2013

PROBLEMS

ZAID YAHYA

∆ U 41=Q 41+W 41

∆ U 41=W 41

11-CH-74 3

W 41=4.597∗10

J Answer mol

Problem 3.10: t An ideal gas, CP= (5/2) R and CV= (3/2) R is changed from P1 = 1bar and V 1 = 12m3 to P2 = 12 bar and t

V 2 = 1 m3 by the following mechanically reversible processes: a) Isothermal compression b) Adiabatic compression followed by cooling at constant pressure. c) Adiabatic compression followed by cooling at constant volume. d) Heating at constant volume followed by cooling at constant pressure. e) Cooling at constant pressure followed by heating at constant volume. Calculate Q, W, change in U, and change in H for each of these processes, and sketch the paths of all processes on a single PV diagram. Given Data:

5 C P= R 2

3 CV = R 2

V t2=1 m3

Q=?

Initial pressure=P1=1 ¯¿ W =?

∆ H=?

Solution:  Since

Temperature=constant  Therefore, for all parts of the problem,

∆ H=0

∆ U =0

(a)

Isothermal compression, 

For an isothermal process, we have

Q=−R T 1 ln

P2 P1

 Since  For an ideal gas, we have 18

∆ U =?

t

3

V 1=12 m

Final pressure=P 2=12 ¯¿

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

P1 V 1=R T 1  Therefore,

Q=−P1 V 1 ln



P2 P1

12 ∗101325 N 1 ∗J 2 1.01325 ¯¿ m ∗1 kJ Nm 3 Q=−1 ¯¿ 12 m ∗ln 1000 J

Q=−2981.88 kJ Answer

According to first law of thermodynamics

∆ U =Q+W

0=Q+ W

W =−Q

W 12=2981.88 kJ Answer

(b)

Adiabatic compression followed by cooling at constant pressure  Since  For an adiabatic process, there is no exchange of heat  Therefore,

Q=0 Answer  The process completes in two steps  First step, an adiabatic compression to final pressure P 2 , intermediate volume can be given as ' γ

P2 ( V ) =P 1 V 1

V ' =V 1

P1 P2

( )

1 γ

 For mono atomic gas, we have

γ =1.67

1 12

( )

V ' =12 m3∗

1 1.67

V ' =2.71 m3

 We know that, '

P V −P 1 V 1 W 1= 2 γ−1

19

August 20, 2013

PROBLEMS

ZAID YAHYA

( 12∗2.71−1∗12 ) ¯¿ m3 ∗101325 N 1.67−1 ∗J 1.01325 ¯¿ m2 ∗1 kJ Nm W 1= 1000 J

11-CH-74

W 1=3063 kJ →(1)

 Second step, cooling at constant pressure P2  We know that, for a mechanically reversible process

m3∗101325 N ∗J 1.01325 ¯¿ m2 ∗1 kJ Nm ¯ W 2=−12 ( 1−2.71 ) 1000 J

W 2=−P2 ( V 2−V ' )

W 2=2052 kJ →( 2)

 Now

W =W 1+ W 2

W = ( 3063+ 2052 ) kJ

W =5115kJ Answer

(c)

Adiabatic compression followed by cooling at constant volume  Since  For an adiabatic process, there is no exchange of heat  Therefore,

Q=0 Answer  First step, an adiabatic compression to volume V2 , intermediate pressure can be given as '

γ

P V 2 =P1 V 1

V P =P1 1 V2 '

( )

 For mono atomic gas, we have

γ =1.67

P' =1

1.67 12¯ 1

( )

 We know that,

20

γ

P' =63.42 ¯¿

August 20, 2013

PROBLEMS

'

W 1=

ZAID YAHYA

P V 2−P 1 V 1 γ −1

( 63.42∗1−1∗12 ) ¯¿ m3 ∗101325 N 1.67−1 ∗J 1.01325 ¯¿ m 2 ∗1 kJ Nm W 1= 1000 J

11-CH-74

W 1=7674.76 kJ

 Second step, cooling at constant Volume,  Therefore, No work will be done

W 2=0  Now

W =W 1+ W 2

W = (7674.76+ 0 ) kJ

W =7674.76 kJ Answer

(d)

Heating at constant volume followed by cooling at constant pressure  The process completes in two steps  Step 1, Heating at constant volume to P2  Therefore no work will be done

W 1=0  Step 2, Cooling at constant pressure P2 To V2  We know that, for a mechanically reversible process

W 2=−P2 ∆ V

W 2=−P2 ( V 2−V 1 )

m3∗101325 N ∗J 1.01325 ¯¿ m2 ∗1 kJ Nm ¯ W 2=−12 ( 1−12 ) 1000 J

 Now

W =W 1+ W 2

W = ( 0+13200 ) kJ

W =13200 kJ Answer

 According to first law of thermodynamics

∆ U =Q+W

0=Q+ W

Q=−W

Q=−13200 kJ Answer

(e)

Cooling at constant pressure followed by heating at constant volume 21

W 2=13200 kJ

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 The process completes in two steps  Step 1, Cooling at constant Pressure P1 to V2  Therefore, for a mechanically reversible process 3

W 1=−P1 ∆ V

W 1=−P1 ( V 2−V 1 )

m ∗101325 N ∗J 2 1.01325 ¯ ¿m ∗1 kJ Nm ¯ W 1=−1 ( 1−12 ) 1000 J

W 1=1100 kJ

 Step 1, Heating at constant Volume V2 to pressure P2  Therefore no work will be done

W 2=0  Now

W =W 1+ W 2

W = (1100 +0 ) kJ

W =1100 kJ Answer

 According to first law of thermodynamics

∆ U =Q+W

0=Q+ W

Q=−W

Q=−1100 kJ Answer

Problem 3.11: The environmental lapse rate

dT dz characterizes the local variation of temperature with elevation in the

earth's atmosphere. Atmospheric pressure varies with elevation according to the hydrostatic formula, dP =−M ρg dz Where M is a molar mass, ρ is molar density and g is the local acceleration of gravity. Assume that the atmosphere is an ideal gas, with T related to P by the polytropic formula equation (3.35 c). Develop an expression for the environmental lapse rate in relation to M, g, R, and δ.

Solution:  Given that

dP =−M ρg → ( 1 ) dz 

22

The polytropic relation is

August 20, 2013

TP

1−δ δ

PROBLEMS

ZAID YAHYA

=Constant

 Or

TP

1−δ δ

1−δ

=T o Po δ

 Where  To =Temperature at sea level, so it is constant  Po = Pressure at sea level, so it is constant

T P

δ−1 δ

T To

( )

=

To P

δ δ−1

δ−1 δ o

P = Po

T P = T o Po

( )

δ−1 δ

T P=Po To

( )

δ δ −1

→(a)

P=

Po To

δ δ−1

δ

∗T δ −1

 Differentiate w.r.t to Temperature on both sides

Po δ δ −1

Po

∗δ

dP T o = ∗T dT δ−1

1 δ−1

δ δ−1

∗δ

1 To δ−1 dP= ∗T dT →(2) δ−1

 We know that, for an ideal gas

ρ=

P RT

 Where 

R=Specific gas constant=R ' /M

 Put (a) in above equation

ρ=

 Put in (1) 23

1 T ∗Po RT To

( )

δ δ −1

11-CH-74

August 20, 2013

PROBLEMS

dP g∗1 T =−M Po dz RT To

( )

δ δ −1

ZAID YAHYA

dP=−M

g∗1 T Po RT To

( )

δ δ−1

11-CH-74

∗dz

 Put (2) in above

Po δ δ−1

−δ −1 ∗M g δ ∗Po δ R ∗T o δ−1 Po

∗δ

To δ−1

∗T

1 δ −1

dT =−M

g∗1 T P RT o T o

( )

δ δ−1

∗dz

−δ−1 ∗M g δ ∗Po δ R ∗T o δ−1 Po dT = dz

To

δ δ−1

T

To

dT = dz

δ δ−1

T∗T

δ

∗T δ −1

1 δ−1

δ

∗T δ −1

δ δ−1

−δ ∗M g dT δ−1 = Proved dz R

Problem 3.12: An evacuated tank is filled with gas from a constant pressure line. Develop an expression relating the temperature of the gas in the tank to temperature T’ of the gas in line. Assume that gas is ideal with constant heat capacities, and ignore heat transfer between the gas and the tank. Mass and energy balances for this problem are treated in Ex. 2.13.

Solution:  

24

Choose the tank as the control volume. There is no work, no heat transfer & kinetic & potential energy changes are assumed negligible. Therefore, applying energy balance

August 20, 2013

PROBLEMS

d ( mU )tank +∆ ( Hm )=0 dt

ZAID YAHYA

d ( mU )tank + H ' ' m' ' −H ' m' =0 dt

 Since  Tank is filled with gas from an entrance line, but no gas is being escaped out,  Therefore,

d ( mU )tank +0−H ' m' =0 dt

d ( mU )tank −H ' m' =0→(1) dt

 Where prime (‘) denotes the entrance stream  Applying mass balance

m' =

d mtank → ( 2) dt

 Combining equation (1) & (2)

d ( mU )tank ' d m tank −H =0 dt dt

1 {d ( mU )tank −H ' d mtank }=0 dt

'

d ( mU )tank =H d mtank

 Integrating on both sides m2

m2

∫ d ( mU )tank=H ' ∫ d mtank m1

∆ ( mU )tank =H ' ( m 2−m 1 )

m1

m 2 U 2 −m1 U 1=H ' ( m 2−m 1 )  Because mass in the tank initially is zero, therefore

m1=0 m2 U 2 =H ' m2 '

U 2=H →(3)  We know that 25

11-CH-74

August 20, 2013

PROBLEMS

U=C V T

ZAID YAHYA

11-CH-74

U 2=C V T 2 → ( a )

 Also

H ' =C P T ' → ( b )  Put (a) & (b) in (3)

CV T =C P T '

T=

CP ' T CV

 Since heat capacities are constant, therefore

γ=

CP CV

T =γ T ' Proved

Problem 3.14: A tank of 0.1-m3 volume contains air at 25 oC and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at the constant conditions of 45oC and 1,500 kPa. A valve in the line is cracked so that air flows slowly into the tank until the pressure equals the line pressure. If the process occurs slowly enough that the temperature in the tank remains at 25 oC, how much heat is lost from the tank? Assume air to be an ideal gas for which CP = (7/2) R and CV = (5/2) R Given Data:

Volume=V =0.1 m3 Heat lost =Q=?

