Problem 31 - A Modern Course in Statistical Physics

Question

A solution of particles A and B has a Gibbs free energy $$ \begin{aligned} G\left(P, T, \mathrm{n}_{\mathrm{A}}, \mathrm{n}_{\mathrm{B}}\right)=& \mathrm{n}_{\mathrm{A}} g_{\mathrm{A}}(P, T)+\mathrm{n}_{\mathrm{B}} g_{\mathrm{B}}(P, T)+\frac{1}{2} \lambda_{\mathrm{AA}} \frac{\mathrm{n}_{\mathrm{A}}^{2}}{\mathrm{n}}+\frac{1}{2} \lambda_{\mathrm{BB}} \frac{\mathrm{n}_{\mathrm{B}}^{2}}{\mathrm{n}} \\ &+\lambda_{\mathrm{AB}} \frac{\mathrm{n}_{\mathrm{A}} \mathrm{n}_{\mathrm{B}}}{\mathrm{n}}+\mathrm{n}_{\mathrm{A}} R T \ln x_{\mathrm{A}}+\mathrm{n}_{\mathrm{B}} R T \ln x_{\mathrm{B}} \end{aligned} $$ Initially, the solution has \(\mathrm{n}_{\mathrm{A}}\) moles of A and \(\mathrm{n}_{\mathrm{B}}\) moles of B. (a) If an amount \(\Delta \mathrm{n}_{\mathrm{B} \text {, of } \mathrm{B} \text { is added }}\) keeping the pressure and temperature fixed, what is the change in the chemical potential of A? (b) For the case \(\lambda_{\mathrm{AA}}=\lambda_{\mathrm{BB}}=\lambda_{\mathrm{AB} \text {, }}\) does the chemical potential of A increase or decrease?

Step-by-Step Solution

Answer

The chemical potential of A increases if \( \lambda > RT \) and decreases if \( \lambda < RT \).

Step 1: Understanding the Gibbs Free Energy

The given Gibbs free energy for the solution is: \[ G\big(P, T, n_A, n_B\big) = n_A g_A(P, T) + n_B g_B(P, T) + \frac{1}{2} \lambda_{AA} \frac{n_A^2}{n} + \frac{1}{2} \lambda_{BB} \frac{n_B^2}{n} + \lambda_{AB} \frac{n_A n_B}{n} + n_A RT \ln x_A + n_B RT \ln x_B \]

Step 2: Expressing the Chemical Potential

The chemical potential of a component (A) in a mixture is given by the partial derivative of the Gibbs free energy with respect to the number of moles of that component, keeping the other parameters constant: \( \mu_A = \left( \frac{\partial G}{\partial n_A} \right)_{P,T,n_B} \).

Step 3: Partial Derivative with Respect to \( n_A \)

Take the partial derivative of the Gibbs free energy with respect to \( n_A \): \[ \mu_A = g_A(P, T) + \lambda_{AA} \frac{n_A}{n} + \lambda_{AB} \frac{n_B}{n} - \lambda_{AA} \frac{n_A^2}{n^2} - \lambda_{AB} \frac{n_A n_B}{n^2} + RT \ln x_A + RT \frac{n_B}{n} \].

Step 4: Change in Chemical Potential \( \Delta\mu_A \) with \( \Delta n_B \)

When \( \Delta n_B \) moles of B are added, keeping the pressure and temperature fixed, the change in the chemical potential of A can be expressed as: \[ \Delta \mu_A = \left( \frac{\partial \mu_A}{\partial n_B} \right) \Delta n_B \]

Step 5: Calculate \( \left( \frac{\partial \mu_A}{\partial n_B} \right) \)

From the expression of \( \mu_A \), the term \(\left( \frac{\partial \mu_A}{\partial n_B} \right) \) is derived as \[ \frac{\partial \mu_A}{\partial n_B} = \lambda_{AB} \times \frac{1}{n} - \left( \lambda_{AB} \times \frac{n_A}{n^2} - \frac{1}{n} \right) - \frac{n_A}{n} RT + \frac{RT}{n} = \left( \lambda_{AB} - RT \right) \frac{1}{n} \].

Step 6: Resulting \( \Delta \mu_A \)

Now multiply \( \left( \frac{\partial \mu_A}{\partial n_B} \right) \) by \( \Delta n_B \) to find the change in \( \mu_A \): \[ \Delta \mu_A = \Delta n_B \times \left( \lambda_{AB} - RT \right) \times \frac{1}{n} \]

Step 7: Analyzing the Case \(\lambda_{AA} = \lambda_{BB} = \lambda_{AB} \)

If \( \lambda_{AA} = \lambda_{BB} = \lambda_{AB} \), then \lambda_{AB} = \lambda. Substitute \( \lambda \) for \( \lambda_{AB} \) in the expression: \( \Delta \mu_A = \Delta n_B \times \left( \lambda - RT \right) \times \frac{1}{n} \). A positive or negative change depends on whether \( \lambda \) is greater or less than \( RT \).

