CHEMICAL ENGINEERING

Solutions Manual

FUNDAMENTALS OF

CHEMICAL ENGINEERING

THERMODYNAMICS 1e

KEVIN DAHM,

DONALD VISCO

Part 1: Solutions Manual (All Chapters Arranged

Reverse: 15-1)

Part 2: Answers to Selected Problems

Part 3: Expanded Solutions to Selected Examples

Part 4: Feedback to Food for Thought Questions

This is the Only Original and Complete Solutions Manual, all Other Files in The Market are Fake/Old/ Wrong Edition. 1 / 4

689 © 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 15: Synthesis of

Thermodynamic Principles 5Cv5d This problem revisits Example 15(3, part B. Re(do the problem three times,

changing the amount of water to:

• 25 moles
• 100 moles
• 500 moles
• Use graphs or tables to compare the results of parts A(C to the original problem

that used 50 moles of water. Give interpretations, based on physical phenomena, for any trends that are evident.

Solution:

Example 15(3 involved the reaction:

689

©20158201

In Example 15(3B, 10 moles of NO

• and 50 moles of H2O are placed in a reactor at P=1

bar and T=25˚C and allowed to reach equilibrium. Both liquid and vapor phases are present. Here, we vary the number of moles of H 2O.

The equations required to model the system in Example 15(3B were as follows:

1) Water mole balance( in part A the total number of moles of water is 25.6C8engaL r 9 i©.Al89 i©.R 2) Nitrogen dioxide balance( two moles of NO

• are consumed to form N2O4, and

either compound can be present in either phase.ht8engaL r 9 s.© Al89 s.© Rl 69 s©.v Al 869 s©.v R 3) Reaction equilibrium( the composition of the vapor phase must satisfy the

equilibrium expression derived in Example 15(3, equations 15.24 and 15.27:

d Mr8 y s©.v y s.© © r

:

s©.v R

s©.v R 1 s.© R 1 i©.R b

:

s.© R

s©.v R 1 s.© R 1 i©.R b © r c,pp

Part 1: Solutions Manual 2 / 4

Chapter 15: Synthesis of Thermodynamic Principles

690 © 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.4) Raoult’s Law( Water is expected to follow Raoult’s Law because in the liquid phase it is the major component of a dilute solution (γ~1) and in the vapor phase

the pressure is low enough to assume ideal gas behavior:

" 2 i©.A 2 s©.v A l 2 s.© A l 2 i©.A g VaxasvhHRG3 pH" 2 i©.y 2 s©.v y l 2 s.© y l 2 i©.y g VvHRG3 5) Henry’s Law( VLE for NO

• and N2O4, as minor components of a dilute solution,

are expected to follow Henry’s Law:

) *+ *pHu *,

Which, when expressed in terms of numbers of moles of each component, is:

+ s.L" 2 s.L A 2 oL.r A l2 s.L A l2 6L.A #p " 2 s.L y 2 oL.r y l2 s.L y l2 6L.y

#V1 -./

+ oL.r" 2 oL.r A 2 oL.r A l2 s.L A l2 6L.A #p" 2 oL.r y 2 oL.r y l2 s.L y l2 6L.y

#V1 -./

The H values from the example are 4695 bar for NO

• and 40 bar for N2O4.

The above are six equations in six unknowns, N H20

L, NH20

V, NN02

L, NN02

V, NN204

L, and N N2O4

• The only differences between cases A, B and C is the total number of moles of

water in the water balance. When these equations were solved with the nonlinear

equation solver in POLYMATH, the result was:

Case A Example 15( 3B Case B Case C Moles Water 25 50 100 500 Initial moles NO2 10 10 10 10 Moles Water in Liquid 24.81 49.82 99.84 499.96 Moles Water in Vapor 0.19 0.18 0.16 0.04 Moles NO2 in Liquid 0.00165 0.0033 0.0066 0.033 Moles NO2 in Vapor 1.83 1.75 1.60 0.41 Moles N2O4 in Liquid 0.19 0.39 0.78 3.90 Moles N2O4 in Vapor 3.89 3.73 3.41 0.88

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Chapter 15: Synthesis of Thermodynamic Principles

691 © 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.D) The amount of N2O4 increases as the amount of water increases, because it is more soluble in water than NO

• As the total amount of water increases, the

amount in the vapor phase actually decreases. This is because the mole fraction of water in the vapor phase is fixed and the total amount of vapor decreases as more N 2O4 dissolves in the liquid.5Cvod This problem revisits Example 15(4. Re(do the problem three times, changing the

volume of the system to:

A) 300 L

B) 1000 L

C) 1500 L

D) Use graphs or tables to compare the results of parts A(C to the original problem that used a volume of 600 L. Give interpretations, based on physical phenomena, for any trends that are evident.

Solution:

Example 15(4 involved the inter(conversions between isomers of xylene:

(R1) o(xylene ↔ m(xylene (R2) m(xylene ↔ p(xylene In a reactor at 100°C and a fixed volume (but unspecified pressure). Both liquid and vapor were present, but the reaction is modeled as occurring in the liquid phase only.

The equations used to model the system are:

1) Total moles of xylene( there are a total of 10 moles of xylene, distributed in an

unknown way between the three isomers and the two phases:

a f Tm a

Tm a 1 Tm a f dm a

dm a 1

di cPh:Synshhhhhhhhhhhhh

2) Reaction equilibrium( the liquid phase compositions must satisfy the equilibrium constants calculated in Step 2 of Example 15(4, and the equilibrium expressions

derived in steps 3 and 5 of Example 15(4:

) 232,56ih a 7 T a 8 T i OBcPhhhhhhhhhhhhhhhh) 7X7,°tih a 9 T a 7 T i PBOxO 3) Raoult’s Law( these three isomers are structurally and chemically similar enough to assume the liquid is an ideal solution, and their vapor pressures are low enough

to assume the vapor is an ideal gas. Therefore we write Raoult’s Law:

) *, *

:;

*,

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