T 1 =25 o. C =298 K 7 C P= R 2

P1=101.33 kPa

T 2 =45 o.C =318 K

5 CV = R 2

Solution: 

According to first law of thermodynamics

∆ U =Q+W →(1)  Since

∆ H=∆ U +∆ ( PV ) 26

∆ U =∆ H−∆ ( PV )

∆ U =∆ H−P ∆ V −V ∆ P→ ( a )

P2=1500 kPa

August 20, 2013

PROBLEMS

ZAID YAHYA

 Also, we know that

W =−P ∆ V → ( b )  Put (a) & (b) in (1)

∆ H−P ∆V −V ∆ P=Q−P ∆ V  Also, we have

∆ H=nC P ∆T

∆ H=nC P ( T 2−T 1 )

 Put in (2)

n C P ( T 2−T 1 )−V ∆ P=Q →(3)

 For “n” 

We know that for an ideal gas,

PV =nRT  Initial number of moles of gas can be obtained as,

P1 V =n1 R T 1

n1=

P1 V RT1

 The final number of moles of gas at temperature T1 are

P2 V =n2 R T 1

n2 =

P2 V RT 1

 Now, Applying molar balance

n=n1−n 2  Put in (3)

27

n=

P1 V P2 V − R T1 R T1

n=

( P1−P 2) V RT1

∆ H−V ∆ P=Q→ ( 2 )

11-CH-74

August 20, 2013

PROBLEMS

ZAID YAHYA

( P 1−P2 ) V R T1

( P1−P2 ) V C ( T 2−T 1 )−V ∆ P=Q P

( P1−P2 ) V T1 2

RT1 2

11-CH-74

∗7 R∗( T 2−T 1 )−V ∆ P=Q

∗7 ∗( T 2 −T 1 )−V ( P2 −P 1 )=Q

( 101.33−1500 ) kPa∗0.1 m3 ∗7 298 K ∗( 318−298 ) K −0.1 m3 ( 1500−101.33 ) kPa=Q 2 kPa∗1 kN ∗1 kJ 2 3 1 kPa∗m Q=−172.717 m 1 kNm

Q=−172.717 kJ Answer

Problem 3.17: A rigid, no conducting tank with a volume of 4 m 3 is divided into two unequal parts by a thin membrane. One side of the membrane, representing 1/3 of the tank, contains nitrogen gas at 6 bars and 100 oC, and the other side, representing 2/3 of the tank, is evacuated. The membrane ruptures and the gas fills the tank. a) What is the final, temperature of the gas? How much work is done? Is the process reversible? b) Describe a reversible process by which the gas can be returned to its initial state, How much work is done Assume nitrogen is an ideal gas for which CP = (7/2) R & CV = (5/2) R Given Data:

3

Volume of thetank=V 1 =4 m V 3=

V 1∗2 8 3 = m 3 3

Solution:

(a)

Finaltemperature=T 2=? 28

V 2=

V 1∗1 4 3 = m 3 3

Pressure=P2=6 ¯¿

o

Temperature=T 1=100 . C

August 20, 2013 

PROBLEMS

ZAID YAHYA

11-CH-74

According to first law of thermodynamics

∆ U =Q+W  Since  No work is done & no heat is transferred  Therefore

Q=W =0

mC V ∆ T =0

∆ U =0 

∆ T =0

T 2 −T 1=0

T 2 =T 1

T 2 =100℃ Answer

No, process is not reversible (b)

 Since  Therefore, the process is isothermal  For an isothermal process we have

W =−R T 2 ln

V2 V1

 As, for an ideal gas

P2 V 2 =R T 2

W =−P2 V 2 ln

V2 V1

4 4 W =−6 ¯¿ m3 ln 3 3∗4

W =8.788 ¯¿

m 3∗101325 N kJ ∗1 2 ¯ 1000 Nm 1.01325 ¿ m

W =878.8 kJ Answer

Problem 3.18: An ideal gas initially at 30 0C and 100 kPa undergoes the following cyclic processes in a closed system: a In mechanically reversible processes, it is first compressed adiabatically to 500 kPa then cooled at a constant pressure of 500 kPa to 30 0C and finally expanded isothermally to its original state b The cycle traverses exactly the same changes of state but each step is irreversible with an efficiency of 80% compared with the corresponding mechanically reversible process NOTE: the initial step can no longer be adiabatic Find Q W ∆ U and ∆ H for each step of the process and for the cycle Take C = (7/2) R and C = p

(5/2) R

29

V

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

Given Data:

T 1 =30 0.C

T 1 =303.15 K

P1=100 kPa

Q=?

W =?

∆ U =?

∆ H=?

7 C P= R 2

5 CV = R 2

lution: (a)

P2=500 kPa 1) Adiabatic Compression from point 1 to point 2

Q12=0

 Now, from first law of thermodynamics,

∆ U 12=Q12 +W 12

∆ U 12 =W 12

W 12=∆U 12=CV ∆ T 12

5 W 12=∆U 12= R ( T 2−T 1 ) → ( 1 ) 2

 For ‘T2’  We know that

T 2 P2 = T 1 P1

( )

γ −1 γ

T 2 =T 1

P2 P1

( )

γ −1 γ

T 2 =303.15 K

500 100

( )

1.4−1 1.4

T 2 =480.13 K

 Put in (1)

5 J kJ kJ W 12=∆U 12= ∗8.314 ( 480.13−303.15 ) K∗1 W 12=∆ U 12=3.679 2 mol∗K 1000 J mol  Also, we have

∆ H 12=C P ( T 2−T 1 )

7 J kJ ∆ H 12= ∗8.314 ( 480.13−303.15 ) K∗1 2 mol∗K 1000 J

2) Cooling at constant pressure from point 2 to point 3  Therefore at constant pressure we have,

Q23=∆ H 23=C P ∆T 23

30

7 Q23=∆ H 23= R ( T 3−T 2 ) 2

∆ H 12=5.15

kJ mol

So

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 Here

T 3 =303.15 K

7 J kJ kJ Q23=∆ H 23= ∗8.314 ( 303.15−480.13 ) K∗1 Q =∆ H 23=−5.15 2 mol∗K 1000 J 23 mol  Also, we have

∆ U 23 =C V ( T 3−T 2) 5 J kJ ∆ U 23 = ∗8.314 ( 303.15−480.13 ) K∗1 2 mol∗K 1000 J ∆ U 23=−3.679

kJ mol

 Now, from first law of thermodynamics,

∆ U 23=Q23 +W 23

W 23=∆ U 23−Q23

W 23=−3.679+5.15

W 23=1.471

3) Isothermal expansion from point 3 to point 1  Since for an isothermal process temperature remains constant  Therefore,

∆ U 31=∆ H 31=0  Here

P3=P2=500 kPa 

For an Isothermal process we have

500 ∗1 kJ P3 J 100 W 31=−R T 3 ln W 31=−8.314 ∗303.15 K∗ln P1 mol∗K 1000 J W 31=−4.056 

31

According to first law of thermodynamics

kJ mol

kJ mol

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

∆ U 31 =Q31 +W 31 0=Q31+W 31

Q31=−W 31

Q31=4.056

kJ mol

 For the complete cycle,

Q=Q12 +Q23+ Q31

Q=0−5.15+ 4.056

W =W 12+W 23+W 31

∆ U =∆ U 12 +∆ U 23 +∆ U 31

kJ Answer mol

W =3.679+ 1.471−4.056

W =1.094

∆ H=∆ H 12+ ∆ H 23+ ∆ H 31

Q=−1.094

kJ Answer mol

∆ H=5.15−5.15+0

∆ H=0

kJ Answer mol

∆ U =3.679−3.679+0

∆ U =0

kJ Answer mol

(b)

 If each step that is 80% accomplishes the same change of state then values of

∆U

in part (a) but values of Q & W will change. 1. Adiabatic Compression from point 1 to point 2

W 12=

W 12 0.8

W 12=

3.679 0.8

W 12=4.598

kJ mol

 According to first law of thermodynamics

∆ U 12 =Q12 +W 12 3.679

kJ kJ =Q12+ 4.598 mol mol

Q12=3.679

kJ kJ −4.598 mol mol

2. Cooling at constant pressure from point 2 to point 3 32

Q12=−0.92

kJ mol

&

∆H

will remain same as

August 20, 2013

PROBLEMS

ZAID YAHYA

W 23=

W 23 0.8

W 23=

1.471 0.8

11-CH-74

W 23=1.839

kJ mol

 According to first law of thermodynamics

∆ U 23=Q23 +W 23

−3.679

kJ kJ =Q23 +1.839 mol mol

Q23=−3.679

kJ kJ −1.839 mol mol

kJ ∗0.8 mol

W 31=3.245

Q23=−5.518

kJ mol

3. Isothermal expansion from point 3 to point 1  Since initial step can no longer be adiabatic , therefore

W 31=W 31∗0.8

W 31=−4.056

kJ mol

 According to first law of thermodynamics

∆ U 31 =Q31 +W 31

Q31=−W 31+ 0

Q31=3.245

kJ mol

 For the complete cycle,

Q=Q12 +Q 23+Q31

W =W 12+W 23+W 31

Q=−0.92−5.518+ 3.245

W =4.598+1.839−3.245

W =3.192

Q=−3.193

kJ Answer mol

kJ Answer mol

Problem 3.19: One cubic meter of an ideal gas at 600 K and 1,000 kPa expands to five times its initial volume as follows: a) By a mechanically reversible, isothermal process b) By a mechanically reversible adiabatic process c) By adiabatic irreversible process in which expansion is against a restraining pressure of 100 kPa For each case calculate the final temperature, pressure and the work done by the gas, Cp=21 J mol-1K-1. Given Data:

33

August 20, 2013

V 1=1 m3

PROBLEMS

T 1 =600 K

ZAID YAHYA

P1=1000 kPa

V 2=5 V 1

V 2=5 m3

C P=21

11-CH-74

J mol K

CV =?

