The pressure can be estimated from the Antoine equation and Raoult’s law to be164 mmHg.) I need understanding how this pressure was calculated. Be thorough to the fullest.
The pressure can be estimated from the Antoine equation and Raoult’s law to be164 mmHg.) I need understanding how this pressure was calculated. Be thorough to the fullest.
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:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
Related questions
Question
(The pressure can be estimated from the Antoine equation and Raoult’s law to be164 mmHg.)
I need understanding how this pressure was calculated. Be thorough to the fullest.
![cal%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdf
Copilot
Q168
The values of nout were determined from the known mole fractions of benzene and toluene in the outlet
streams and the calculated values of ny and nL. We do not know the feed-stream pressure and so we assume
that AH for the change from 1 atm to Pfeed is negligible, and since the process is not running at an unusually
low temperature or high pressure, we neglect the effects of pressure on enthalpy in the calculations of H₁
through H4. (The pressure can be estimated from the Antoine equation and Raoult's law to be 164 mm Hg.)
The heat capacity and latent heat data needed to calculate the outlet enthalpies are obtained from Tables B.1
and B.2.
-
Q Search
The formulas (including the APEx formulas) and values of the unknown specific enthalpies are given
below. Convince yourself that the formulas represent AH for the transitions from the reference states to the
process states.
H₁ =
=
(Cp)c,H₁1dT[=Enthalpy(“benzene”,10,50,“C”,“I”)] = 5.341 kJ/mol
50°C
H₂ = √ (C₂)CHT [=Enthalpy ("toluene", 10,50,"C","1")] = 6.341 kJ/mol
10°C
(Cp)C6H6(v)dT
50°C
H4 =
VITY EXERCISE
10°C
-80.1°C
H3 = √°C (Cp)C¡Há(1)ďT +(AĤv )C.H₂ (80.1°C) + +
10°C
451
110.62°C
of 695 Q
10°℃
80.1°C
[=Enthalpy("benzene", 10,80.1,"C","1") + Hv("benzene")
+Enthalpy("benzene",80.1,50,"C","g")] = 37.53 kJ/mol
The energy balance is
50°C
(C₂)CH(1)dT + (A‚)c‚µ‚(110.62°C) + fo
110.62°C
50°C
[=Enthalpy("toluene", 10,110.62,"C","1") + Hv("toluene")
+Enthalpy("toluene",110.62,50,"C","g")] = 42.94 kJ/mol
Q
Ση
Q = AH = Σn₁A₁ - Σnit₁ =
out
C
in
(Cp)C₂H₂(v) dT
Q=17.7 kJ
A gas emerges from a stack at 1200°C. Rather than being released directly to the atmosphere, it can
he nassed through one or several heat exchangers and the heat it loses can he nut to use in a variety
D
Ka
€
ENG](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9ba88ecd-654b-4f6b-9783-e93c33ca6a16%2Ffd12b65a-6603-4aa8-82fc-881b9d1b9602%2Fsielfl4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:cal%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdf
Copilot
Q168
The values of nout were determined from the known mole fractions of benzene and toluene in the outlet
streams and the calculated values of ny and nL. We do not know the feed-stream pressure and so we assume
that AH for the change from 1 atm to Pfeed is negligible, and since the process is not running at an unusually
low temperature or high pressure, we neglect the effects of pressure on enthalpy in the calculations of H₁
through H4. (The pressure can be estimated from the Antoine equation and Raoult's law to be 164 mm Hg.)
The heat capacity and latent heat data needed to calculate the outlet enthalpies are obtained from Tables B.1
and B.2.
-
Q Search
The formulas (including the APEx formulas) and values of the unknown specific enthalpies are given
below. Convince yourself that the formulas represent AH for the transitions from the reference states to the
process states.
H₁ =
=
(Cp)c,H₁1dT[=Enthalpy(“benzene”,10,50,“C”,“I”)] = 5.341 kJ/mol
50°C
H₂ = √ (C₂)CHT [=Enthalpy ("toluene", 10,50,"C","1")] = 6.341 kJ/mol
10°C
(Cp)C6H6(v)dT
50°C
H4 =
VITY EXERCISE
10°C
-80.1°C
H3 = √°C (Cp)C¡Há(1)ďT +(AĤv )C.H₂ (80.1°C) + +
10°C
451
110.62°C
of 695 Q
10°℃
80.1°C
[=Enthalpy("benzene", 10,80.1,"C","1") + Hv("benzene")
+Enthalpy("benzene",80.1,50,"C","g")] = 37.53 kJ/mol
The energy balance is
50°C
(C₂)CH(1)dT + (A‚)c‚µ‚(110.62°C) + fo
110.62°C
50°C
[=Enthalpy("toluene", 10,110.62,"C","1") + Hv("toluene")
+Enthalpy("toluene",110.62,50,"C","g")] = 42.94 kJ/mol
Q
Ση
Q = AH = Σn₁A₁ - Σnit₁ =
out
C
in
(Cp)C₂H₂(v) dT
Q=17.7 kJ
A gas emerges from a stack at 1200°C. Rather than being released directly to the atmosphere, it can
he nassed through one or several heat exchangers and the heat it loses can he nut to use in a variety
D
Ka
€
ENG
Transcribed Image Text:*Elementary Principles of Chemic X +
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|
loud a
- 8.4-4
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Solution Basis: 1 mol Feed
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Ask Copilot
+
450 of 695
via capy changes for the unce stups.
college/felder
Partial Vaporization of a Mixture
An equimolar liquid mixture of benzene (B) and toluene (T) at 10°C is fed continuously to a vessel in which
the mixture is heated to 50°C. The liquid product is 40.0 mole% B, and the vapor product is 68.4 mole% B.
How much heat must be transferred to the mixture per g-mole of feed?
We start with a degree-of-freedom analysis:
mol, 10°C
0.500 mol B/mol
0.500 mol T/mol
Q Search
A
EVAPORATOR
QD
0000
ny(mol), 50°C
0.684 mol B/mol
0.316 mol T/mol
QU)
(mol), 50°C
0.400 mol B/mol
0.600 mol T/mol
3 unknown variables (ny, nL, Q)
-2 material balances
−1
-1 energy balance
= 0 degrees of freedom
We could count each specific enthalpy to be determined as an unknown variable, but then we would also
degrees of freedom unchanged.
count the equations for each of them in terms of heat capacities and latent heats, leaving the number of
Bal
Q 16
ENG
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