Chapter 3

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Full text

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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 44.18∗10−6 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 V

(

dV dT

)

P →(1) Isothermal Compressibilty=κ=−1 V

(

∂ V ∂ P

)

T

Or

κ=−1 V

(

dV dP

)

T → (2) Temperature=T =50 0

C

Pressure=P1=1 ¯¿ ¯ ¿−1 κ=44.18∗10−6¿ Density of water=ρ1=1 kg m3 ρ2=(1+1 ) kg m3 ρ2=1.01 kg m3 P2=?

Solution:

(a)  We know that

(2)

ρ=1 V V = 1 ρ  Put in (1) & (2) β=ρ

(

d dT 1 ρ

)

P β=ρ ρ2

(

dT

)

P β=−1 ρ

(

dT

)

P Proved  Now, κ=−ρ

(

d dP 1 ρ

)

T κ= ρ ρ2

(

dP

)

T κ=1 ρ

(

dP

)

T Proved

(b)  As κ=1 ρ

(

dP

)

T κdP=dρ ρ

 Integrating on both sides

κ

P1 P2 dP=

ρ1 ρ2 ρ κ|P|P1 P2 =|ln ρ|ρ1 ρ2 κ

(

P2P1

)

=

(

ln ρ2−ln ρ1

)

κ

(

P2P1

)

=lnρ2 ρ1  Putting values ¯ ¿

(

P2−1

)

=ln1.01 1 44.18∗10−6 ¿ ¯¿ 44.18∗10−6 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

(

∂ β

(3)

Solution:

 We know that Volume expansivity=β=1 V

(

∂V ∂ T

)

P

Since β is very small

V = 1 ∂ β

(

∂ V ∂ T

)

P → (1) Isothermal Compressibilty=κ=−1 V

(

∂ V ∂ P

)

T

Since κ is very small

V =−1 ∂κ

(

∂ V ∂ P

)

T 1 ∂ β

(

∂V ∂ T

)

P =−1 ∂ κ

(

∂V ∂ P

)

T 1 ∂ β

(

1 ∂T

)

P ∂V =−1 ∂ κ

(

1 ∂ P

)

T ∂V

(

∂ κ ∂T

)

P =

(

∂ β ∂ P

)

T

(

∂ β ∂ P

)

T =−

(

∂κ ∂T

)

P Proved

Problem 3.3:

The Tait equation for liquids is written for an isotherm as: V =V0

(

1− AP

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=−1 V

(

∂ V ∂ P

)

T → (1)

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Given that

V =V0

(

1− AP

B+P

)

 Where

 V0 = Hypothetical molar/specific volume at zero pressure, so it is constant

 V = Molar/specific volume  Now, V =VoAP B+PVo V −Vo= −AP B+PVo V −Vo Vo =−AP B+ P

Differentiate w.r.t Pressure 1 Vo ∂ P

(

V −Vo

)

= − ∂ P

(

AP B +P

)

1 Vo

(

∂ V ∂ P−0

)

=−

[

A (B+P )− AP(1) (B+P )2

]

−1 Vo

(

∂V ∂ P

)

= AB+ AP− AP (B+P)2 −1 Vo

(

∂V ∂ P

)

= AB (B+P)2  Since, Temperature is constant

 Therefore, −1 Vo

(

∂V ∂ P

)

T = AB (B+P)2  Or, From (1) κ= AB (B+P )2Proved

Problem 3.4:

For liquid water the isothermal compressibility is given by:

κ= c

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?

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Given Data:

Isothermal compressibility=κ= c

V ( P+b) Mass of water=m=1 kg Pressure=P1=1 ¯¿

P2=500 bars Temperature=T =60 0C b=2700 bars c=0.125 cm3 /g Work=W =?

Solution:

 We know that W =−

PdV →(1) κ= c V ( P+b )→(2)  Also κ=−1 V

(

dV dP

)

T → (3)  Comparing (2) & (3) −1 V dV dP= c V (P+b)dV = c dP P+b  Put in (1) W=−

P c dP P+b W =c

P 1 P2 P P+bdP W =c

P 1 P2 P+b−b P+b dP W =c

P 1 P2 P+b P+bdP−c

P 1 P2 b P+bdP W=c

P1 P2 dP−b c

P1 P2 1 P+bdP W =c|P|P1 P2 −bc

|

ln(P+b)

|

P1 P2 W=c

(

P2−P1

)

bc

[

ln

(

P2+b

)

−ln

(

P1+b

)

]

W=c

(

P2P1

)

bc

[

lnP2+b P1+b

]

 Putting values

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W=0.125cm 3 g ∗(500−1 )−2700¯¯ ¿0.125 cm3 g ∗ln 500+2700 1+2700 cm3∗¯¿ g cm3∗¯¿ g −57.216¿ W =62.375¿ cm3∗¯¿ g W=5.16¿ cm3 ∗¯¿ g ∗1 m 3 1003cm3 ∗101325 N 1.01325 ¯¿m2 ∗J Nm W =5.16¿ W=0.516J g Answer

Problem 3.5:

Calculate the reversible work done in compressing 1 ft3 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 ft3 Temperature=T =32 F Pressure=P1=1 atm

Pressure=P2=3000 atm κ /atm−1=3.9∗10−6−0.1∗10−9P (atm)

 Where

 Term, 3.9*10-6 has unit of atm-1 & 0.1*10-9 has units of atm-2

Solution:

 We know that, work done for a reversible process is

W=−

PdV →(1)  Also κ=−1 V

(

dV dP

)

dV =−κVdP

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 Put in (1), W=−

P (−κVdp) W=V

P1 P2 κ P dP W=V

P1 P2

(

3.9∗10−6 −0.1∗10−9 P

)

P dP W=V

P1 P2 3.9∗10−6P dP−V

P1 P2 0.1∗10−9P2dP W=3.9∗10−6 V

1 3000 P dP−0.1∗10−9V

1 3000 P2dP W=3.9∗10−6 V

|

P 2 2

|

1 3000 −0 .1∗10−9V

|

P 3 3

|

1 3000 W=1.95∗10 −6 ∗1 ft3 atm

(

3000 2 −12

)

atm2−3.333∗10 −11 ∗1 ft3 atm2 ∗

(

3000 3 −13

)

atm3 W =(17.55−0.8991) atm∗ft3 W=16.65atm∗ft3Answer

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 0oC to 20oC. Determine ΔVt, W, Q, and ΔUt. The properties for liquid carbon tetrachloride at 1 bar & 0oC 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 ¯¿ Temperature=T1=0 0

C

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Temperature=T2=20 0

C

T2=(20+273.15) K T2=293.15 K β=1.2∗10−3 K−1 C P=0.84 kJ kg∗K ρ=1590 kg m3 ∆ V t =? W=? Q=? ∆ Ut=?

Solution:

 As V =1 ρ V1= 1 ρ1 V1= 1 1590 m3 kg  Also,  we know that Volume expansivity=β=1 V

(

dV dT

)

P βdT =1 V dV

 Integrating on both sides,

β

T1 T2 dT =

V1 V2 dV V β|T|T1 T2 =|lnV|V1 V2 β

(

T 2−T1

)

=

(

lnV2−ln V1

)

β

(

T2−T1

)

=ln V2 V1

Putting values 1.2∗10−3 K ∗(293.15−273.15 ) K =ln V2∗1590 kg m3 e 0.024 =V2∗1590 kg m3 V2= 1.024 1590∗m 3 kg V2=0.000644m3 kg  Now, ∆ V =V2V1 ∆ V =

(

0.000644− 1 1590

)

m3 kg ∆ V =15.28∗10 −6m3 kg

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∆ Vt=∆ V ∗m ∆ Vt=

(

15.28∗10−6∗5

)

m3

kgkg ∆ V

t

=7.638∗10−5m3Answer

 Now,

 We know that for a reversible process,

Work done=W=−P ∆ Vt W=−1 ¯¿7.638∗10−5 m3∗101325 N 1.01325 ¯¿m2 ∗J Nm ∗1 kJ 1000 J W=−7.638∗10−3kJ Answer  Now,

 For a reversible process at constant pressure,we have

Q=∆ H Q=m CP∆ T Q=5 kg∗0.84 kJ

kg∗K∗(293.15−273.15) K Q=84 kJ Answer

 Now,

 According to first law of thermodynamics,

∆ Ut

=Q+W ∆ Ut

=

(

84−7.368∗10−3

)

kJ ∆ Ut=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=−1 V

(

dV dP

)

T dV V =−κdP

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dVV =−κ

dP lnV =−κP+lnA (T )

 Where ln A (T) is constant of integration & A depends on T only

lnV −lnAT =−κP

ln V

A (T )=−κP

 Taking anti log on both sides,

V AT=e −κP 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 +VdPPdV =VdP−d (PV )  Put in (1) dW =VdP−d ( PV )→ (2)  We know that Isothermal compressibilty=κ=−1 V

(

dV dP

)

TdV κ =VdP  Put in (2) dW =dV κd ( PV )

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dW =−1

κ

dV −

d ( PV ) W=−1

κ ΔV −Δ( PV )

 Since volume changes from V1 to V2 & pressure changes from P1 to P2 ,  Therefore, W=−1 κ

(

V2−V1

)

(

P2V2−P1V1

)

W=

(

V1V2

)

κ +P1V1−P2V2Proved

Problem 3.8:

One mole of an ideal gas with CV = 5/2 R, CP = 7/2 R expands from P1 = 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: CV=5 2R CP= 7 2R P1=8 ¯¿ T1=600 K P2=1 ¯¿ W=? Q=? ∆ U =? ∆ H=?

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)

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 For T2 , We know that for an ideal gas T1 P1 =T2 P2 T2= T1 P1 ∗P2 8 ¯¿∗1¯¿ T2=600 K¿ T2=75 K  Put in (2), Q=∆U =5 2R (75−600 ) K Q=∆U = −5 2 ∗8.314 J mol∗K∗525 K Q=∆U =−10912 J mol Q=∆U =−10.912 kJ mol Answer  Also

 For a mechanically reversible process we have,

∆ H=CP∆ T ∆ H=7 2R

(

T2−T1

)

∆ H= 7 2∗8.314 J mol∗K∗(75−600 ) K ∆ H=−15277 J mol ∆ H=−15.277 kJ molAnswer (b)

 For a constant temperature process,

∆ U =0 ∆ H=0

 We know that at constant temperature, work done is

W=R T1lnP2 P1 W=8.314 J mol∗K∗600 K∗ln 1 8 W=−10373 J mol W=−10.373 kJ molAnswer

 Now, according to first law of thermodynamics,

∆ U =Q+W 0=Q+W Q=−W

 Or

Q=10.373 kJ

mol Answer (c)