T 2 =?

W =?

Solution: 

We know that,

C P−C V =R

CV =C P −R

CV =( 21−8.314 )

J mol∗K

CV =12.686

J mol∗K

 As

γ=

CP CV

γ =1.6554

(a)  Since, for an isothermal process  Temperature remains constant, therefore

T 2 =T 1=600 K Answer 

For an ideal gas we have

P1 V 1 P 2 V 2 = T1 T2



P1 V 1 ∗T 2 T1 P 2= V2

3

1000 kPa∗1 m ∗600 K 600 K P 2= 3 5m

We know that, for an isothermal process

W =−R T 1 ln  Since

P1 V 1=R T 1  Therefore,

34

V2 V1

P2=200 kPa Answer

P2=?

August 20, 2013

PROBLEMS

ZAID YAHYA

5 ∗N 1 ∗J Pa∗m2 3 W =−1000 kPa∗1 m ln Nm

V W =−P1 V 1 ln 2 V1

11-CH-74

W =−1609.43 kJ Answer

(b) 

We know that, for an adiabatic process γ

P1 V 1 =P2 V 2



V P2=P1 1 V2

1 5

1.6554

()

P2=1000 kPa∗

P2=69.65 kPa Answer

For an ideal gas we have

P1 V 1 P2 V 2 = T1 T2 

γ

( )

γ

P V T 2 = 2 2 ∗T 1 P1 V 1

69.65 kPa∗5 m3 T2= ∗600 K 1000 kPa∗1 m3

T 2 =208.95 K Answer

For an adiabatic process work done is

P V −P1 V 1 W= 2 2 γ −1

N ∗J ( 69.65∗5−1000∗1 ) kPa∗m Pa∗m 2 W= 1.6554−1 Nm 3

W =−994.43 kJ Answer

(c)

Pr=100 kPa 

Since, for an adiabatic process

Q=0 

According to first law of thermodynamics

∆ U =Q+W

∆ U =W

∆ U =W =−Pr dV

∆ U =W =−Pr ( V 2−V 1)

3

kPa∗m ∗N ∗J 2 Pa∗m ∆ U =W =−100 ( 5−1 ) Nm

∆ U =−400 kJ

T2=

35

n CV ∆T =−400 kJ

−400 kJ + T 1 →( 1) n CV

n CV ( T 2−T 1 )=−400 kJ

August 20, 2013 

PROBLEMS

ZAID YAHYA

11-CH-74

For an ideal gas we have, 3

P1 V 1=nR T 1



n=

P1 V 1 RT1

1000 kPa∗1 m ∗mol∗K ∗kN 8.314 J∗600 K ∗kJ kPa∗m2 n= kNm

Put in (1)

−400 kJ∗mol∗K ∗1000 J 0.2005 mol∗12.686 J T2= +600 K 1 kJ 

n=0.2005 mol

T 2 =−157.26 K +600 K

T 2 =442.74 K Answer

For an ideal gas we have

P1 V 1 P2 V 2 = T1 T2

P1 V 1 ∗T 2 T1 P2= V2

3

1000 kPa∗1 m ∗442.74 K 600 K P2= 3 5m

P2=147.58 kPa Answe r

Problem 3.20: One mole of air, initially at 150 0C and 8 bars undergoes the following mechanically reversible changes. It expands isothermally to a pressure such that when it is cooled at constant volume to 50 0C its final pressure is 3 bars. Assuming air is an ideal gas for which C P = (7/2) R and CV = (5/2) R, calculate W, Q, ∆ U , and ∆ H Given Data:

Mole of air=n=1mol

0

Initial Temperature=T 1=150 .C =423.15 K

Finaltemperature=T 3=50 0.C =323.15 K

Final pressure=P 3=3 ¯¿

Initial pressure=P1=8 ¯¿ 7 C P= R 2

5 CV = R 2

Solution:  Since process is reversible  Two different steps are used in this case to reach final state of the air.

Step 12:  36

For step 12 temperatures is constant,

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

T 1 =T 2 

Therefore

∆ U 12=∆ H 12=0  For an isothermal process we have

W 12=R T 1 ln

V1 V2

 As

V 2=V 3

W 12=R T 1 ln

V1 →(1) V3

 We know that

P1 V 1 P 3 V 3 = T1 T3

V 1 P3 ¿T 1 = V 3 T 3∗P1

8.314 J∗423.15 K ∗1 kJ mol∗K 3∗423.15 W 12= ∗ln 1000 J 8∗323.15

P ¿T W 12=R T 1 ln 1 3 T 1∗P3 W 12=−2.502



kJ mol

According to first law of thermodynamics

∆ U 12=Q12 +W 12

0=Q12+W 12

Q12=−W 12

Q12=2.502

kJ mol

Step 23:  For step 23 volume is constant,  Therefore,

W 23=0 

According to first law of thermodynamics

∆ U 23=Q23 +W 23

∆ U 23=Q23 +0

Q23=∆ U 23

Q23=∆ U 23=CV ∆ T

5 Q23=∆ U 23= R ( 323.15−423.15 ) K 2

37

Q23=∆ U 23=CV ( T 3−T 2 )

August 20, 2013

PROBLEMS

ZAID YAHYA

❑❑ ¿❑ ❑

11-CH-74

¿❑ ❑

¿❑❑ K ❑❑

()

¿ 8.314 ❑ ❑

¿❑❑

¿−2.0785 ❑ ❑ 

W e know that

∆ H 23=C P ∆ T

∆ H 23=C P ( T 3−T 2 )

J ∗1 kJ 7 mol∗K ∆ H 23= ∗8.314 ( 423.15−323.15 ) K 2 1000 J

∆ H 23=2.91

kJ mol

 For the complete cycle,

Work=W =W 12 +W 23

Q=Q 12 +Q23

W = (−2.502+0 )

Q=( 2.502−2.0785 )

∆ U =∆ U 12 +∆ U 23

∆ H=∆ H 12+ ∆ H 23

kJ mol

kJ mol

W =−2.502

Q=0.424

kJ Answe r mol

kJ Answer mol

∆ U =( 0−2.0785 )

kJ mol

∆ U =−2.0785

∆ H= ( 0−2.91 )

kJ mol

∆ H=−2.91

kJ Answe r mol

kJ Answe r mol

Problem 3.21: An ideal gas flows through a horizontal tube at steady state. No heat is added and no shaft work is done. The cross-sectional area of the tube changes with length, and this causes the velocity to change. Derive an equation relating the temperature to the velocity of the gas. If nitrogen at 150 0C flows past one section of the tube with a velocity of 2.5 m/s, what is the temperature at another section where its velocity is 50 m/s? Let CP = (7/2) R Given Data:

38

August 20, 2013

PROBLEMS

ZAID YAHYA 0

Velocity=u 1=2.5

Temperature=T 1=150 .C =423.15 K

m sec

Molecualr weight of Nitrogen=28

11-CH-74

T 2 =?

u2=50

m sec

7 C P= R 2

g mol

Solution: 

Applying energy balance for steady state flow process

∆ H+

∆ u2 + g ∆ z=Q+W S 2

 Since

∆ z=W S=Q=0  Therefore, 2

∆u ∆ H+ =0 2

−∆u CP ∆ T = 2

2

−u22−u12 T2= + T1 2C P

−u22−u12 C P ( T 2 −T 1 )= 2

−( 50 2−2.52 )∗2∗m2∗mol∗K ∗28 g Nitrogen 2∗7∗8.314 J∗sec 2 ∗J 1 mol Nitrogen ∗N∗sec 2 N∗m ∗1 kg kg∗m T2= + 423.15 K 1000 g T 2 =421.95 K

0

T 2 =( 421.95−273.15 ) .C

0

T 2 =148.8 .C

T 2 =−1.199 K + 423.15 K

Answe r

Problem 3.22: One mole of an ideal gas, initially at 30 0C and 1 bar, is changed to 130 0C and 10 bars by three different mechanically reversible processes: a) The gas is first heated at constant volume until its temperature is 130 0C; then it is compressed isothermally until its pressure is 10 bar b) The gas is first heated at constant pressure until its temperature is 130 0C; then it is compressed isothermally to 10 bar c) The gas is first compressed isothermally to 10 bar; then it is heated at constant pressure to 130 0C 39

August 20, 2013

PROBLEMS

Calculate Q, W,

ZAID YAHYA

∆ U ∧∆ H

11-CH-74

in each case. Take CP = (7/2) R and CV = (5/2) R. alternatively, take CP =

(5/2) R and CV = (3/2) R Given Data:

T 1 =30 0.C

T 1 =( 30+273.15 ) K T 3 =403.15 K

P1=1 ¯¿

T 1 =303.15 K P3=10 ¯¿

Q=?

W =?

T 2 =130 0.C ∆ U =?

T 3 =( 130+273.15 ) K ∆ H=?