Step 8: Conclusion

If \( \lambda > RT \), \( \Delta \mu_A \) will be positive (increase). If \( \lambda < RT \), \( \Delta \mu_A \) will be negative (decrease).

Key Concepts

Gibbs free energy

Gibbs free energy is a thermodynamic potential that measures the maximum reversible work that a system can perform at a constant temperature and pressure. It is a useful quantity for understanding spontaneous processes and chemical reactions. The formula for Gibbs free energy is given by: \[ G = H - TS \], where

• \( G \) is the Gibbs free energy
• \( H \) is the enthalpy
• \( T \) is the temperature
• \( S \) is the entropy

In the context of mixtures, Gibbs free energy helps us determine how components interact and change state. For example, the Gibbs free energy of a solution can include terms that consider the interactions between different particles. This is crucial in thermodynamics as it gives insight into the stability and spontaneity of the system. In the provided exercise, the Gibbs free energy equation includes contributions from both components A and B, as well as their interactions. The formula given is: \[ G\big(P, T, n_A, n_B\big) = n_A g_A(P, T) + n_B g_B(P, T) + \frac{1}{2} \frac{\text{interaction terms}}{n} + \text{entropy terms} \] This detailed expression helps in analyzing the contributions of each factor when determining the chemical potential.

Partial derivative

A partial derivative represents how a function changes as one of the variables changes, while keeping the other variables constant. It is a fundamental concept in calculus and is extensively used in thermodynamics to understand how specific properties of a system change. In the context of the exercise, we use the partial derivative to express the chemical potential of a component in a mixture. For a component A in a mixture: \[ \text{Partial derivative of } \frac{\text{G}}{\text{n}_A} = \text{Chemical Potential of A} \text{i.e., } \, \text{\textmu}_A = \bigg(\frac{\text{dG}}{\text{d}n_A} \bigg)_{P,T,n_B} \] This means we take the derivative of the Gibbs free energy with respect to the number of moles of A, keeping the pressure, temperature, and moles of B constant. The calculation involves differentiating each term in the Gibbs energy expression that contains \text{n}_A. This technique shows how adding a small amount of one component affects the overall system, crucial for understanding chemical equilibrium and reactions.

Mixtures in thermodynamics

Mixtures are systems composed of more than one substance, and their study in thermodynamics involves understanding how the components interact and affect one another. Key properties to consider include:

• Gibbs free energy
• Chemical potential
• Entropy
• Enthalpy

Mixtures can exhibit complex behavior due to the interactions between particles. For example, in the provided exercise, the terms involving \(\text{n}_A\) and \(\text{n}_B\) indicate how particles A and B influence the system's Gibbs free energy. When we add component B to the mixture, we observe how the chemical potential of component A changes. This is assessed using the partial derivatives of the Gibbs free energy function. Changes like this help us understand phase equilibria, solutions, and chemical reactions involving mixtures. Ultimately, these in-depth analyses aid in developing practical applications such as material synthesis, pharmacology, and chemical engineering processes.

Chemical potential change

Chemical potential is a measure of the potential for a substance to undergo a change in composition or phase. It is denoted by \text{\textmu} and in the context of a mixture can be thought of as the 'driving force' for mass transfer and reaction processes. In a mixture, the chemical potential of a component can change if more of that component is added, or if the conditions (like pressure or temperature) are altered. Mathematically, the change in chemical potential with respect to the number of moles of component B is represented as: \text{\textDelta\textmu}_A = \bigg(\frac{\text{\textpartial\textmu}_A}{\text{\textpartial}\text{n}_B}\bigg)\text{\textDelta}\text{n}_B. This partial derivative assessment tells us how responsive the chemical potential is to changes in the amounts of other components. In the given problem when we add \text{\textDelta}\text{n}_B moles of B, we need to understand how \text{\textmu}_A reacts. Depending on this reactivity, we see whether the solution becomes more or less favorable for component A. This understanding helps in controlling chemical processes and predicting the outcomes of mixing components under various conditions. The derived formula in the exercise: \[ \text{\textDelta\textmu}_A = \frac{\text{\textDelta n}_B \times (\text{\textlambda}_{AB} - \text{RT})}{\text{n}} \] allows us to determine whether \text{\textmu}_A will increase or decrease, giving insights into the interactions and energetics of the mixture.