INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<
INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<
8th Edition
ISBN: 9781260940961
Author: SMITH
Publisher: INTER MCG
Question
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Chapter 5, Problem 5.27P

(a)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  dS=CPdTT(VT)PdPat constant pressure

  ΔS=RT1T2CPRdTT......(1)

(a)

Expert Solution
Check Mark

Answer to Problem 5.27P

  ΔS=548.12JK

Explanation of Solution

Given information:

It is given that

  n=10mol of SO2

  T0=200C+273.15=473.15K

  T=1100C+273.15=1373.15K

From equation (1)

  ΔS=RT1T2CPRdTT

Where

  T0TΔCPRdTT=[A+[BT0+(CT02+Dτ2T02)(τ+12)](τ1lnτ)]lnτ......(2)

And

  τ=TT0

The 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

Hence

  τ=TT0τ=1373.15K473.15K=2.902

and

  T0TΔCPRdTT=[A+[BT0+(CT02+D τ 2 T 0 2)(τ+12)](τ1lnτ)]lnτT0TΔCPRdTT=[5.699+[0.801×103×473.15+(0×473.152+1.015× 105 2.9022× 473.152)(2.902+12)](2.9021ln2.902)]ln2.902T0TΔCPRdTT=6.59279

Therefore, entropy change is

  ΔS=RT1T2CPRdTTΔS=8.314Jmol K×6.59279ΔS=54.8124Jmol K

For n=10mol of SO2

  ΔS=54.8124Jmol K×10molΔS=548.12JK

(b)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  dS=CPdTT(VT)PdPat constant pressure

  ΔS=RT1T2CPRdTT......(1)

(b)

Expert Solution
Check Mark

Answer to Problem 5.27P

  ΔS=2018.79JK

Explanation of Solution

Given information:

It is given that

  n=12mol of Propane

  T0=250C+273.15=523.15K

  T=1200C+273.15=1473.15K

From equation (1)

  ΔS=RT1T2CPRdTT

Where

  T0TΔCPRdTT=[A+[BT0+(CT02+Dτ2T02)(τ+12)](τ1lnτ)]lnτ......(2)

And

  τ=TT0

The 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

Hence

  τ=TT0τ=1473.15K523.15K=2.816

and

  T0TΔCPRdTT=[A+[BT0+(CT02+D τ 2 T 0 2)(τ+12)](τ1lnτ)]lnτT0TΔCPRdTT=[1.213+[28.785×103×523.15+(8.824×106×523.152)(2.816+12)](2.8161ln2.816)]ln2.816T0TΔCPRdTT=20.2349

Therefore, entropy change is

  ΔS=RT1T2CPRdTTΔS=8.314Jmol K×20.2349ΔS=168.23Jmol K

For n=12mol of Propane

  ΔS=168.23Jmol K×12molΔS=2018.79JK

(c)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  dS=CPdTT(VT)PdPat constant pressure

  ΔS=RT1T2CPRdTT......(1)

(c)

Expert Solution
Check Mark

Answer to Problem 5.27P

  ΔS=73168.8JK

Explanation of Solution

Given information:

It is given that

  n=20kg of methane

  T0=100C+273.15=373.15K

  T=800C+273.15=1073.15K

  M=16.043gmmol of methane

From equation (1)

  ΔS=RT1T2CPRdTT

Where

  T0TΔCPRdTT=[A+[BT0+(CT02+Dτ2T02)(τ+12)](τ1lnτ)]lnτ......(2)

And

  τ=TT0

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

  iA103B106C105DMathane1.7029.0812.1640

Hence

  τ=TT0τ=1073.15K373.15K=2.876

and

  T0TΔCPRdTT=[A+[BT0+(CT02+D τ 2 T 0 2)(τ+12)](τ1lnτ)]lnτT0TΔCPRdTT=[1.702+[9.081×103×373.15+(2.164×106×373.152)(2.876+12)](2.8761ln2.876)]ln2.876T0TΔCPRdTT=7.05946

Therefore, entropy change is

  ΔS=RT1T2CPRdTTΔS=8.314Jmol K×7.05946ΔS=58.69Jmol K

For n=20kg of methane and M=16.043gmmol of methane

  ΔS=58.69Jmol K×20kg of methane×1000gm1kg×mol16.043gmΔS=73168.8JK

(d)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  dS=CPdTT(VT)PdPat constant pressure