Solution: 7 C P= R 2

5 CV = R 2

 Each part consist of two steps, 12 & 23  For the overall processes

∆ U =∆ U 12=∆U 23=CV ∆T

5 ∆ U =∆ U 12=∆U 23= R ( T 3−T 1 ) 2

5 J K∗1 kJ ∆ U =∆ U 12=∆U 23= ∗8.314 ( 403.15−303.15 ) 2 mol∗K 1000 J ∆ U =∆ U 12=∆U 23=2.079

kJ →(a) Answe r mol

 Now

∆ H=∆ H 12=∆ H 23=C P ∆ T 7 ∆ H=∆ H 12=∆ H 23= R ( T 2−T 1) 2 7 J kJ ∆ H=∆ H 12=∆ H 23= ∗8.314 ( 403.15−303.15 ) K∗1 2 mol∗K 1000 J ∆ H=∆ H 12=∆ H 23=2.91

40

kJ → ( b ) Answe r mol

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

(a)

Step 12:  For step “12” volume is constant  Therefore

W 12=0  Here

T 2 =T 3 

According to first law of thermodynamics

∆ U 12=Q12 +W 12

∆ U 12=Q12

Q12=∆ U 12=C V ∆ T

Q12=∆ U 12=2.079

kJ [¿ ( a ) ] mol

 Also we have 

∆ H 12=2.91

kJ [ ¿( b)] mol

Step 23:  Since for step “23” process is isothermal  Therefore

∆ U 23 =∆ H 23=0  Here

T 2 =T 3  Now, intermediate pressure can be calculated as

P1 P2 = T1 T 2



P P2= 1∗T 2 T1

¯¿ ∗403.15 K 303.15 K P2=¿

P2=1.329 b ar

For an isothermal process we have

W 23=R T 2 ln 41

1

P3 P2

W 23=8.314

J K∗1 kJ 10 ∗403.15 ∗ln mol∗K 1000 J 1.329

August 20, 2013

PROBLEMS

ZAID YAHYA

W 23=6.764



11-CH-74

kJ mol

According to first law of thermodynamics

∆ U 23=Q23 +W 23

0=Q23+W 23

Q23=−W 23

Q23=−6.764

kJ mol

 For the complete cycle,

Work=W =W 12 +W 23

Q=Q12 +Q23

W = ( 0+6.764 )

Q=( 2.079−6.764 )

∆ U =∆ U 12 +∆ U 23

kJ mol

kJ mol

W =6.764

Q=−4.685

kJ Answe r mol

∆ U =( 2.079+0 )

kJ mol

∆ U =2.079

∆ H= ( 2.91+ 0 )

kJ mol

∆ H=2.91

∆ H=∆ H 12+ ∆ H 23

kJ Answe r mol

kJ Answe r mol

kJ Answe r mol

(b)

Step 12:  For step “12” volume is constant  Therefore, at constant pressure we have

Q12 ¿ ∆ H 12=2.91

kJ mol

[ ¿(b)]

 Also,

∆ U 12 =2.079



[ ¿(a)]

According to first law of thermodynamics

∆ U 12=Q12 +W 12

42

kJ mol

W 12=∆U 12−Q12

W 12=( 2.079−2.91 )

kJ mol

W 12=−0.831

kJ mol

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

Step 23:  Since for step “23” process is isothermal ( T = Constant)  Therefore

∆ U 23 =∆ H 23=0  Here

T 2 =T 3∧P 1=P2 

For an isothermal process we have

W 23=R T 2 ln



P3 P2

W 23=8.314

J K∗1 kJ 10 ∗403.15 ∗ln mol∗K 1000 J 1

W 23=7.718

kJ mol

According to first law of thermodynamics

∆ U 23 =Q23 +W 23

0=Q23+W 23

Q23=−W 23

Q23=−7.718

kJ mol

 For the complete cycle,

Work=W =W 12 +W 23

Q=Q12 +Q23

W = (−0.831+7.718 )

Q=( 2.91−7.718 )

∆ U =∆ U 12 +∆ U 23

∆ H=∆ H 12+ ∆ H 23

kJ mol

kJ mol

Q=−4.808

kJ mol

∆ U =2.079

∆ H= ( 2.91+ 0 )

kJ mol

∆ H=2.91

 Since for step “12” process is isothermal ( T = Constant)  Therefore

∆ U 12=∆ H 12=0

43

kJ Answer mol

kJ Answe r mol

∆ U =( 2.079+0 )

(c)

 Here

W =6.887

kJ Answe r mol

kJ Answe r mol

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

P2=P3 

For an isothermal process we have

W 12=R T 1 ln

P2 P1

W 12=8.314

W 12=5.8034



J K∗1 kJ 10 ∗303.15 ∗ln mol∗K 1000 J 1 kJ mol

According to first law of thermodynamics

∆ U 12=Q12 +W 12

0=Q12+W 12

Q12=−W 12

Q12=−5.8034

kJ mol

Step 23:  For step “23” volume is constant  Therefore, at constant pressure we have

Q23=∆ H 23=2.91

kJ [¿ (b)] mol

 Here

T 2 =T 3  Now

∆ U 23=2.079



kJ [ ¿(a)] mol

According to first law of thermodynamics

∆ U 23=Q23 +W 23

W 23=∆ U 23−Q23

W 23=( 2.079−2.91 )

kJ mol

W 23=−0.831

 For the complete cycle,

Work=W =W 12 +W 23

44

W = (5.8034−0.831 )

kJ mol

W =4.972

kJ Answer mol

kJ mol

August 20, 2013

PROBLEMS

ZAID YAHYA

Q=Q 12 +Q23

Q=(−5.8034+2.91 )

∆ U =∆ U 12 +∆ U 23

∆ H=∆ H 12+ ∆ H 23

kJ mol

Q=−2.894

11-CH-74

kJ Answe r mol

∆ U =( 0+2.079 )

kJ mol

∆ U =2.079

∆ H= ( 0+2.91 )

kJ mol

∆ H=2.91

kJ Answe r mol

kJ Answe r mol

Solution: 5 C P= R 2

3 CV = R 2

 Each part consist of two steps, 12 & 23  For the overall processes

∆ U =∆ U 12=∆U 23=CV ∆T

3 ∆ U =∆ U 12=∆U 23= R ( T 3 −T 1 ) 2

3 J K∗1 kJ ∆ U =∆ U 12=∆U 23= ∗8.314 ( 403.15−303.15 ) 2 mol∗K 1000 J ∆ U =∆ U 12=∆U 23=1.247

kJ →(a) Answe r mol

 Now

∆ H=∆ H 12=∆ H 23=C P ∆ T 5 ∆ H=∆ H 12=∆ H 23= R ( T 2−T 1 ) 2 5 J kJ ∆ H=∆ H 12=∆ H 23= ∗8.314 ( 403.15−303.15 ) K∗1 2 mol∗K 1000 J ∆ H=∆ H 12=∆ H 23=2.079

45

kJ → ( b ) Answer mol

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

(a)

Step 12:  For step “12” volume is constant  Therefore

W 12=0  Here

T 2 =T 3 

According to first law of thermodynamics

∆ U 12 =Q12 +W 12

∆ U 12 =Q12

Q12=∆ U 12=C V ∆ T

Q12=∆ U 12=1.247

 Also we have

∆ H 12=2.079

kJ [ ¿( b)] mol

Step 23:  Since for step “23” process is isothermal  Therefore

∆ U 23 =∆ H 23=0  Here

T 2 =T 3  Now, intermediate pressure can be calculated as

P1 P2 = T1 T 2



46

P P2= 1∗T 2 T1

For an isothermal process we have

1

¯¿ ∗403.15 K 303.15 K P2=¿

P2=1.329 b ar

kJ [ ¿( a) ] mol

August 20, 2013

PROBLEMS

ZAID YAHYA

W 23=R T 2 ln

P3 P2

W 23=8.314

J K∗1 kJ 10 ∗403.15 ∗ln mol∗K 1000 J 1.329

W 23=6.764



11-CH-74

kJ mol

According to first law of thermodynamics

∆ U 23=Q23 +W 23

0=Q23+W 23

Q23=−W 23

Q23=−6.764

kJ mol

 For the complete cycle,

Work=W =W 12 +W 23

Q=Q 12 +Q23

W = ( 0+6.764 )

Q=( 1.247−6.764 )

kJ mol

kJ mol

W =6.764

Q=−5.516

kJ Answe r mol

kJ Answe r mol

∆ U =∆ U 12 +∆ U 23

∆ U =( 1.247+0 )

kJ mol

∆ U =1.247

kJ Answer mol

∆ H=∆ H 12+ ∆ H 23

∆ H= ( 2.079+ 0 )

kJ mol

∆ H=2.079

kJ Answe r mol

(b)

Step 12:  For step “12” volume is constant  Therefore, at constant pressure we have

Q12 ¿ ∆ H 12=2.079

kJ mol

[ ¿(b)]

 Also,

∆ U 12 =1.247

 47

kJ mol

According to first law of thermodynamics

[ ¿(a)]

August 20, 2013

PROBLEMS

ZAID YAHYA

∆ U 12 =Q12 +W 12

W 12=∆U 12−Q12

11-CH-74

W 12=( 1.247−2.079 )

kJ mol

W 12=−0.832

Step 23:  Since for step “23” process is isothermal ( T = Constant)  Therefore

∆ U 23 =∆ H 23=0  Here

T 2 =T 3∧P 1=P2 

For an isothermal process we have

W 23=R T 2 ln



P3 P2

W 23=8.314

J K∗1 kJ 10 ∗403.15 ∗ln mol∗K 1000 J 1

W 23=7.718

kJ mol

According to first law of thermodynamics

∆ U 23 =Q23 +W 23

0=Q23+W 23

Q23=−W 23

Q23=−7.718

kJ mol

 For the complete cycle,

Work=W =W 12 +W 23

Q=Q12 +Q23

(c)

Step 12: 48

W = (−0.832+7.718 )

Q=( 2.079−7.718 )

kJ mol

kJ mol

W =6.886

Q=−5.639

kJ Answer mol

kJ Answer mol

∆ U =∆ U 12 +∆ U 23

∆ U =( 1.247+0 )

kJ mol

∆ U =1.247

kJ Answer mol

∆ H=∆ H 12+ ∆ H 23

∆ H= ( 2.079+ 0 )

kJ mol

∆ H=2.079

kJ Answe r mol

kJ mol

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 Since for step “12” process is isothermal ( T = Constant)  Therefore