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Q=0

 Now, according to first law of thermodynamics,

∆ U =Q+W ∆ U =W →(1)

∆ U =CV∆T

 Put in (1)

W=∆ U=CV∆ T W=∆ U=CV

(

T2−T1

)

→(2)  For T2 , We know that for an adiabatic process

T1P1 (1−γ) γ =T 2P2 (1−γ) γ T2=T1

(

P1 P2

)

(1−γ) γ T 2=600 K

(

8 1

)

(1−1.4) 1.4 T 2=331.23 K  Put in (2) W=∆ U=5 2R∗(331.23−600) K W=∆ U= −5 2 ∗8.314 J mol∗K∗268.77 K W=∆ U=−5586.4 J mol W=∆ U=5.5864 J mol Answer  For a mechanically reversible adiabatic process we have

∆ H=CP∆ T ∆ H=7 2R

(

T2−T1

)

∆ H= 7 2∗8.314 J mol∗K∗(331.23−600) K ∆ H=−7.821 J mol Answer

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:

(14)

Initial Temperature=T1=600 K Initial Pressure=P1=10 ¯¿ CP= 7 2R CV= 5 2R

Solution:

(b)

Step 12, an Isothermal process,

 Since

 For an isothermal process, temperature is constant  Therefore,

T2=T1=600 K P2=3 ¯¿

 We know that, for an ideal gas

P2V2=R T2 V2=RT2 P2 mol∗K∗3 ¯ ¿∗1.01325 ¯¿m2 101325 NN∗m J V2=8.314∗J∗600 K ¿ V2=0.0166 m 3 mol

Step 23, an Isochoric process,

 Since

 For an isochoric process, Volume is constant  Therefore,

V3=V2=0.0166 m3

mol P3=2 ¯¿

 We know that, for an ideal gas

P3V3=RT3 T3=P3V3 R T 3= 2¯¿0.0166 m3 ∗mol∗K mol∗8.314 JJ N∗m ∗101325 N 1.01325 ¯¿m2 T3=400 K

Step 34, an Isobaric process,

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 Therefore,

P4=P3=2 ¯¿

 For T4 , we will use an adiabatic relation of temperature and pressure

 As T4 T1=

(

P4 P1

)

R CP T 4=T1

(

P4 P1

)

R CP T 4=600 K∗

(

2 10

)

2∗R 7R T 4=378.83 K

 We know that, for an ideal gas

P4V4=RT4 V4=R T4 P4 mol∗K∗2 ¯¿∗1.01325 ¯¿m2 101325 NNm J V4= 8.314 J∗378.83 K ¿ V4=0.0157 m3 mol

Step 41, an adiabatic process,

 Since

 Gas returns to its initial state adiabatically  Therefore,

T1=600 K P1=10 ¯¿

 We know that, for an ideal gas

P1V1=R T1 V1=RT1 P1 mol∗K∗10 ¯¿∗1.01325 ¯¿m2 101325 NNm J V1=8.314 J∗600 K ¿ V1=4.988∗10 −3 m3 mol (c)

Step 12, an Isothermal process,

 Since

 For an isothermal process, temperature is constant  Therefore

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∆ U12=0 ∆ H12=0

 For an isothermal process, we have

Q=−R T1lnP2 P1 Q=−8.314 J mol∗K∗600 K∗ln 3 10 Q=6006 J mol Q=6.006∗10 3 J mol Answer

 According to first law of thermodynamics

∆ U12=Q12+W12 0=Q12+W12 W12=−Q12 W12=−6.006∗10 3 J

mol Answer

Step 23, an Isochoric process,

 Since

 For an isochoric process, Volume is constant  Therefore,

W23=0

 At constant volume we have

Q23=∆ U23=CV∆ T Q23=∆ U23=CV

(

T3T2

)

Q23=∆ U23=5 2R ( 400−600 ) K Q23=∆ U23= 5 2∗8.314 J mol∗K ∗(−200) K Q23=∆ U23=−4157 J mol Q23=∆ U23=−4.157∗10 3 J molAnswer  We know that ∆ H23=CP∆ T ∆ H23=CP

(

T3−T2

)

∆ H23= 7 2R ( 400−600) K ∆ H23= 7 2∗8.314 J mol∗K∗(−200) K ∆ H23=−5820 J mol ∆ H23=−5.82∗10 3 J molAnswer

Step 34, an Isobaric process,

 Since

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Q34=∆ H34=CP∆ T Q34=∆ H34= 7 2R

(

T4−T3

)

Q34=∆ H34= 7 2∗8.314 J mol∗K∗(378.83−400) K Q34=∆ H34=−616 J molAnswer

 For an Isobaric process we have

W34=−R ∆ T W34=−R

(

T4T3

)

W34=−8.314 J mol∗K∗(378.83−400) K W34=176 J mol Answer  We know that, ∆ U34=CV∆ T ∆ U34=5 2R

(

T4−T3

)

∆ U34= 5 2∗8.314 J mol∗K∗(378.83−400) K ∆ U34=−440 J mol Answer

Step 41, an adiabatic process,

 Since

 For an adiabatic process there is no exchange of heat  Therefore, Q41=0  We know that, ∆ U41=CV∆ T ∆ U41=5 2R

(

T1−T4

)

