One expression that has been suggested for the excess Gibbs energy of a binary| mixture that is asymmetric in composition is GO = Ax,¤2(X1 – x2) a. Find expressions for the activity coefficients in which y, is specified in terms of æ2 and y2 in terms of x1. b. Does this excess Gibbs energy model satisfy the Gibbs-Duhem equation?
One expression that has been suggested for the excess Gibbs energy of a binary| mixture that is asymmetric in composition is GO = Ax,¤2(X1 – x2) a. Find expressions for the activity coefficients in which y, is specified in terms of æ2 and y2 in terms of x1. b. Does this excess Gibbs energy model satisfy the Gibbs-Duhem equation?
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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Please complete part B).
![**Excess Gibbs Energy of a Binary Mixture**
One expression that has been suggested for the excess Gibbs energy of a binary mixture that is asymmetric in composition is given by:
\[ G^{ex} = Ax_1 x_2 (x_1 - x_2) \]
### Problems:
**a.** Find expressions for the activity coefficients in which \(\gamma_1\) is specified in terms of \(x_2\) and \(\gamma_2\) in terms of \(x_1\).
**b.** Does this excess Gibbs energy model satisfy the Gibbs-Duhem equation?
**Note:** Attached contains part a) solution and part of part b) solution. I need help finishing part b (taking the partial derivatives).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44dce5c9-bcec-4e4b-a98e-07662a0f3f15%2F66310ae4-c7da-4757-ac9e-d28a98bdba0d%2Fcpjrz4n.png&w=3840&q=75)
Transcribed Image Text:**Excess Gibbs Energy of a Binary Mixture**
One expression that has been suggested for the excess Gibbs energy of a binary mixture that is asymmetric in composition is given by:
\[ G^{ex} = Ax_1 x_2 (x_1 - x_2) \]
### Problems:
**a.** Find expressions for the activity coefficients in which \(\gamma_1\) is specified in terms of \(x_2\) and \(\gamma_2\) in terms of \(x_1\).
**b.** Does this excess Gibbs energy model satisfy the Gibbs-Duhem equation?
**Note:** Attached contains part a) solution and part of part b) solution. I need help finishing part b (taking the partial derivatives).
![### Detailed Transcription and Explanation
---
#### Thermodynamics Equations and Derivations
1. **Equation Definitions and Relationships**
- \( G^{ex} = A \cdot x_1 x_2 (x_1 - x_2) \)
- \( RT \ln \gamma_2 = \left( \frac{\partial (N G^{ex})}{\partial N_2} \right)_{N_1TP} \)
- \( RT \ln \gamma_1 = \left( \frac{\partial (N G^{ex})}{\partial N_1} \right)_{N_2TP} \)
---
2. **Derivation Steps**
- Given: \( G^{ex} = A \cdot x_1 x_2 (x_1 - x_2) \)
- \[ G^{ex} = c A \left(\frac{N_1}{N_1 + N_2}\right) \left(\frac{N_2}{N_1 + N_2}\right) \left( \frac{N_1 - N_2}{N_1 + N_2} \right) \]
- Simplifying: \( G^{ex} = A \frac{(N_1 N_2) (N_1 - N_2)}{(N_1 + N_2)^3} \)
- Use the relationship of \( \frac{N_1}{N} = x_1 \)
- And \( \frac{N_2}{N} = x_2 \)
- \[ RT \ln \gamma_2 = \left( \frac{\partial (N G^{ex})}{\partial N_1} \right)_{N_2TP} = \frac{\partial}{\partial N_1} \left( A \frac {N_1 N_2 (N_1 - N_2)}{(N_1 + N_2)^2} \right) \]
---
3. **Gibbs-Duhem Equation**
- \( \sum \left( x_i d \ln \gamma_i \right) = 0 \)
---
4. **Partial Derivatives and Gibbs-Duhem Simplification**
- For \( x_1 \):
\[ x_1 \frac{d \ln \gamma_](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44dce5c9-bcec-4e4b-a98e-07662a0f3f15%2F66310ae4-c7da-4757-ac9e-d28a98bdba0d%2Fkgxq2bi_reoriented.jpeg&w=3840&q=75)
Transcribed Image Text:### Detailed Transcription and Explanation
---
#### Thermodynamics Equations and Derivations
1. **Equation Definitions and Relationships**
- \( G^{ex} = A \cdot x_1 x_2 (x_1 - x_2) \)
- \( RT \ln \gamma_2 = \left( \frac{\partial (N G^{ex})}{\partial N_2} \right)_{N_1TP} \)
- \( RT \ln \gamma_1 = \left( \frac{\partial (N G^{ex})}{\partial N_1} \right)_{N_2TP} \)
---
2. **Derivation Steps**
- Given: \( G^{ex} = A \cdot x_1 x_2 (x_1 - x_2) \)
- \[ G^{ex} = c A \left(\frac{N_1}{N_1 + N_2}\right) \left(\frac{N_2}{N_1 + N_2}\right) \left( \frac{N_1 - N_2}{N_1 + N_2} \right) \]
- Simplifying: \( G^{ex} = A \frac{(N_1 N_2) (N_1 - N_2)}{(N_1 + N_2)^3} \)
- Use the relationship of \( \frac{N_1}{N} = x_1 \)
- And \( \frac{N_2}{N} = x_2 \)
- \[ RT \ln \gamma_2 = \left( \frac{\partial (N G^{ex})}{\partial N_1} \right)_{N_2TP} = \frac{\partial}{\partial N_1} \left( A \frac {N_1 N_2 (N_1 - N_2)}{(N_1 + N_2)^2} \right) \]
---
3. **Gibbs-Duhem Equation**
- \( \sum \left( x_i d \ln \gamma_i \right) = 0 \)
---
4. **Partial Derivatives and Gibbs-Duhem Simplification**
- For \( x_1 \):
\[ x_1 \frac{d \ln \gamma_
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