∆ U 12=∆ H 12=0  Here

P2=P3 

For an isothermal process we have

W 12=R T 1 ln

P2 P1

W 12=8.314

W 12=5.8034



J K∗1 kJ 10 ∗303.15 ∗ln mol∗K 1000 J 1 kJ mol

According to first law of thermodynamics

∆ U 12=Q12 +W 12

0=Q12+W 12

Q12=−W 12

Q12=−5.8034

kJ mol

Step 23:  For step “23” volume is constant  Therefore, at constant pressure we have

Q23=∆ H 23=2.079

kJ [ ¿(b)] mol

 Here

T 2 =T 3  Now

∆ U 23=1.247



According to first law of thermodynamics

∆ U 23=Q23 +W 23  For the complete cycle, 49

kJ [ ¿(a)] mol

W 23=∆ U 23−Q23

W 23=( 1.247−2.079 )

kJ mol

W 23=−0.832

kJ mol

August 20, 2013

PROBLEMS

ZAID YAHYA

Work=W =W 12 +W 23

Q=Q 12 +Q23

W = (5.8034−0.832 )

Q=(−5.8034+2.079 )

kJ mol

11-CH-74

W =4.9714

kJ mol

Q=−3.724

kJ Answe r mol

kJ Answe r mol

∆ U =∆ U 12 +∆ U 23

∆ U =( 0+1.247 )

kJ mol

∆ U =1.247

kJ Answer mol

∆ H=∆ H 12+ ∆ H 23

∆ H= ( 0+2.079 )

kJ mol

∆ H=2.079

kJ Answe r mol

Problem 3.23: One mole of an ideal gas, initially at 30

℃

and 1 bars, undergoes the following mechanically

reversible changes. It is compressed isothermally to point such that when it is heated at constant volume to 120 ℃ its final pressure is 12 bars. Calculate Q, W, ∆ U ∧∆ H for the process. Take C (7/2) R and P=

CV = (5/2) R. Given Data:

T 1 =30℃ T 3 =393.15 K

T 1 =( 30+273.15 ) K P3=12 ¯¿

Q=?

T 1 =303.15 K W =?

P1=1 ¯¿

∆ U =?

∆ H=?

Solution: 

The process consist of two steps, 12 & 23

Step 12:  Since for step “12” process is isothermal ( T = Constant)  Therefore

∆ U 12=∆ H 12=0  Now, intermediate pressure can be calculated as 50

T 3 =120℃ 7 C P= R 2

T 3 =( 120+273.15 ) K 5 CV = R 2

August 20, 2013

PROBLEMS

P2 T2



ZAID YAHYA

P ¿ 3 T3

P P2= 3 ∗T 2 T3

12

¯¿ ∗303.15 K 393.15 K P2=¿

P2=9.25 b ar

For an isothermal process we have

W 12=R T 1 ln

P2 P1

W 12=8.314

J K∗1 kJ 9.25 ∗303.15 .15 ∗ln mol∗K 1000 J 1

W 12=5.607



11-CH-74

kJ mol

According to first law of thermodynamics

∆ U 12=Q12 +W 12

0=Q12+W 12

Q12=−W 12

Q12=−5.607

kJ mol

Step 23:  Since for step “23” volume is constant  Therefore

W 23=0 

According to first law of thermodynamics

∆ U 23 =Q23 +W 23

∆ U 23=Q23

Q23=∆ U 23=CV ∆ T

5 Q23=∆ U 23= R ( T 3−T 1 ) 2

5 J K∗1 kJ kJ Q23=∆ U 23= ∗8.314 ( 393.15−303.15 ) Q23=∆ U 23=1.871 2 mol∗K 1000 J mol  Now

∆ H 23=C P ∆ T 7 ∆ H 23= R ( T 2−T 1 ) 2

51

7 J kJ ∆ H 23= ∗8.314 (393.15−303.15 ) K∗1 2 mol∗K 1000 J

∆ H 23=2.619

kJ mol

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 For the complete cycle,

Work=W =W 12 +W 23

Q=Q 12 +Q23

W = (5.607 +0 )

Q=(−5.607+1.871 )

kJ mol

kJ mol

W =5.607

Q=−3.736

kJ Answe r mol

kJ Answe r mol

∆ U =∆ U 12 +∆ U 23

∆ U =( 0+1.871 )

kJ mol

∆ U =1.871

kJ Answe r mol

∆ H=∆ H 12+ ∆ H 23

∆ H= ( 0+2.691 )

kJ mol

∆ H=2.691

kJ Answe r mol

Problem 3.24: A process consists of two steps: (1) One mole of air at T = 800 K and P = 4 bars are cooled at constant volume to T = 350 K. (2) The air is then heated air constant pressure until its temperature reaches 800 K. If this two step process is replaced by a single isothermal expansion of the air from 800 K and 4 bar to some final pressure P, what is the value of P that makes the work of two step processes the same? Assume mechanical reversibility and treat air as an ideal gas with CP = (7/2) R and CV = (5/2) T. Given Data:

T 1 =800 K

P1=4 ¯¿

T 2 =350 K

P=?

Solution:  For the first step volume is constant  Therefore,

W 12=0  For the work done is

W =W 23=−P 2 ∆V →(1) 

For one mole of an ideal gas we have,

P ∆ V =R ∆ T 52

P2 ∆ V =R ∆ T

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 Put in (1)

W =−R ( T 3−T 2 )

W =−R ∆ T  Since

T 3 =T 1  Therefore

W =−R ( T 1−T 2) →(2) 

For an isothermal process we have

W =R T 1 ln

P → (3) P1

 Compare (2) and (3)

P −R ( T 1−T 2 )=R T 1 ln P1

T 2−T 1 P =ln T1 P1

P T 2 −T 1=T 1 ln P1 4 ¯¿ 0.5698=P

P 4 ¯¿ ( 350−800 ) K =ln ¿ 800 K

4 ¯¿ P e−0.5625= ¿

¯ P=2.279 Answer

Problem 3.25: A scheme for finding the internal volume

V tB

of the gas cylinder consists of the following steps. The

cylinder is filled with a gas to low pressure P 1, and connected through a small line and valve to an t

evacuated reference tank of known volume V A . The valve is opened, and the gas flows through the line into the reference tank. After the system returns to its initial temperature, a sensitive pressure transducer provides a valve for the pressure change from the following data: t 3 a) V A =256 cm b)

∆ P/ P1=−0.0639

Given Data: 53

∆P

in the cylinder. Determine the cylinder volume

t

VB

August 20, 2013

PROBLEMS

V tB =?

ZAID YAHYA

V tA =256 cm3

11-CH-74

∆P =−0.0639 P1

Solution: ∆P =−0.0639 P1

P2 −P 1 =−0.0639 P1

P2 −1=−0.0639 P1

P2 =−0.0639+1 P1

P2 =0.9361 →(1) P1

 Assume that gas is ideal & P2 is the pressure of the tank  When gas flows through the line into the tank then tank’s total volume becomes 

t

t

V A +V B

Now, by applying condition for an ideal gas t B

t A

P1 V =P2 ( V +V

t B

)

P2 Vt = t B t P1 V A +V B

 Put in (1)

V tB =0.9361 V tA +V tB

V tB =0.9361 ( V tA +V tB )

V tB ( 1−0.9361 ) =0.9361V tA

V tB =0.9361V tA +0.9361V tB

0.0639 V tB =256 cm 3∗0.9361

t

V B=

V tB −0.9361V tB=0.9361V tA

239.6461 3 cm 0.0639

V tB =3750.26 cm 3 Answe r

Problem 3.26: A closed, non-conducting, horizontal cylinder is fitted with non-conducting, frictionless, floating piston which divides the cylinder in two Sections A & B. The two sections contains equal masses of air, initially at the same conditions, T1 = 300 K and P1 = 1 atm. An electrical heating element in section A is activated, and the air temperature slowly increases: TA in section A because of heat transfer, and T B in section B because of adiabatic compression by slowly moving piston. Treat air as an ideal gas with C P = (7/2) R and let nA be the number of moles of air in section A. For the process as described, evaluate one of the following sets of quantities: a) TA, TB, and Q/ nA, if P (final) = 1.25 atm b) TB, Q/ nA, and P (final), if TA = 425 K c) TA, Q/nA, and P (final), if TB = 325 K d) TA, TB, and P (final), if Q/nA = 3 kJ mol-1. Given Data:

54

August 20, 2013

T 1 =300 K

PROBLEMS

P1=1 atm

ZAID YAHYA

11-CH-74

7 C P= R 2

Solution: 

According to ideal gas equation,

PV =nRT  Applying ideal gas equation for initial conditions  On section “A”

P1 V A =n A R T 1

V A=

nART1 P1

 On section “B”

P1 V B=n B RT 1  Since

n A =n B  Therefore,

P1 V B=n A R T 1

V B=

n A RT 1 P1

 Total initial volume can be given as

V i=V A + V B

V i=

nA R T1 nA R T1 + P1 P1

V i=2.

n A RT 1 P1

Let P2 be the final pressure & TA & TB are the final temperatures of section A & section B respectively  Applying ideal gas equation for final conditions  On section “A” 

P2 V A =n A R T A V A=

nART A P2

 On section “B”

P2 V B=n A R T B  Total Final volume can be given as 55

V B=

n A RT B P2

August 20, 2013

V f =V A +V B



PROBLEMS

Vf=

ZAID YAHYA

nA R T A nA R T B + P2 P2

Vf=

11-CH-74

nA R (T A+ T B) P2

Since the total volume is constant, therefore

V i=V f

2.

n A R T 1 n A R ( T A +T B ) 2. T 1 ( T A +T B ) = = →(1) P1 P2 P1 P2

(a)

P2=1.25 atm  

Since the process occurring in section B is reversible adiabatic compression Therefore, for an adiabatic compression we have

T 1 ( P1 )

1−γ γ

=T B ( P2 )

T B=T 1



1−γ γ

P2 P1

( )

T B=

T 1 ( P1 )

( P 2) γ−1 γ

1−γ γ

1− γ γ

→(2)