∆ U41= 5 2∗8.314 J mol∗K∗(600−378.83) K ∆ U41=4597 J mol ∆ U41=4.597∗103 J mol Answer  We know that ∆ H41=CP∆T ∆ H41=7 2R

(

T1−T4

)

∆ H41= 7 2∗8.314 J mol∗K∗(600−378.83) K ∆ H41=6435.8 J mol ∆ H41=6.4358∗103 J mol Answer

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∆ U41=Q41+W41 ∆ U41=W41 W41=4.597∗10 3 J

molAnswer

Problem 3.10:

An ideal gas, CP= (5/2) R and CV= (3/2) R is changed from P1 = 1bar and V1

t

= 12m3 to P

2 = 12 bar and V2t = 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:

CP=5

2R CV=

3

2R Initial pressure=P1=1 ¯¿ V1t=12 m3 Final pressure=P2=12¯¿

V2 t =1 m3 Q=? W=? ∆ H=? ∆ U =?

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 T1lnP2

P1 Since

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P1V1=R T1  Therefore, Q=−P1V1lnP2 P1 Q=−1¯¿12 m3∗ln 12 1 ∗101325 N 1.01325 ¯¿m2 ∗J Nm ∗1 kJ 1000 J Q=−2981.88 kJ Answer

 According to first law of thermodynamics

∆ U =Q+W 0=Q+W W=−Q W12=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 P2 , intermediate volume can be given as

P2

(

V'

)

γ=P1V1 V'=V1

(

P1 P2

)

1 γ

 For mono atomic gas, we have

γ=1.67 V'=12 m3∗

(

1 12

)

1 1.67 V' =2.71 m3  We know that, W1= P2V'P1V1 γ−1

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W1= (12∗2.71−1∗12)¯¿m3 1.67−1 ∗101325 N 1.01325 ¯¿m2 ∗J Nm ∗1 kJ 1000 J W1=3063 kJ →(1)

 Second step, cooling at constant pressure P2

 We know that, for a mechanically reversible process

W2=−P2

(

V2−V '

)

W2=−12(1−2.71)¯ m3∗101325 N 1.01325 ¯¿m2 ∗J Nm ∗1 kJ 1000 J W2=2052 kJ →(2)  Now W=W1+W2 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 P'V2 γ =P1V1 P ' =P1

(

V1 V2

)

γ

 For mono atomic gas, we have

γ=1.67 P'=1

(

12¯ 1

)

1.67 P'=63.42¯¿  We know that,

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W1=P ' V2−P1V1 γ−1 W1= (63.42∗1−1∗12)¯¿m3 1.67−1 ∗101325 N 1.01325 ¯¿m2 ∗J Nm ∗1 kJ 1000 J W1=7674.76 kJ

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

W2=0

 Now

W=W1+W2 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

W1=0

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

W2=−P2∆ V W2=−P2

(

V2V1

)

W2=−12(1−12)¯ m3∗101325 N 1.01325 ¯¿m2 ∗J Nm ∗1 kJ 1000 J W2=13200 kJ  Now W=W1+W2 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)

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 The process completes in two steps

 Step 1, Cooling at constant Pressure P1 to V2  Therefore, for a mechanically reversible process

W1=−P1∆ V W1=−P1

(

V2V1

)

W1=−1(1−12)¯ m3∗101325 N 1.01325 ¯¿m2 ∗J Nm ∗1 kJ 1000 J W1=1100 kJ

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

W2=0

 Now

W=W1+W2 W=(1100+0) kJ W=1100kJ Answer

 According to first law of thermodynamics

∆ U =Q+W 0=Q+W Q=−W Q=−1100kJ Answer

Problem 3.11:

The environmental lapse rate dTdz characterizes the local variation of temperature with elevation in the earth's atmosphere. Atmospheric pressure varies with elevation according to the hydrostatic formula,

dP

dz =−M ρg

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

dz=−M ρg →(1)

(23)

T P 1−δ δ =Constant  Or T P 1−δ δ =T oPo 1−δ 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 = To Po δ−1 δ T To =

(

P Po

)

δ−1 δ

(

T To

)

δ δ−1= P Po P=Po

(

T To

)

δ δ −1→(a) P= Po To δ δ−1T δ δ −1

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

dP dT= Po To δ δ −1δ δ−1T 1 δ−1 dP= Po To δ δ−1δ δ−1T 1 δ−1dT →(2)

 We know that, for an ideal gas

ρ= P RT

 Where

R=Specific gas constant=R'/M  Put (a) in above equation

ρ= 1 RTPo

(

T To

)

δ δ −1  Put in (1)

(24)

dP dz=−M g∗1 RT Po

(

T To

)

δ δ −1 dP=−Mg∗1 RT Po

(

T To

)

δ δ−1dz  Put (2) in above Po To δ δ−1δ δ−1T 1 δ −1dT =−Mg∗1 RT Po

(

T To

)

δ δ−1dz dT dz= −δ−1 δM g RPo PoTo δ δ−1 To δ δ−1T δ δ −1 T∗T 1 δ−1 dT dz= −δ−1 δM g RPo PoTo δ δ−1 To δ δ−1T δ δ −1 T δ δ−1 dT dz= −δ δ−1M g R Proved

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:

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

(25)

d (mU )tank dt +∆ ( Hm)=0 d (mU )tank dt +H '' m''H'm'=0  Since

 Tank is filled with gas from an entrance line, but no gas is being escaped out,  Therefore, d (mU )tank dt +0−H 'm'=0 d (mU )tank dtH 'm'=0→(1)

