#4 with a liquid at 2.0 atm (1520 mmHg) pressure and 85 °C. The liquid consists of 35 mol% benzene (CeHe) and 65 mol% toluene (CHe) Estimate the gas composition using Raoult's law and assuming ideal gas behavior. The Antoine equations for benzene and toluene are (with Po in mmHg, and T in °C): A gas containing nitrogen, benzene and toluene is in equilibrium B=Benz 1203.531 log10(Penzene) = 6.8972– TEToulene T+219.888 1346.773 log10(Proluene) = 6.95805– T+219.693
#4 with a liquid at 2.0 atm (1520 mmHg) pressure and 85 °C. The liquid consists of 35 mol% benzene (CeHe) and 65 mol% toluene (CHe) Estimate the gas composition using Raoult's law and assuming ideal gas behavior. The Antoine equations for benzene and toluene are (with Po in mmHg, and T in °C): A gas containing nitrogen, benzene and toluene is in equilibrium B=Benz 1203.531 log10(Penzene) = 6.8972– TEToulene T+219.888 1346.773 log10(Proluene) = 6.95805– T+219.693
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
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![**Educational Content: Gas-Liquid Equilibrium involving Nitrogen, Benzene, and Toluene**
### Problem Statement:
A gas containing nitrogen, benzene, and toluene is in equilibrium with a liquid at 2.0 atm (1520 mmHg) pressure and 85°C. The liquid consists of 35 mol% benzene (C₆H₆) and 65 mol% toluene (C₇H₈). Estimate the gas composition using Raoult's law and assuming ideal gas behavior. The Antoine equations for benzene and toluene are:
- \( \log_{10}(P^o_{benzene}) = 6.8972 - \frac{1203.531}{T + 219.888} \)
- \( \log_{10}(P^o_{toluene}) = 6.9580 - \frac{1346.773}{T + 219.693} \)
### Given Data:
- \(B =\) Benzene
- \(T =\) Toluene
**Temperature (\(T\)):** 85°C
**Pressure (\(P\)):** 2.00 atm = 1520 mmHg
### Calculations:
#### Vapor Pressure of Each Component
**For Benzene:**
1. Antoine equation:
\[
\log_{10}(P^o_{B}) = 6.8972 - \frac{1203.531}{85 + 219.888}
\]
2. Solving gives:
\[
\log_{10}(P^o_{B}) = 2.9497
\]
3. Vapor pressure \(P^o_{B} = 890 \text{ mmHg}\)
**For Toluene:**
1. Antoine equation:
\[
\log_{10}(P^o_{T}) = 6.95805 - \frac{1346.773}{85 + 219.693}
\]
2. Solving gives:
\[
\log_{10}(P^o_{T}) = 2.5379
\]
3. Vapor pressure \(P^o_{T} = 345.10 \text{ mmHg}\)
#### Mole Fraction:
**In the liquid phase:**
- Benzene (B): \( X_B = \frac{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff4dfc165-1571-419f-99fc-a4587d76c340%2F770cd8bf-ad53-47db-afcf-0156439d6b6b%2Fyjg71ge_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Educational Content: Gas-Liquid Equilibrium involving Nitrogen, Benzene, and Toluene**
### Problem Statement:
A gas containing nitrogen, benzene, and toluene is in equilibrium with a liquid at 2.0 atm (1520 mmHg) pressure and 85°C. The liquid consists of 35 mol% benzene (C₆H₆) and 65 mol% toluene (C₇H₈). Estimate the gas composition using Raoult's law and assuming ideal gas behavior. The Antoine equations for benzene and toluene are:
- \( \log_{10}(P^o_{benzene}) = 6.8972 - \frac{1203.531}{T + 219.888} \)
- \( \log_{10}(P^o_{toluene}) = 6.9580 - \frac{1346.773}{T + 219.693} \)
### Given Data:
- \(B =\) Benzene
- \(T =\) Toluene
**Temperature (\(T\)):** 85°C
**Pressure (\(P\)):** 2.00 atm = 1520 mmHg
### Calculations:
#### Vapor Pressure of Each Component
**For Benzene:**
1. Antoine equation:
\[
\log_{10}(P^o_{B}) = 6.8972 - \frac{1203.531}{85 + 219.888}
\]
2. Solving gives:
\[
\log_{10}(P^o_{B}) = 2.9497
\]
3. Vapor pressure \(P^o_{B} = 890 \text{ mmHg}\)
**For Toluene:**
1. Antoine equation:
\[
\log_{10}(P^o_{T}) = 6.95805 - \frac{1346.773}{85 + 219.693}
\]
2. Solving gives:
\[
\log_{10}(P^o_{T}) = 2.5379
\]
3. Vapor pressure \(P^o_{T} = 345.10 \text{ mmHg}\)
#### Mole Fraction:
**In the liquid phase:**
- Benzene (B): \( X_B = \frac{
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