We know that,

C P−C V =R

CV =C P −R

7 CV = R−R 2

CV =

7 R−2 R 2

5 CV = R 2

 As

γ=

CP CV

γ=

7∗R∗2 2∗5∗R

γ =1.4

 Put in (2)

T B=300 K

1.25 1

( )

1.4 −1 1.4

T B=300 K∗1.0658

T B=319.74 K Answe r

56

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 Put in (1)

2∗300 K ( T A +319.74 K ) = 1 atm 1.25 atm 600 K∗1.25=T A +319.74 K 

T A =750 K−319.74 K

T A =430.26 K Answe r

According to first law of thermodynamics

∆ U =Q+W  Since volume is constant, therefore

∆ U =Q→( a)  For section A & B

∆ U =∆ U A +∆ U B  Put in (a)

Q=∆U A + ∆U B

Q=n A CV ∆ T +n A C V ∆ T

Q=n A CV [ T A +T B−2T 1 ]

Q=n A CV ( T A−T 1 ) +n A C V ( T B −T 1 )

Q =C V [ T A +T B −2 T 1 ] →(3) nA

Q 5 J K∗1 kJ = ∗8.314 ∗150.02 nA 2 mol∗K 1000 J

Q 5 = R ( 430.26+ 319.74−2∗300 ) K nA 2 Q kJ =3.118 Answe r nA mol

(b)

T A =425 K  From equation (1)

2. T 1 ( T A +T B ) = P1 P2  Put in (2)

57

Q=n A CV [ T A−T 1 +T B−T 1 ]

P2 ( T A +T B ) = P1 2.T 1

August 20, 2013

PROBLEMS

ZAID YAHYA

T B=T 1

 Assume

)

1.4−1 1.4

T B=300 K∗( 1.0634 )  Since

)

γ −1 γ

T B=319 K

425+ 319 T B=300 K 2∗300

(

(

T A +T B 2. T 1

11-CH-74

T B=319.02 K

319 ≈ 319.02 , therefore

T B=319.02 K Answe r  Put in (1)

2∗300 K ( 425+319.02 ) K = 1 atm P2

P2=

744.02 atm 600

P2=1.24 atm Answe r

 From equation (3)

Q =C V [ T A +T B −2 T 1 ] nA Q 5 = R ( 425+319.02−2∗300 ) K nA 2 Q 5 J K∗1 kJ Q kJ = ∗8.314 ∗144.02 =2.993 Answer nA 2 mol∗K 1000 J n A mol (c)

T B=325 K  Put in (2)

58

August 20, 2013

T B=T 1

PROBLEMS

P2 P1

( )

γ−1 γ

ZAID YAHYA

P2 325 K=300 K 1 atm

( )

γ −1 γ

P2 325 = 300 1 atm

( )

γ −1 γ

11-CH-74

325 300

( )

γ γ −1

=

P2 1 atm

325 300

( )

P2=

1.4 1.4−1

P2=1.323 atm Answe r  Put in (1)

2. T 1 ( T A +T B ) = P1 P2

2∗300 K T A +325 K = 1 atm 1.323 atm

T A +325 K =600 K∗1.323

T A =793.9 K−325 K

T A =468.9 K Answe r  From equation (3)

Q =C V [ T A +T B −2 T 1 ] nA Q 5 = R ( 468.9+325−2∗300 ) K nA 2 Q 5 J K∗1 kJ Q kJ = ∗8.314 ∗193.9 =4.0302 Answe r nA 2 mol∗K 1000 J n A mol (d)

Q kJ =3 nA mol  From equation (1)

2. T 1 ( T A +T B ) = P1 P2

T A +T B=

2.T 1∗P2 → (b ) P1

 From equation (3)

Q =C V [ T A +T B −2 T 1 ] nA  Comparing (b) and (c) 59

T A +T B−2 T 1 =

Q n A∗C V

T A +T B=

Q + 2T 1 →(c ) n A∗C V

atm

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

2. T 1∗P2 Q = +2 T 1 P1 n A∗CV

P 2=

P1 Q +2 T 1 2.T 1 n A∗C V

[

]

P 2=

[

1 atm 2∗3 kJ +2∗300 K 2∗300 K 5 R∗mol

P 2=

]

[

6 kJ∗mol∗K ∗1000 J 1 atm 5∗mol∗8.314 J P 2= +600 K 600 K 1 kJ

]

1 atm [ 144.335+ 600 ] K P2=1.2406 atm Answe r 600 K

 Put in (2)

T B=T 1

P2 P1

( )

γ−1 γ

T B=300 K

(

1.2406 1

)

1.4 −1 1.4

T B=300 K∗1.0635

T B=319.06 K Answer  Put in (1)

2. T 1 ( T A +T B ) = P1 P2 2∗300 K ( T A + 319.06 K ) = 1 atm 1.2406 atm 600 K∗1.2406=T A +319.06 K

T A =744.36 K−319.06 K

T A =425.3 K Answe r

Problem 3.27: One mole of an ideal gas with constant heat capacities undergoes an arbitrary mechanically reversible process. Show that: 1 ∆U= ∆ ( PV ) γ−1 60

August 20, 2013

PROBLEMS

ZAID YAHYA

Given Data:

Number of moles=n=1

Solution: 

We know that

∆ U =nCV ∆ T 

∆ U =1.C V ∆ T

∆ U =CV ( T 2−T 1 ) →(1)

For an ideal gas we have,

C P−C V =R

C P CV R − = C V CV C V

CP R −1= CV CV

 Since

γ=

CP CV

 Therefore,

γ −1=

R CV

CV =

R γ −1

 Put in (1)

∆U=



R ( T −T ) γ−1 2 1

∆U=

For one mole of an ideal gas we have

P1 V 1=R T 1 →( a)  Put (a) & (b) in (2)

∆U=

1 ( P V −P1 V 1 ) γ −1 2 2

Problem 3.28: 61

1 ( R T 2−R T 1 ) →( 2) γ−1

∆U=

1 ∆ ( PV ) Proved γ−1

P2 V 2 =R T 2 →( b)

11-CH-74

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

Derive an equation for the work of a mechanically reversible, isothermal compression of 1 mole of a gas from an initial pressure p1 to a final pressure p2 when the equation of state is the virial expansion truncated to: Z =1+ B ' P How does the result compare with the corresponding equation for an ideal gas?

Solution: 

For a mechanically reversible process we have, V2

W =−∫ PdV →(1) V1



Given that

Z =1+ B ' P  Also

Z=

PV RT

 Therefore,

PV ' =1+ B P RT

V=

RT ( 1+ B' P ) P

V =RT

( P1 + B' )

 Differentiate both sides w.r.t to pressure

dV −1 =RT 2 +0 dP P

(

)

dV =

−RT dP P2

 Put in (1) V2

W =−∫ PdV V1

P2

RT W =−∫ −P 2 dP P P 1

P2

W =RT ∫ P1

W =RT ln

62

1 dP P

P2

W =RT |ln P|P

P2 Proved P1

1

W =RT ( ln P 2−ln P1 )

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

Problem 3.30: For methyl chloride at 100 ℃ the second and third virial coefficients are: B=−242.5 cm3 mol−1 ; C=25 200 cm 6 mol−2 . Calculate the work of mechanically reversible, isothermal compression of 1 mol of methyl chloride 1 bar to 55 bars at 100 ℃ . Base calculations on the following forms of virial equations B C Z =1+ + 2 a) V V b)

Z =1+ B' P+C ' P2 '

B=

Where

B C−B 2 ' ∧C = RT ( RT )2

Why don’t both equations give exactly the same result? Given Data:

Temperature=T =100 ℃ 6

C=25200 cm mol

−2

T =( 100+273.15 ) K

T =373.15 K

P1=1 ¯¿

B '=

P2=55 ¯ ¿

B RT

B=−242.5 cm3 mol−1 '

C=

C−B2 ( RT )2

Solution:  As

B'=

B RT

¯¿∗1 m3 100 3 cm3 −242.5 cm3∗mol∗K ∗J mol∗8.314 J∗373.15 K ∗101325 N N∗m ' B= ¿ m2∗1.01325

 Now,

63

¯¿

1 B =−7.817∗10−3 ¿ '

W =?

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

¿¯2∗1 m6 1006 cm 6 [ 25200−(−242.5 )2 ] cm6∗K 2∗mol 2 ∗J 2 mol2∗8.3142 J 2∗373.152 K 2 2 2 ∗101325 N 2 2 N ∗m C' = ¿ m4∗1.013252

2

C−B C= ( RT )2 '

¿¯2 C' =−3.492∗10−5

(a)

Z =1+



B C + V V2

PV B C =1+ + 2 RT V V

P=

RT B C 1+ + 2 →(1) V V V

(

)

For a mechanically reversible process we have, V2

W =−∫ PdV →(2) V1

 Put (1) in (2) V2

RT B C W =−∫ 1+ + 2 dV V V V V 1

(

)

V2

(

W =−RT ∫ 1+ V1

B C 1 + dV →(3) V V2 V

)

 Again using

V 1=

RT B C 1+ + 2 → ( 4 ) P1 V1 V1

(

)

 Assume that 3

V 1=30780

cm →(a) mol ¯¿∗N∗m ∗1.01325 ¯¿ m2 J ∗1003∗cm 3 101325 N RT cm3 mol∗K∗1 =31023.6 →(b) P1 mol 1 m3 RT 8.314 J∗373.15 K = ¿ P1

64

1 ¿

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 Put (a) and (b) in (4)

30780

cm3 cm3 242.5 25200 =31023.6 ∗ 1− + mol mol 30780 307802

(

)

cm3 cm3 ( 30780 =31023.6 1−0.007878+ 0.000026598 ) mol mol

cm3 cm3 30780 =31023.6∗0.99215 mol mol

cm3 cm3 30780 =30780 mol mol  Since

L. H . S=R . H . S  Therefore 3

Initial volume=V 1=30780

cm mol

 Again using

V 2=

RT B C 1+ + 2 → (5 ) P2 V2 V2

(

)