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

m'=d mtank

dt → (2)

 Combining equation (1) & (2)

d(mU)tank dtH 'd mtank dt =0 1 dt

{

d (mU )tankH '

d mtank

}

=0 d (mU )tank=H '

d mtank

 Integrating on both sides

m1 m2 d (mU )tank=H '

m1 m2

d mtank ∆ (mU )tank=H '

(

m2−m1

)

m2U2−m1U1=H'

(

m2−m1

)

 Because mass in the tank initially is zero, therefore

m1=0

m2U2=H'm2

U2=H'→(3)  We know that

(26)

U=CVT U2=CVT2→( a)

 Also

H'

=CPT'→(b )  Put (a) & (b) in (3)

CVT =CPT' T =

CP

CV

T'

 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 T1=25 C. o =298 K P1=101.33 kPa T2=45 C. o =318 K P2=1500 kPa Heat lost =Q=? CP=7 2R CV= 5 2R

Solution:

 According to first law of thermodynamics

∆ U =Q+W →(1)

(27)

 Also, we know that

W=−P ∆ V →(b )

 Put (a) & (b) in (1)

∆ H−P ∆V −V ∆ P=Q−P ∆ V ∆ H−V ∆ P=Q→ (2)  Also, we have ∆ H=nCP∆T ∆ H=nCP

(

T2T1

)

 Put in (2) n CP

(

T2T1

)

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,

P1V =n1R T1 n1=P1V

R T1

 The final number of moles of gas at temperature T1 are P2V =n2R T1 n2=

P2V RT1

 Now, Applying molar balance

n=n1n2 n=P1V R T1 −P2V R T1 n=

(

P1−P2

)

V R T1  Put in (3)

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(

P1P2

)

V R T1 CP

(

T2T1

)

V ∆ P=Q

(

P1P2

)

V R T1 ∗7 2 R∗

(

T2−T1

)

V ∆ P=Q

(

P1P2

)

V T1 ∗7 2 ∗

(

T2−T1

)

V

(

P2−P1

)

=Q (101.33−1500 )kPa∗0.1 m3 298 K ∗7 2 ∗(318−298) K −0.1 m 3(1500−101.33 )kPa=Q Q=−172.717 m3 kPa∗1 kN 1 kPa∗m2∗1 kJ 1 kNm Q=−172.717 kJ Answer

Problem 3.17:

A rigid, no conducting tank with a volume of 4 m3 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: Volume of thetank=V1=4 m3 V2= V1∗1 3 = 4 3m 3 Pressure=P 2=6 ¯¿ Temperature=T1=100 Co. V3= V1∗2 3 = 8 3m 3

Solution:

(a)

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 According to first law of thermodynamics

∆ U =Q+W

 Since

No work is done & no heat is transferred

 Therefore

Q=W =0

∆ U =0 mCV∆ T =0 ∆ T =0 T2T1=0 T2=T1 T2=100℃ Answer

No, process is not reversible

(b)

 Since

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

W=−R T2lnV2

V1  As, for an ideal gas

P2V2=R T2 W=−P2V2lnV2 V1 W=−6 ¯¿4 3m 3 ln 4 3∗4 W =8.788¯¿ m3∗101325 N 1.01325¯¿m2 ∗1 kJ 1000 Nm 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 Cp = (7/2) R and CV = (5/2) R

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Given Data: T1=30 C0. T1=303.15 K P1=100 kPa Q=? W=? ∆ U =? ∆ H=? CP= 7 2R CV= 5 2R

So

lution:

(a) P2=500 kPa

1) Adiabatic Compression from point 1 to point 2

Q12=0

 Now, from first law of thermodynamics,

∆ U12=Q12+W12 ∆ U12=W12 W12=∆U12=CV∆ T12 W12=∆U12= 5 2R

(

T2−T1

)

→ (1)  For ‘T2’  We know that T2 T1=

(

P2 P1

)

γ −1 γ T 2=T1

(

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) W12=∆U12=5 2∗8.314 J mol∗K (480.13−303.15) K∗1 kJ 1000 J W12=∆ U12=3.679 kJ mol  Also, we have ∆ H12=CP

(

T2−T1

)

∆ H12= 7 2∗8.314 J mol∗K (480.13−303.15) K∗1 kJ 1000 J ∆ H12=5.15 kJ mol

2) Cooling at constant pressure from point 2 to point 3

 Therefore at constant pressure we have,

Q23=∆ H23=CP∆T23 Q23=∆ H23=

7

(31)

 Here T3=303.15 K Q23=∆ H23=7 2∗8.314 J mol∗K (303.15−480.13) K∗1 kJ 1000 J Q23=∆ H23=−5.15 kJ mol  Also, we have ∆ U23=CV

(

T3T2

)

∆ U23=5 2∗8.314 J mol∗K (303.15−480.13 ) K∗1 kJ 1000 J ∆ U23=−3.679 kJ mol

 Now, from first law of thermodynamics,

∆ U23=Q23+W23 W23=∆ U23−Q23 W23=−3.679+5.15 W23=1.471

kJ mol

3) Isothermal expansion from point 3 to point 1

 Since for an isothermal process temperature remains constant  Therefore,

∆ U31=∆ H31=0

 Here

P3=P2=500 kPa

 For an Isothermal process we have

W31=−R T3lnP3 P1W31=−8.314 J mol∗K∗303.15 K∗ln 500 100∗1 kJ 1000 J W31=−4.056 kJ mol