  ΔS=RT1T2CPRdTT......(1)

(d)

Expert Solution
Check Mark

Answer to Problem 5.27P

  ΔS=2388.95JK

Explanation of Solution

Given information:

It is given that

  n=10 mol of n-butane

  T0=150C+273.15=423.15K

  T=1150C+273.15=1423.15K

From equation (1)

  ΔS=RT1T2CPRdTT

Where

  T0TΔCPRdTT=[A+[BT0+(CT02+Dτ2T02)(τ+12)](τ1lnτ)]lnτ......(2)

And

  τ=TT0

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

  iA103B106C105DnButane1.93536.91511.4020

Hence

  τ=TT0τ=1423.15K423.15K=3.363

and

  T0TΔCPRdTT=[A+[BT0+(CT02+D τ 2 T 0 2)(τ+12)](τ1lnτ)]lnτT0TΔCPRdTT=[1.935+[36.915×103×423.15+(11.402×106×423.152)(3.363+12)](3.3631ln3.363)]ln3.363T0TΔCPRdTT=28.7341

Therefore, entropy change is

  ΔS=RT1T2CPRdTTΔS=8.314Jmol K×28.7341ΔS=238.895Jmol K

For n=10 mol of n-butane

  ΔS=238.895Jmol K×10 mol of n-butaneΔS=2388.95JK

(e)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  dS=CPdTT(VT)PdPat constant pressure

  ΔS=RT1T2CPRdTT......(1)

(e)

Expert Solution
Check Mark

Answer to Problem 5.27P

  ΔS=1562580.153JK

Explanation of Solution

Given information:

It is given that

  n=1000kg of air

  T0=25C+273.15=298.15K

  T=1000C+273.15=1273.15K

  M=28.851gmmol of air

From equation (1)

  ΔS=RT1T2CPRdTT

Where

  T0TΔCPRdTT=[A+[BT0+(CT02+Dτ2T02)(τ+12)](τ1lnτ)]lnτ......(2)

And

  τ=TT0

The 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

Hence

  τ=TT0τ=1273.15K298.15K=4.2702

and

  T0TΔCPRdTT=[A+[BT0+(CT02+D τ 2 T 0 2)(τ+12)](τ1lnτ)]lnτT0TΔCPRdTT=[3.355+[0.575×103×298.15+(0+0.016× 105 4.27022× 298.152)(4.2702+12)](4.27021ln4.2702)]ln4.2702T0TΔCPRdTT=5.42245

Therefore, entropy change is

  ΔS=RT1T2CPRdTTΔS=8.314Jmol K×5.42245ΔS=45.082Jmol K

For n=1000kg of air and M=28.851gmmol of air

  ΔS=45.082Jmol K×1000kg of air×1000gm1kg×mol28.851gmΔS=1562580.153JK

(f)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  dS=CPdTT(VT)PdPat constant pressure

  ΔS=RT1T2CPRdTT......(1)

(f)

Expert Solution
Check Mark

Answer to Problem 5.27P

  ΔS=970.272JK

Explanation of Solution

Given information:

It is given that

  n=20 mol of ammonia

  T0=100C+273.15=373.15K

  T=800C+273.15=1073.15K

From equation (1)

  ΔS=RT1T2CPRdTT

Where

  T0TΔCPRdTT=[A+[BT0+(CT02+Dτ2T02)(τ+12)](τ1lnτ)]lnτ......(2)

And

  τ=TT0

The 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

Hence

  τ=TT0τ=1073.15K373.15K=2.876

and

  T0TΔCPRdTT=[A+[BT0+(CT02+D τ 2 T 0 2)(τ+12)](τ1lnτ)]lnτT0TΔCPRdTT=[3.578+[3.020×103×373.15+(0+0.186× 105 2.8762× 373.152)(2.876+12)](2.8761ln2.876)]ln2.876T0TΔCPRdTT=5.83517

Therefore, entropy change is

  ΔS=RT1T2CPRdTTΔS=8.314Jmol K×5.83517ΔS=48.514Jmol K

For n=20 mol of ammonia

  ΔS=48.514Jmol K×20 mol of ammoniaΔS=970.272JK

(g)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  dS=CPdTT(VT)PdPat constant pressure