 Assume that

V 2=241.33

cm3 →( c) mol ¯¿∗N∗m ∗1.01325 ¯¿ m2 J ∗100 3∗cm3 101325 N RT cm 3 mol∗K∗55 =564.067 →(d ) P2 mol 1 m3 RT 8.314 J∗373.15 K = ¿ P2

 Put (c) and (d) in (4)

65

August 20, 2013

PROBLEMS

ZAID YAHYA

3

3

241.33

cm cm 242.5 25200 =564.067 ∗ 1− + mol mol 241.33 241.332

241.33

cm3 cm3 =564.067∗0.4278 mol mol

(

241.33

)

11-CH-74 3

3

cm cm ( 241.33 =564.067 1−1.0048+ 0.4327 ) mol mol

cm3 cm3 =241.33 mol mol

 Since

L. H . S=R . H . S  Therefore

Finalvolume=V 2=241.33

cm 3 mol

 Now from equation (3) V2

W =−RT ∫ V1

(

[

B C 1 1+ + 2 dV V V V

)

V2

W =−RT |ln V|V −B

[

W =−RT ln

1

V2

|| 1 V

W =−RT

| |]

1 1 − C 2 2 V V 1

V2

[

V2

V2

V2

1

1

1

∫ V1 dV + B ∫ V12 dV +C ∫ V13 dV V V V

V1

V2 1 1 1 1 1 −B − − C − V1 V 2 V 1 2 V 22 V 12

(

[

W =−RT ( ln V 2−ln V 1 )−B

) (

(

]

1 1 1 1 1 − − C − 2 2 V2 V1 2 V2 V1

) (

)]

−8.314∗J∗373.15 K ∗1 kJ mol∗K 241.33 1 1 25200 1 1 W= ln +242.5 − − − 2 2 1000 J 30780 241.33 30780 2 241.33 30780

[

W =−3.102

)]

(

)

(

)]

kJ kJ [ −4.848+0.996−0.2163 ] W =12.62 Answer mol mol

(b)

Z =1+ B' P+C' P 2

PV =1+ B' P+C ' P2 RT

V=

 Differentiate w.r.t to pressure on both sides 66

RT ( 1+ B' P+ C' P2 ) P

V =RT

( P1 + B +C ' P) '

August 20, 2013

PROBLEMS

ZAID YAHYA

dV −1 =RT 2 +0+C ' dP P

(

)

11-CH-74

( P1 +C ' ) dP

dV =−RT

2

 Put in (2) P2

1 W =−∫ −PRT 2 +C ' dP P P

(

1

)

[

[

P2

P2

1

1

1 W =RT ∫ dP+C ' ∫ PdP P P P '

C W =RT |lnP|P + |P2|P

[

W =RT ln

P2 +C' ( P22−P12 ) P1

]

P2 1

2

P2 1

]

¿¯2 ¿¯2 ( 55 2−1 ) ¿ 55 3.492∗10−5 1 ln − ¿ 1 2 8.314∗J∗373.15 K ∗1 k J mol∗K W= ¿ 1000 J W =12.268

]

W=

3.102 kJ ( 3.9545 ) mol

kJ Answer mol

The answers for part (a) and (b) differ because the relations between the two sets of parameters are exact only for infinite series

Problem 3.32: Calculate Z and V for ethylene at 25 oC and 12 bars by the following equations: a) The truncated virial equation [Eq. (3.40)] with the following experimental values of virial coefficients: B=−140 cm 3 mol−1 C=7200 cm 6 mol−2 b) The truncated virial equation [Eq.(3.38)], with a value of B from the generalized Pitzer correlations [Eq. (3.63)] c) The Redlich/Kwong equation d) The Soave/Redlich/Kwong equation e) The Peng/Robinson equation. Given Data:

67

August 20, 2013

PROBLEMS

ZAID YAHYA

Temperature=T =25 ℃

11-CH-74

T =298.15 K

T =( 25+273.15 ) K

Pressure=P=12 ¯¿

Solution: (a) 3

B=−140 cm mol 

6

−1

C=7200 cm mol

−2

Given equation is

Z =1+

B C + V V2

PV B C =1+ + 2 RT V V

V=

RT B C 1+ + 2 → ( 1 ) P V V

(

)

mol∗K∗12∗¯¿∗N∗m ∗1.01325 ¯¿ m2 J ∗1003∗cm 3 101325 N 1 m3 RT 8.314∗J∗298.15 K = ¿ P

RT cm3 =2065.68 V mol  Put in (1) 3

cm B C V =2065.68 1+ + 2 mol V V

(

)

 Assume

V =1919

cm3 mol

 Therefore 3

1919

 Since

68

3

cm cm 140 7200 =2065.68 1− + mol mol 1919 19192

(

)

3

1919

3

cm cm ( =2065.68 0.9290 ) mol mol

3

1919

3

cm cm =1919 mol mol

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

L. H . S=R . H . S  Therefore

V =1919



cm3 Answer mol

We know that

Z=

PV RT

12∗¯¿ 1919∗cm3∗mol∗K ∗J 8.314∗J∗298.15∗K∗mol ∗101325 N N∗m ∗1m 3 2 ¯ 1.01325 ¿ m Z= 1003 cm3

Z =0.929 Answer

(b) 

Given equation 3.38 is

Z =1+



BP →( 2) RT

We know that

Reduced temperature=T r =

T Tc

Reduced Pressure=Pr =

T =T r T c → ( a )

 Put (a) and (b) in (2)

Z =1+

B Pr Pc R T r Tc

 Since

^ B Pc Reduced second virial coefficient =B= R Tc  Therefore

Z =1+  Also 69

^B P r →(3) Tr

P Pc

P=Pr Pc →(b)

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

O ^ B=B +ω B1

 Put in (3)

Z =1+ ( B O +ω B1 )

Pr →( 4) Tr

 For ethylene (From Table B.1 Appendix B)

ω=0.087 →(i)

Critical temperature=T c =282.3 K

Critical pressure=Pc =50.40 ¯ ¿

Tr=

T 298.15 K = =1.056→(ii) T c 282.3 K

50.40 ¯¿ =0.238 →(iii) ¯ 12 ¿¿ P Pr = =¿ Pc

 Also we have

B o=0.083−

0.422 T r1.6

B o=0.083−

0.422 1.6 1.056

B o=−0.304 → ( iv )

B 1=0.139−

0.172 4.2 1.056

B 1=2.183∗10−3 →( v)  Put (i), (ii), (iii), (iv) & (v) in (4)

Z =1+ (−0.304+0.087∗2.183∗10−3 )

0.238 1.056

Z =0.931 Answer

 We know that

Z=

PV RT

V=

ZRT P

¯¿∗N∗m ∗1.01325 ¯¿ m2 J 3 3 ∗100 cm 101325 N mol∗K∗12 1m3 0.931∗8.314 J∗298.15 K V= ¿

(c)  70

The Redlich/Kwong equation is

V =1923.

cm3 Answer mol

August 20, 2013

PROBLEMS

ZAID YAHYA

Z =1+ β−qβ

11-CH-74

Z−β → ( 5) ( Z + ϵ β ) ( Z +σβ )

 Where

β=Ω

Pr Tr

 From table 3.1

σ =1, Ω=0.08664, ε=0,ψ=0.42748,α ( T r )=T r

−1 2

 Now

β=

0.08664∗0.238 1.056

β=0.0195 →(i ' )

 Also

q=

ψα ( T r ) ΩT r

−1

0.42748∗1.056 2 q= 0.08664∗1.056

q=4.547 →(i i' )

 Put (i’) and (ii’) in (5)

Z =1+ 0.0195−4.547∗0.0195

Z−β ( Z + ϵ β ) ( Z +σβ )

 Assume

Z =0.928

0.928=1.0195−0.08866

0.928−0.0195 0.928 ( 0.928+ 0.0195 )

0.928=1.0195−0.08866∗1.033

 Since

L. H . S=R . H . S  Therefore,

Z =0.928 Answer  71

We know that

0.928=0.928

August 20, 2013

Z=

PROBLEMS

PV RT

V=

ZAID YAHYA

11-CH-74

¯ ¿∗N∗m 2 ∗1.01325 ¯ ¿m J ∗1003 cm3 101325 N mol∗K∗12 1m3 0.928∗8.314 J∗298.15 K V= ¿

ZRT P

3

V =1916.8

cm Answer mol

(d)  From table 3.1

σ =1, Ω=0.08664, ε=0,ψ=0.42748,α ( T r )=α SRK ( T r ; ω )

[ {

(

α SRK ( T r ; ω )= 1+ ( 0.480+ 1.574 ω−0.176 ω2 ) 1−T r

[ {

(

2

α SRK ( T r ; ω )= 1+ ( 0.480+1.574∗0.087−0.176∗0.087 ) 1−1.056

1 2

1 2

2

)}]

2

) }]

2

α SRK ( T r ; ω )=[ 0.9828 ]

α SRK ( T r ; ω )=0.966 →(ii i ' )  Now

q=

ψα ( T r ) ΩT r

q=

0.42748∗0.966 0.08664∗1.056

'

q=4.515 →(ii i )

 Put (i’) and (iii’) in (5)

Z =1+ 0.0195−4.515∗0.0195

Z−β ( Z + ϵ β )( Z +σβ )

 Assume

Z =0.928

0.928=1.0195−0.08803

0.928−0.0195 0.9 ( 0.9+ 0.0195 )

0.928=1.0195−0.2383∗1.36

 Since

L. H . S=R . H . S

72

0.928=0.928

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 Therefore,

Z =0.928 Answer 

We know that

Z=

PV RT

V=

¯ ¿∗N∗m ∗1.01325 ¯ ¿ m2 J ∗1003 cm3 101325 N mol∗K∗12 1m3 0.928∗8.314 J∗298.15 K V= ¿

ZRT P

cm3 V =1917 Answer mol

(e)  From table 3.1

σ =1+ √2=2.41, Ω=0.07780, ε=1−√ 2=−0.414,ψ =0.45724, α ( T r )=α SRK ( T r ; ω )  Now