(32)

∆ U31=Q31+W31

0=Q31+W31 Q31=−W31 Q31=4.056 kJ

mol

 For the complete cycle,

Q=Q12+Q23+Q31 Q=0−5.15+4.056 Q=−1.094molkJ Answer W=W12+W23+W31 W=3.679+1.471−4.056 W=1.094 kJ mol Answer ∆ H=∆ H12+∆ H23+∆ H31 ∆ H=5.15−5.15+0 ∆ H=0 kJ mol Answer ∆ U =∆ U12+∆ U23+∆ U31 ∆ U =3.679−3.679+0 ∆ U =0 kJ mol Answer (b)

If each step that is 80% accomplishes the same change of state then values of ∆ U & ∆ H will remain same as in part (a) but values of Q & W will change.

1. Adiabatic Compression from point 1 to point 2

W12=W12 0.8 W12= 3.679 0.8 W12=4.598 kJ mol

 According to first law of thermodynamics

∆ U12=Q12+W12 3.679 kJ mol=Q12+4.598 kJ mol Q12=3.679 kJ mol−4.598 kJ mol Q12=−0.92 kJ mol

(33)

W23=W23 0.8 W23= 1.471 0.8 W23=1.839 kJ mol

 According to first law of thermodynamics

∆ U23=Q23+W23 −3.679 kJ mol=Q23+1.839 kJ mol Q23=−3.679 kJ mol−1.839 kJ mol Q23=−5.518 kJ mol

3. Isothermal expansion from point 3 to point 1

 Since initial step can no longer be adiabatic , therefore

W31=W31∗0.8 W31=−4.056 kJ

mol∗0.8 W31=3.245

kJ mol

 According to first law of thermodynamics

∆ U31=Q31+W31 Q31=−W31+0

Q31=3.245 kJ

mol

 For the complete cycle,

Q=Q12+Q23+Q31 Q=−0.92−5.518+3.245 Q=−3.193 kJ

mol Answer W=W12+W23+W31 W=4.598+1.839−3.245 W=3.192 kJ

mol Answer

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.

(34)

V1=1 m 3 T 1=600 K P1=1000 kPa V2=5 V1 V2=5 m 3 C P=21 J mol K CV=? T2=? P2=? W=?

Solution:

 We know that, CPCV=R CV=CPR CV=(21−8.314)mol∗KJ CV=12.686mol∗KJ  As γ=CP CV γ=1.6554 (a)

 Since, for an isothermal process

 Temperature remains constant, therefore

T2=T1=600 K Answer

 For an ideal gas we have

P1V1 T1 =P2V2 T2 P2= P1V1 T1 ∗T2 V2 P2= 1000 kPa∗1 m3 600 K ∗600 K 5 m3 P2=200 kPa Answer

 We know that, for an isothermal process

W=−R T1lnV2

V1  Since

P1V1=R T1

(35)

W=−P1V1lnV2 V1 W=−1000 kPa∗1 m3 ln 5 1∗N Pa∗m2∗J Nm W=−1609.43 kJ Answer (b)

 We know that, for an adiabatic process

P1V1γ=P2V2γ P2=P1

(

V1 V2

)

γ P2=1000 kPa∗

(

1 5

)

1.6554 P2=69.65 kPa Answer

 For an ideal gas we have

P1V1 T1 =P2V2 T2 T2=P2V2 P1V1 ∗T1 T2=69.65 kPa∗5 m 3 1000 kPa∗1 m3∗600 K T2=208.95 K Answer  For an adiabatic process work done is

W=P2V2−P1V1 γ−1 W=(69.65∗5−1000∗1) kPa∗m 3 1.6554−1 N Pa∗m2∗J Nm 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=−PrdV ∆ U =W=−Pr

(

V2V1

)

∆ U =W=−100(5−1) kPa∗m3∗N Pa∗m2 ∗J Nm ∆ U =−400 kJ n CV∆T =−400 kJ n CV

(

T2T1

)

=−400 kJ T2=−400 kJ n CV +T1→(1)

(36)

 For an ideal gas we have, P1V1=nR T1 n=P1V1 R T1 n= 1000 kPa∗1 m3 ∗mol∗K 8.314 J∗600 KkN kPa∗m2 ∗kJ kNm n=0.2005 mol  Put in (1) T2= −400 kJ∗mol∗K 0.2005 mol∗12.686 J∗1000 J 1 kJ +600 K T2=−157.26 K +600 K T2=442.74 K Answer

 For an ideal gas we have

P1V1 T1 =P2V2 T2 P2= P1V1 T1 ∗T2 V2 P2= 1000 kPa∗1 m3 600 K ∗442.74 K 5 m3 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 CP = (7/2) R and CV = (5/2) R, calculate W, Q,

∆ U , and ∆ H Given Data:

Mole of air=n=1mol Initial Temperature=T1=150 C. 0 =423.15 K Initial pressure=P1=8 ¯¿ Finaltemperature=T3=50 C. 0 =323.15 K Final pressure=P3=3 ¯¿ CP=7 2R CV= 5 2R

Solution:

 Since process is reversible

 Two different steps are used in this case to reach final state of the air.