  ΔS=RT1T2CPRdTT......(1)

(g)

Expert Solution
Check Mark

Answer to Problem 5.27P

  ΔS=2388.95JK

Explanation of Solution

Given information:

It is given that

  n=10 mol of water

  T0=150C+273.15=423.15K

  T=300C+273.15=573.15K

From equation (1)

  ΔS=RT1T2CPRdTT

Where

  T0TΔCPRdTT=[A+[BT0+(CT02+Dτ2T02)(τ+12)](τ1lnτ)]lnτ......(2)

And

  τ=TT0

The 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

Hence

  τ=TT0τ=573.15K423.15K=1.354

and

  T0TΔCPRdTT=[A+[BT0+(CT02+D τ 2 T 0 2)(τ+12)](τ1lnτ)]lnτT0TΔCPRdTT=[3.47+[1.45×103×423.15+(0+0.121× 105 1.3542× 423.152)(1.354+12)](1.3541ln1.354)]ln1.354T0TΔCPRdTT=1.28419

Therefore, entropy change is

  ΔS=RT1T2CPRdTTΔS=8.314Jmol K×1.28419ΔS=10.677Jmol K

For n=10 mol of water

  ΔS=10.677Jmol K×10 mol of waterΔS=106.77JK

(h)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  dS=CPdTT(VT)PdPat constant pressure

  ΔS=RT1T2CPRdTT......(1)

(h)

Expert Solution
Check Mark

Answer to Problem 5.27P

  ΔS=89.783JK

Explanation of Solution

Given information:

It is given that

  n=5mol of Cl2

  T0=200C+273.15=473.15K

  T=500C+273.15=773.15K

From equation (1)

  ΔS=RT1T2CPRdTT

Where

  T0TΔCPRdTT=[A+[BT0+(CT02+Dτ2T02)(τ+12)](τ1lnτ)]lnτ......(2)

And

  τ=TT0

The Values of above constants for Cl2 in equation (2) are given in appendix C table C.1 and noted down below:

  iA103B106C105DChlorine4.4420.08900.344

Hence

  τ=TT0τ=773.15K473.15K=1.634

and

  T0TΔCPRdTT=[A+[BT0+(CT02+D τ 2 T 0 2)(τ+12)](τ1lnτ)]lnτT0TΔCPRdTT=[4.442+[0.089×103×473.15+(0+0.344× 105 1.6342× 473.152)(1.634+12)](1.6341ln1.634)]ln1.634T0TΔCPRdTT=2.15980

Therefore, entropy change is

  ΔS=RT1T2CPRdTTΔS=8.314Jmol K×2.15980ΔS=17.956Jmol K

For n=5mol of Cl2

  ΔS=17.956Jmol K×5mol of Cl2ΔS=89.783JK

(i)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  dS=CPdTT(VT)PdPat constant pressure

  ΔS=RT1T2CPRdTT......(1)

(i)

Expert Solution
Check Mark

Answer to Problem 5.27P

  ΔS=13340.52JK

Explanation of Solution

Given information:

It is given that

  n=10kg of ethylbenzene

  T0=300C+273.15=573.15K

  T=700C+273.15=973.15K

  M=106.167gmmol of ethylbenzene

From equation (1)

  ΔS=RT1T2CPRdTT

Where

  T0TΔCPRdTT=[A+[BT0+(CT02+Dτ2T02)(τ+12)](τ1lnτ)]lnτ......(2)

And

  τ=TT0

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

  iA103B106C105Dethylbenzene1.12455.3818.4760

Hence

  τ=TT0τ=973.15K573.15K=1.698

and

  T0TΔCPRdTT=[A+[BT0+(CT02+D τ 2 T 0 2)(τ+12)](τ1lnτ)]lnτT0TΔCPRdTT=[1.124+[55.38×103×573.15+(18.476×106×573.152)(1.698+12)](1.6981ln1.698)]ln1.698T0TΔCPRdTT=17.0354

Therefore, entropy change is

  ΔS=RT1T2CPRdTTΔS=8.314Jmol K×17.0354ΔS=141.632Jmol K

For n=10kg of ethylbenzene and M=106.167gmmol of ethylbenzene

  ΔS=141.632Jmol K×10kg of ethylbenzene×1000gm1kg×mol106.167gmΔS=13340.52JK

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