β=Ω

β=

0.07780∗0.238 1.056

Pr Tr

β=0.0175 →(iv ' )

[ {

(

α SRK ( T r ; ω )= 1+ ( 0.37464 +1.54226 ω−0.26992 ω2 ) 1−T r

[ {

2

(

1 2

2

)}]

α SRK ( T r ; ω )= 1+ ( 0.37464+ 1.54226∗0.087−0.26992∗0.087 ) 1−1.056

2

α SRK ( T r ; ω )=[ 0.986 ]

α SRK ( T r ; ω )=0.9722

 Now

q=

73

ψα ( T r ) ΩT r

q=

0.42748∗0.9722 0.08664∗1.056

q=4.542 →(v ' )

1 2

2

) }]

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 Put (iv’) and (v’) in (5)

Z =1+ 0.0175−4.542∗0.0175

Z−β ( Z + ϵ β )( Z+ σβ )

 Assume

0.92=1.0175−0.07949

Z =0.92

0.92−0.0175 ( 0.92−0.414∗0.0175 ) ( 0.92+ 2.41∗0.0175 )

0.92=1.0175−0.07949∗1.027

0.92 ≈ 0.93  Since

L. H . S ≈ R . H . S  Therefore,

Z =0.92 Answer 

We know that

Z=

PV RT

V=

ZRT P

¯¿∗N∗m ∗1.01325 ¯¿ m2 J ∗1003 cm3 101325 N mol∗K∗12 1m3 0.92∗8.314 J∗298.15 K V= ¿

3

V =1900.4

cm Answer mol

Problem 3.33: Calculate Z and V for ethane at 50 oC and 15 bars by the following equations: a) The truncated virial equation [Eq. (3.40)] with the following experimental values of virial coefficients: 3 −1 6 −2 B=−156.7 cm mol C=9650 cm mol b) The truncated virial equation [Eq.(3.38)], with a value of B from the generalized Pitzer correlations [Eq. (3.63)] c) The Redlich/Kwong equation d) The Soave/Redlich/Kwong equation e) The Peng/Robinson equation. Given Data:

74

August 20, 2013

PROBLEMS

ZAID YAHYA

Temperature=T =50 ℃

11-CH-74

T =323.15 K

T =( 50+273.15 ) K

Pressure=P=15 ¯¿

Solution: (a) 3

B=−156.7 cm mol 

6

−1

−2

C=9650 cm mol

Given equation is

Z =1+

B C + V V2 PV B C =1+ + 2 RT V V

V=

RT B C 1+ + 2 → ( 1 ) P V V

(

)

mol∗K∗15∗¯¿∗N∗m ∗1.01325 ¯¿ m2 J ∗1003∗cm 3 101325∗N 1 m3 RT 8.314∗J∗323.15 K = ¿ P

RT cm3 =2065.68 V mol

 Put in (1)

V =1791

cm 3 B C 1+ + 2 mol V V

(

)

 Assume

cm3 V =1625 mol  Therefore

1625

75

cm3 cm3 156.7 9650 =1791 1− + mol mol 1625 16252

(

)

1625

cm3 cm3 ( =1791 0.9072 ) mol mol

1625

cm3 cm3 =1625 mol mol

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 Since

L. H . S=R . H . S  Therefore 3

V =1625



cm Answer mol

We know that

Z=

PV RT

15∗ ¯¿ 1625∗cm3∗mol∗K ∗J 8.314∗J∗323.15∗K∗mol ∗101325 N N∗m ∗1m3 2 1.01325 ¯¿ m Z= 1003 cm3

Z =0.907 Answer

(b) 

Given equation 3.38 is

Z =1+



BP →( 2) RT

We know that

Reduced temperature=T r =

T Tc

T =T r T c → ( a )

Reduced Pressure=Pr =

 Put (a) and (b) in (2)

Z =1+

B Pr Pc R T r Tc

 Since

^ B Pc Reduced second virial coefficient =B= R Tc  Therefore

76

P Pc

P=Pr Pc →(b)

August 20, 2013

PROBLEMS

ZAID YAHYA

Z =1+

11-CH-74

^B P r →(3) Tr

 Also O ^ B=B +ω B1

 Put in (3)

Z =1+ ( B O +ω B1 )

Pr →( 4) Tr

 For ethylene (From Table B.1 Appendix B)

ω=0.1 →(i)

Critical temperature=T c =305.3 K

Critical pressure=Pc =48.72 ¯¿

Tr=

T 323.15 K = =1.058 →(ii) T c 305.3 K

48.72 ¯¿=0.308 →(iii ) ¯ 15 ¿¿ P Pr = =¿ Pc

 Also we have o

B =0.083−

0.422 T r1.6

B o=0.083−

0.422 1.6 1.058

B o=−0.303 → ( iv )

B 1=0.139−

0.172 4.2 1.058

B 1=3.266∗10−3 →(v)  Put (i), (ii), (iii), (iv) & (v) in (4)

Z =1+ (−0.303+ 0.1∗3.266∗10−3 )

0.308 1.058

Z =0.912 Answer

 We know that

Z=

77

PV RT

V=

ZRT P

¯¿∗N∗m ∗1.01325 ¯¿ m2 J 3 3 ∗100 cm 101325 N mol∗K∗15 1 m3 0.912∗8.314 J∗323.15 K V= ¿

V =1633.36

cm3 Answer mol

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

(c) 

The Redlich/Kwong equation is

Z =1+ β−qβ

Z−β → ( 5) ( Z + ϵ β ) ( Z +σβ )

 Where

β=Ω

Pr Tr

 From table 3.1

σ =1, Ω=0.08664, ε=0,ψ=0.42748,α ( T r )=T r

−1 2

 Now

β=

0.08664∗0.308 1.058

'

β=0.0252 →(i )

 Also

ψα ( T r ) q= ΩT r

q=

−1 2

0.42748∗1.058 0.08664∗1.058

q=4.533 →(ii ' )

 Put (i’) and (ii’) in (5)

Z =1+ 0.0252−4.533∗0.0252

Z−β ( Z+ ϵ β )( Z+ σβ )

 Assume

Z =0.906

0.906=1.0252−0.114

0.906−0.0252 0.906 ( 0.906+ 0.0252 )

0.906=1.0252−0.114∗1.044

 Since

L. H . S=R . H . S  Therefore, 78

0.906=0.906

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

Z =0.906 Answer 

We know that

Z=

PV RT

V=

¯¿∗N∗m 2 ∗1.01325 ¯¿ m J 3 3 ∗100 cm 101325 N mol∗K∗15 1 m3 0.906∗8.314 J∗323.15 K V= ¿

ZRT P

3

V =1622.7

cm Answer mol

(d)  From table 3.1

σ =1, Ω=0.08664, ε=0,ψ=0.42748,α ( T r )=α SRK ( T r ; ω )

[ {

(

α SRK ( T r ; ω )= 1+ 0.480+1.574 ω−0.176 ω2 1−T r

[ {

2

(

α SRK ( T r ; ω )= 1+ 0.480+1.574∗0.1−0.176∗0.1 1−1.058

1 2

1 2

2

)}]

2

) }]

2

α SRK ( T r ; ω )=[ 1+0.6374 ]

α SRK ( T r ; ω )=2.681 →(ii i' )  Now

q=

ψα ( T r ) ΩT r

q=

0.42748∗2.681 0.08664∗1.058

'

q=12.50 →(iii )

 Put (i’) and (iii’) in (5)

Z =1+ 0.0252−12.50∗0.0252

Z−β ( Z + ϵ β ) ( Z +σβ )

 Assume

Z =0.907

79

0.907=1.0252−0.315

0.907−0.0252 0.907 ( 0.907+0.0252 )

0.907=1.0252−0.315∗1.0429

0.695=0.695

August 20, 2013

PROBLEMS

ZAID YAHYA

11-CH-74

 Since

L. H . S=R . H . S  Therefore,

Z =0.695 Answer 

We know that

Z=

PV RT

V=

ZRT P

¯¿∗N∗m 2 ∗1.01325 ¯¿ m J ∗1003 cm3 101325 N mol∗K∗12 1m3 0.695∗8.314 J∗298.15 K V= ¿

3

V =1435.6

cm Answer mol

(e)  From table 3.1

σ =1, Ω=0.07780, ε=0,ψ=0.45724,α ( T r )=α SRK ( T r ; ω )  Now

β=Ω

β=

0.07780∗0.238 1.056

Pr Tr

β=0.0175 →(iv ' )

[ {

(

α SRK ( T r ; ω )= 1+ 0.37464 +1.54226 ω−0.26992ω 2 1−T r

1 2

)}]

2

2 2 α SRK ( T r ; ω )=[ 1+ {0.37464+1.54226∗0.087−0.26992∗0.087 ( 1−1.056 ) } ]

2

α SRK ( T r ; ω )=[ 1+0.509 ]  Now

80

α SRK ( T r ; ω )=2.277

2

August 20, 2013

PROBLEMS

ZAID YAHYA

q=

ψα ( T r ) ΩT r

q=

0.42748∗2.277 0.08664∗1.056

11-CH-74

q=10.64 →( v ' )

 Put (iv’) and (v’) in (5)

Z =1+ 0.0175−10.64∗0.0175

Z−β ( Z + ϵ β )( Z+ σβ )

 Assume

Z =0.793

0.793=1.0175−0.1862

0.793−0.0175 0.793 ( 0.793+0.0175 )

0.793=1.0175−0.1862∗1.2066

0.793=0.793

 Since

L. H . S=R . H . S  Therefore,

Z =0.793 Answer 

We know that

Z=

81

PV RT

V=

ZRT P

¯ ¿∗N∗m ∗1.01325 ¯ ¿ m2 J ∗1003 cm3 101325 N mol∗K∗12 1m3 0.793∗8.314 J∗298.15 K V= ¿

V =1638.1

cm 3 Answer mol