(37)

T1=T2

 Therefore

∆ U12=∆ H12=0

 For an isothermal process we have

W12=R T1lnV1 V2  As V2=V3 W12=R T1ln V1 V3 →(1)  We know that P1V1 T1 =P3V3 T3 V1 V3 =P3¿T1 T3∗P1 W12=R T1lnP1¿T3 T1∗P3 W12= 8.314 J∗423.15 K mol∗K ∗1 kJ 1000 J ∗ln 3∗423.15 8∗323.15 W12=−2.502 kJ mol

 According to first law of thermodynamics

∆ U12=Q12+W12 0=Q12+W12 Q12=−W12 Q12=2.502 kJ

mol

Step 23:

 For step 23 volume is constant,  Therefore,

W23=0

 According to first law of thermodynamics

∆ U23=Q23+W23 ∆ U23=Q23+0 Q23=∆ U23 Q23=∆ U23=CV∆ T Q23=∆ U23=CV

(

T3T2

)

Q23=∆ U23=5

(38)

¿❑ ¿❑ ❑ ¿8.314❑❑ ¿❑ ❑ () K❑❑ ¿❑❑ ¿−2.0785❑ ❑  W e know that ∆ H23=CP∆ T ∆ H23=CP

(

T3T2

)

∆ H 23= 7 2∗8.314 J mol∗K∗1 kJ 1000 J (423.15−323.15) K ∆ H23=2.91 kJ mol

 For the complete cycle,

Work=W =W12+W23 W=(−2.502+0 ) kJ mol W=−2.502 kJ mol Answe r Q=Q12+Q23 Q=(2.502−2.0785) kJ mol Q=0.424 kJ mol Answer

∆ U =∆ U12+∆ U23 ∆ U =(0−2.0785 )molkJ ∆ U =−2.0785molkJ Answe r

∆ H=∆ H12+∆ H23 ∆ H=(0−2.91) kJ

mol ∆ H=−2.91 kJ

mol Answe r

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

(39)

Temperature=T1=150 C0. =423.15 K Velocity=u1=2.5 m sec T2=? u2=50 m sec CP= 7 2R

Molecualr weight of Nitrogen=28 g mol

Solution:

 Applying energy balance for steady state flow process

∆ H +∆ u 2 2 +g ∆ z=Q+WS  Since ∆ z=WS=Q=0  Therefore, ∆ H +∆ u2 2 =0 CP∆ T =∆u2 2 CP

(

T2−T1

)

= −u2 2 −u1 2 2 T2= −u22−u12 2CP +T1 T2= −

(

502−2.52

)

∗2∗m2∗mol∗K 2∗7∗8.314 J∗sec2 ∗28 g Nitrogen 1 mol NitrogenJ N∗mN∗sec 2 kg∗m ∗1 kg 1000 g +423.15 K T2=−1.199 K +423.15 K T2=421.95 K T2=(421.95−273.15) C. 0 T2=148.8 C. 0 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

(40)

Calculate Q, W, ∆ U ∧∆ H 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: T1=30 C. 0 T 1=(30+273.15) K T1=303.15 K P1=1 ¯¿ T2=130 C. 0 T 3=(130+273.15) K T3=403.15 K P3=10 ¯¿ Q=? W=? ∆ U =? ∆ H=?

Solution:

CP= 7 2R CV= 5 2R

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

∆ U =∆ U12=∆U23=CV∆T ∆ U =∆ U12=∆U23=5

2R

(

T3−T1

)

∆ U =∆ U12=∆U23=52∗8.314mol∗KJ (403.15−303.15)K∗1 kJ1000 J

∆ U =∆ U12=∆U23=2.079 kJ

mol→(a) Answe r

 Now ∆ H=∆ H12=∆ H23=CP∆ T ∆ H=∆ H12=∆ H23=7 2R

(

T2−T1

)

∆ H=∆ H12=∆ H23=7 2∗8.314 J mol∗K(403.15−303.15) K∗1 kJ 1000 J ∆ H=∆ H12=∆ H23=2.91 kJ mol→ (b) Answe r

(41)

(a)

Step 12:

 For step “12” volume is constant  Therefore

W12=0

 Here

T2=T3

 According to first law of thermodynamics

∆ U12=Q12+W12 ∆ U12=Q12 Q12=∆ U12=CV∆ T Q12=∆ U12=2.079 kJ mol

[

¿(a)

]

 Also we have  ∆ H12=2.91 kJ mol

[

¿(b)

]

Step 23:

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

∆ U23=∆ H23=0

 Here

T2=T3

 Now, intermediate pressure can be calculated as

P1 T1 =P2 T2 P2=P1 T1 ∗T2 1 ¯¿ 303.15 K∗403.15 K P2=¿ P2=1.329 b ar

 For an isothermal process we have

W23=R T2lnP3 P2 W23=8.314 J mol∗K∗403.15 K∗1 kJ 1000 J∗ln 10 1.329

(42)

W23=6.764

kJ mol

 According to first law of thermodynamics

∆ U23=Q23+W23 0=Q23+W23 Q23=−W23 Q23=−6.764 kJ

mol

 For the complete cycle,

Work=W =W12+W23 W=(0+6.764 ) kJ mol W=6.764 kJ mol Answe r Q=Q12+Q23 Q=(2.079−6.764) kJ mol Q=−4.685 kJ mol Answe r ∆ U =∆ U12+∆ U23 ∆ U =(2.079+0) kJ mol ∆ U =2.079 kJ molAnswe r ∆ H=∆ H12+∆ H23 ∆ H=(2.91+0) kJ mol ∆ H=2.91 kJ mol Answe r (b)

