Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
Question
Chapter 4, Problem 4.1P

(a)

Interpretation Introduction

Interpretation:

Heat transferred ( Q ) by SO2 on rise of temperature from 2000C to 11000C

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  Q=nΔH=RT0TΔCPRdT......(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)......(2)

Where τ=TT0

(a)

Expert Solution
Check Mark

Answer to Problem 4.1P

  Q=470034.765J

Explanation of Solution

Given information:

  T1=2000CT1=2000C+273.15=473.15K

  T2= 11000CT2= 11000C+273.15=1373.15K

  n=10mol

Values of above constants for SO2 in equation (2) are given in appendix C table C.1 and noted down below:

  iA103B106C105DSO25.6990.80101.015

  gas constant R in SI unit is 8.314JmolK

  τ=TT0=1373.15473.15=2.902

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)473.151373.15ΔCPRdT=5.699×473.15×(2.9021)+0.801×1032473.152(2.90221)+03473.153(2.90231)+1.015×105473.15(2.90212.902)473.151373.15ΔCPRdT=5653.533K

Now from equation (1),

  Q=RT0TΔ C PRdTQ=8.314JmolK×5653.533KQ=47003.4765Jmol 

Now,

  n=10molQ=47003.4765Jmol×10mol Q=470034.765J

(b)

Interpretation Introduction

Interpretation:

Heat transferred ( Q ) by propane on rise of temperature from 250 to 12000C

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  Q=nΔH=RT0TΔCPRdT......(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)......(2)

Where τ=TT0

(b)

Expert Solution
Check Mark

Answer to Problem 4.1P

  1942106.848J

Explanation of Solution

Given information:

  T1=2000CT1=2500C+273.15=523.15K

  T2= 11000CT2= 12000C+273.15=1473.15K

  n=12mol

Values of above constants for propane in equation (2) are given in appendix C table C.1 and noted down below:

  iA103B106C105DPropane1.21328.7858.8240

  gas constant R in SI unit is 8.314JmolK

  τ=TT0=1473.15523.15=2.816

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)T0TΔCPRdT=1.213×523.15×(2.8161)+28.785×1032523.152(2.81621)+8.824×1063×523.153(2.81631)+0523.15(2.81612.816)T0TΔCPRdT=19466.23013K

Now from equation (1),

  Q=RT0TΔ C PRdTQ=8.314JmolK×19466.23013KQ=161842.2373Jmol 

Now,

  n=12molQ=161842.2373Jmol×12mol Q=1942106.848J

(c)

Interpretation Introduction

Interpretation:

Heat transferred ( Q ) by methane on rise of temperature from 100 to 8000C

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  Q=nΔH=RT0TΔCPRdT......(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)......(2)

Where τ=TT0

(c)

Expert Solution
Check Mark

Answer to Problem 4.1P

  51153886.48J

Explanation of Solution

Given information:

  T1=1000CT1=1000C+273.15=373.15K

  T2= 8000CT2= 8000C+273.15=1073.15K

  m=20kg

Values of above constants for methane in equation (2) are given in appendix C table C.1 and noted down below:

  iA103B106C105Dmethane1.7029.0812.1640

  gas constant R in SI unit is 8.314JmolK

  τ=TT0=1073.15373.15=2.876

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)T0TΔCPRdT=1.702×373.15×(2.8761)+9.081×1032373.152(2.87621)+2.164×1063×373.153(2.87631)+0373.15(2.87612.876)T0TΔCPRdT=4934.49807K

Now from equation (1),

  Q=RT0TΔ C PRdTQ=8.314JmolK×4934.49807KQ=41025.41696Jmol 

Now,

  m=20kgn=20000g16.04g/mol=1246.883molQ=41025.41696Jmol×1246.883mol Q=51153886.48J

(d)

Interpretation Introduction

Interpretation:

Heat transferred ( Q ) by n butane on rise of temperature from 150 to 11500C

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  Q=nΔH=RT0TΔCPRdT......(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)......(2)

Where τ=TT0

(d)

Expert Solution
Check Mark

Answer to Problem 4.1P

  2107019.862J

Explanation of Solution

Given information:

  T1=1500CT1=1500C+273.15=423.15K

  T2= 11500CT2= 11500C+273.15=1423.15K

  n=10mol

Values of above constants for n butane in equation (2) are given in appendix C table C.1 and noted down below:

  iA103B106C105Dn-butane1.93536.91511.4020

  gas constant R in SI unit is 8.314JmolK

  τ=TT0=1423.15423.15=3.363

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)T0TΔCPRdT=1.935×423.15×(3.3631)+36.915×1032423.152(3.36321)+11.402×1063×423.153(3.36331)+0423.15(3.36313.363)T0TΔCPRdT=25343.03418K

Now from equation (1),

  Q=RT0TΔ C PRdTQ=8.314JmolK×25343.03418KQ=210701.9862Jmol 

Now,

  n=10molQ=210701.9862Jmol×10mol Q=2107019.862J

(e)

Interpretation Introduction

Interpretation:

Heat transferred ( Q ) by air on rise of temperature from 25 to 10000C

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  Q=nΔH=RT0TΔCPRdT......(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)......(2)

Where τ=TT0

(e)

Expert Solution
Check Mark

Answer to Problem 4.1P

  1064048522J

Explanation of Solution

Given information:

  T1=250CT1=250C+273.15=298.15K

  T2= 10000CT2= 10000C+273.15=1273.15K

  m=1000kg

Values of above constants for air in equation (2) are given in appendix C table C.1 and noted down below:

  iA103B106C105Dair3.3550.57500.016

  gas constant R in SI unit is 8.314JmolK

  τ=TT0=1073.15373.15=4.27017

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)T0TΔCPRdT=3.355×298.15×(4.27011)+0.575×1032298.152(4.270121)+03×298.153(4.270131)+0.016298.15×(4.270114.2701)T0TΔCPRdT=3711.4995K

Now from equation (1),

  Q=RT0TΔ C PRdTQ=8.314JmolK×3711.4995KQ=30857.40715Jmol 

Now,

  m=1000kgn=1000×1000g29g/mol=34482.75862molQ=30857.40715Jmol ×34482.75862molQ=1064048522J

(f)

Interpretation Introduction

Interpretation:

Heat transferred ( Q ) by ammonia on rise of temperature from 100 to 8000C

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  Q=nΔH=RT0TΔCPRdT......(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)......(2)

Where τ=TT0

(f)

Expert Solution
Check Mark

Answer to Problem 4.1P

  665290.495J

Explanation of Solution

Given information:

  T1=1000CT1=1000C+273.15=373.15K

  T2= 8000CT2= 8000C+273.15=1073.15K

  n=20mol

Values of above constants for ammonia in equation (2) are given in appendix C table C.1 and noted down below:

  iA103B106C105Dammonia3.5783.02000.186

  gas constant R in SI unit is 8.314JmolK

  τ=TT0=1073.15373.15=2.876

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)T0TΔCPRdT=3.578×373.15×(2.8761)+3.020×1032373.152(2.87621)+03×373.153(2.87631)+0.186×105373.15(2.87612.876)T0TΔCPRdT=4001.025K

Now from equation (1),

  Q=RT0TΔ C PRdTQ=8.314JmolK×4001.025KQ=33264.525Jmol 

Now,

  n=20molQ=33264.525Jmol ×20mol Q=665290.495J

(g)

Interpretation Introduction

Interpretation:

Heat transferred ( Q ) by water on rise of temperature from 150 to 3000C

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  Q=nΔH=RT0TΔCPRdT......(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)......(2)

Where τ=TT0

(g)

Expert Solution
Check Mark

Answer to Problem 4.1P

  52983.298J

Explanation of Solution

Given information:

  T1=1500CT1=1500C+273.15=423.15K

  T2= 3000CT2= 3000C+273.15=573.15K

  n=10mol

Values of above constants for water in equation (2) are given in appendix C table C.1 and noted down below:

  iA103B106C105Dwater3.4701.45000.121

  gas constant R in SI unit is 8.314JmolK

  τ=TT0=573.15423.15=1.355

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)T0TΔCPRdT=3.470×423.15×(1.3551)+1.45×1032423.152(1.35521)+03×423.153(1.35531)+0.121×105423.15(1.35511.355)T0TΔCPRdT=637.278K

Now from equation (1),

  Q=RT0TΔ C PRdTQ=8.314JmolK×637.278KQ=5298.3298Jmol 

Now,

  n=10molQ=5298.3298Jmol ×10mol Q=52983.298J

(h)

Interpretation Introduction

Interpretation:

Heat transferred ( Q ) by chlorine on rise of temperature from 2000C to 5000C

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  Q=nΔH=RT0TΔCPRdT......(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)......(2)

Where τ=TT0

(h)

Expert Solution
Check Mark

Answer to Problem 4.1P

  54910.861J

Explanation of Solution

Given information:

  T1=2000CT1=2000C+273.15=473.15K

  T2= 5000CT2= 5000C+273.15=773.15K

  n=5mol

Values of above constants for chlorine in equation (2) are given in appendix C table C.1 and noted down below:

  iA103B106C105DChlorine4.4420.08900.344

  gas constant R in SI unit is 8.314JmolK

  τ=TT0=773.15473.15=1.634

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)473.15773.15ΔCPRdT=4.442×473.15×(1.6341)+0.089×1032473.152(1.63421)+03473.153(1.63431)+0.344×105473.15(1.63411.634)473.15773.15ΔCPRdT=1320.925K

Now from equation (1),

  Q=RT0TΔ C PRdTQ=8.314JmolK×1320.925KQ=10982.172Jmol 

Now,

  n=5molQ=10982.172Jmol ×5mol Q=54910.861J

(i)

Interpretation Introduction

Interpretation:

Heat transferred ( Q ) by ethylbenzene on rise of temperature from 300 to 7000C

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  Q=nΔH=RT0TΔCPRdT......(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)......(2)

Where τ=TT0

(i)

Expert Solution
Check Mark

Answer to Problem 4.1P

  10228945.09J

Explanation of Solution

Given information:

  T1=3000CT1=3000C+273.15=573.15K

  T2= 7000CT2= 7000C+273.15=973.15K

  m=10kg

Values of above constants for ethylbenzene in equation (2) are given in appendix C table C.1 and noted down below:

  iA103B106C105Dethylbenzene1.12455.38018.4760

  gas constant R in SI unit is 8.314JmolK

  τ=TT0=973.15573.15=1.698

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)T0TΔCPRdT=1.124×573.15×(1.6981)+55.38×1032573.152(1.69821)+18.476×1063×573.153(1.69831)+0573.15(1.69811.698)T0TΔCPRdT=13062.39K

Now from equation (1),

  Q=RT0TΔ C PRdTQ=8.314JmolK×13062.39KQ=108600.71Jmol 

Now,

  m=10kgn=10000g106.17g/mol=94.189molQ=108600.71Jmol×94.189molQ=10228945.09J

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