Step 12:

 For step “12” volume is constant

 Therefore, at constant pressure we have

Q12¿∆ H12=2.91 kJ

mol

[

¿(b)

]

 Also,

∆ U12=2.079 kJ

mol

[

¿(a)

]

 According to first law of thermodynamics

∆ U12=Q12+W12 W12=∆U12Q12 W12=(2.079−2.91)

kJ

mol W12=−0.831

kJ mol

(43)

Step 23:

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

∆ U23=∆ H23=0

 Here

T2=T3∧P1=P2

 For an isothermal process we have

W23=R T2lnP3 P2 W23=8.314 J mol∗K∗403.15 K∗1 kJ 1000 J ∗ln 10 1 W23=7.718 kJ mol

 According to first law of thermodynamics

∆ U23=Q23+W23 0=Q23+W23 Q23=−W23 Q23=−7.718

kJ mol

 For the complete cycle,

Work=W =W12+W23 W=(−0.831+7.718) kJ mol W=6.887 kJ mol Answer Q=Q12+Q23 Q=(2.91−7.718) kJ mol Q=−4.808 kJ mol Answe r ∆ U =∆ U12+∆ U23 ∆ U =(2.079+0) kJ mol ∆ U =2.079 kJ molAnswe r ∆ H=∆ H12+∆ H23 ∆ H=(2.91+0) kJ mol ∆ H=2.91 kJ mol Answe r (c)

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

∆ U12=∆ H12=0  Here

(44)

P2=P3

 For an isothermal process we have

W12=R T1lnP2 P1 W12=8.314 J mol∗K∗303.15 K∗1 kJ 1000 J ∗ln 10 1 W12=5.8034 kJ mol

 According to first law of thermodynamics

∆ U12=Q12+W12 0=Q12+W12 Q12=−W12 Q12=−5.8034molkJ

Step 23:

 For step “23” volume is constant

 Therefore, at constant pressure we have

Q23=∆ H23=2.91 kJ mol

[

¿(b)

]

 Here T2=T3  Now ∆ U23=2.079 kJ mol

[

¿(a)

]

 According to first law of thermodynamics

∆ U23=Q23+W23 W23=∆ U23Q23 W23=(2.079−2.91) kJ

mol W23=−0.831

kJ mol

 For the complete cycle,

(45)

Q=Q12+Q23 Q=(−5.8034+2.91) kJ mol Q=−2.894 kJ molAnswe r ∆ U =∆ U12+∆ U23 ∆ U =(0+2.079) kJ mol ∆ U =2.079 kJ molAnswe r ∆ H=∆ H12+∆ H23 ∆ H=(0+2.91)molkJ ∆ H=2.91molkJ Answe r

Solution:

CP=5

2R CV=

3 2R

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

∆ U =∆ U12=∆U23=CV∆T ∆ U =∆ U12=∆U23=3 2R

(

T3−T1

)

∆ U =∆ U12=∆U23=3 2∗8.314 J mol∗K(403.15−303.15) K∗1 kJ 1000 J ∆ U =∆ U12=∆U23=1.247 kJ

mol→(a) Answe r

 Now ∆ H=∆ H12=∆ H23=CP∆ T ∆ H=∆ H12=∆ H23=5 2R

(

T2−T1

)

∆ H=∆ H12=∆ H23=5 2∗8.314 J mol∗K(403.15−303.15) K∗1 kJ 1000 J ∆ H=∆ H12=∆ H23=2.079 kJ mol→ (b) Answer

(46)

(a)

Step 12:

 For step “12” volume is constant  Therefore

W12=0

 Here

T2=T3

 According to first law of thermodynamics

∆ U12=Q12+W12 ∆ U12=Q12 Q12=∆ U12=CV∆ T Q12=∆ U12=1.247 kJ mol

[

¿(a)

]

 Also we have ∆ H12=2.079 kJ mol

[

¿(b)

]

Step 23:

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

∆ U23=∆ H23=0

 Here

T2=T3

 Now, intermediate pressure can be calculated as

P1 T1 =P2 T2 P2=P1 T1 ∗T2 1 ¯¿ 303.15 K∗403.15 K P2=¿ P2=1.329 b ar

(47)

W23=R T2lnP3

P2

W23=8.314mol∗KJ ∗403.15K∗1 kJ1000 J∗ln1.32910

W23=6.764 kJ

mol

 According to first law of thermodynamics

∆ U23=Q23+W23 0=Q23+W23 Q23=−W23 Q23=−6.764

kJ mol

 For the complete cycle,

Work=W =W12+W23 W=(0+6.764 ) kJ mol W=6.764 kJ mol Answe r Q=Q12+Q23 Q=(1.247−6.764 ) kJ mol Q=−5.516 kJ molAnswe r ∆ U =∆ U12+∆ U23 ∆ U =(1.247+0) kJ mol ∆ U =1.247 kJ mol Answer ∆ H=∆ H12+∆ H23 ∆ H=(2.079+0) kJ mol ∆ H=2.079 kJ mol Answe r (b)

Step 12:

 For step “12” volume is constant

 Therefore, at constant pressure we have

Q12¿∆ H12=2.079 kJ

mol

[

¿(b)

]

 Also,

∆ U12=1.247 kJ

mol

[

¿(a)

]

Figure

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References

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