/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 A solution of particles A and B ... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

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

Step by step solution

01

Understanding the Gibbs Free Energy

The given Gibbs free energy for the solution is: \[ G\big(P, T, n_A, n_B\big) = n_A g_A(P, T) + n_B g_B(P, T) + \frac{1}{2} \lambda_{AA} \frac{n_A^2}{n} + \frac{1}{2} \lambda_{BB} \frac{n_B^2}{n} + \lambda_{AB} \frac{n_A n_B}{n} + n_A RT \ln x_A + n_B RT \ln x_B \]
02

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: \( \mu_A = \left( \frac{\partial G}{\partial n_A} \right)_{P,T,n_B} \).
03

Partial Derivative with Respect to \( n_A \)

Take the partial derivative of the Gibbs free energy with respect to \( n_A \): \[ \mu_A = g_A(P, T) + \lambda_{AA} \frac{n_A}{n} + \lambda_{AB} \frac{n_B}{n} - \lambda_{AA} \frac{n_A^2}{n^2} - \lambda_{AB} \frac{n_A n_B}{n^2} + RT \ln x_A + RT \frac{n_B}{n} \].
04

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: \[ \Delta \mu_A = \left( \frac{\partial \mu_A}{\partial n_B} \right) \Delta n_B \]
05

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 \[ \frac{\partial \mu_A}{\partial n_B} = \lambda_{AB} \times \frac{1}{n} - \left( \lambda_{AB} \times \frac{n_A}{n^2} - \frac{1}{n} \right) - \frac{n_A}{n} RT + \frac{RT}{n} = \left( \lambda_{AB} - RT \right) \frac{1}{n} \].
06

Resulting \( \Delta \mu_A \)

Now multiply \( \left( \frac{\partial \mu_A}{\partial n_B} \right) \) by \( \Delta n_B \) to find the change in \( \mu_A \): \[ \Delta \mu_A = \Delta n_B \times \left( \lambda_{AB} - RT \right) \times \frac{1}{n} \]
07

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 \( RT \).
08

Conclusion

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Compute the heat capacity at constant magnetic field \(C_{H, n}\), the susceptibilities \(X_{T}, \mathrm{n}\) and \(X_{S, n}\) ' and the thermal expansivity \(\alpha_{H, n}\) for a magnetic system, given that the mechanical equation of state is \(M=\mathrm{n} D H / T\) and the heat capacity is \(C_{M, n}=\pi c_{\prime}\) where \(M\) is the magnetization, \(H\) is the magnetic field, \(\mathrm{n}\) is the number of moles, \(D\) is a constant, \(c\) is the molar heat capacity, and \(T\) is the temperature.

Compute the entropy, enthalpy, Helmholtz free energy, and Gibbs free energy of a paramagnetic substance and write them explicitly in terms of their natural variables when possible. Assume that the mechanical equation of state is \(m=(D H / T)\) and that the molar heat capacity at constant magnetization is \(c_{\mathrm{m}}=c\), where \(m\) is the molar magnetization, \(H\) is the magnetic field, \(D\) is a constant, \(c\) is a constant, and \(T\) is the temperature.

For a low-density gas the virial expansion can be terminated at first order in the density and the equation of state is \( P=\frac{N k_{\mathrm{B}} T}{V}\left[1+\frac{N}{V} B_{2}(T)\right] $$ where \)B_{2}(T)\( is the second virial coefficient. The heat capacity will have corrections to its ideal gas value. We can write it in the form $$ C_{V, N}=\frac{3}{2} N k_{\mathrm{B}}-\frac{N^{2} k_{\mathrm{B}}}{V} F(T) $$ (a) Find the form that \)F(T)\( must have in order for the two equations to be thermodynamically consistent. (b) Find \)C_{P, N}$. (c) Find the entropy and internal energy.

A van der Waals gas can be cooled by free expansion. Since no work is done and no heat is added during free expansion, the internal energy remains constant. An infinitesimal change in volume \(\mathrm{d} V\) causes an infinitesimal temperature change in \(\mathrm{d} T\), where $$ \mathrm{d} T=\left(\frac{\partial T}{\partial V}\right)_{U, \mathrm{n}} \mathrm{d} V $$ (a) Compute the Joule coefficient \({ }^{(\partial T / \partial V)}_{U, \mathrm{H}}\) for a van der Waals gas (note that the heat capacity \(C_{V, n}\) is independent of volume and use \(C_{V, n}=3 / 2 \mathrm{n} R\) ) (b) Compute the change in temperature of one mole of oxygen \(\left(\mathrm{O}_{2}\right)\) and one mole of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) if they each expand from an initial volume \(V_{\mathrm{i}}=10^{-3} \mathrm{~m}^{3}\) at temperature \(T_{\mathrm{i}}=300 \mathrm{~K}\) to a final volume \(V_{\mathrm{f}}=\infty\). (For \(\mathrm{O}_{2}\) the van der Waals constant is \(a=0.1382 \mathrm{~Pa} \mathrm{~m}^{6} / \mathrm{mol}^{2}\) and for \(\mathrm{CO}_{2}\) it is \(a=0.3658 \mathrm{~Pa} \mathrm{~m}^{6} / \mathrm{mol}^{2}\).)

A biological molecule of unknown mass can be prepared in pure powdered form. If 15 \(\mathrm{g}\) of this powder is added to a container with \(1 \mathrm{~L}\) of water at \(T=300 \mathrm{~K}\), which is initially at atmospheric pressure, the pressure inside the container increases to \(P=1.3 \mathrm{~atm}\). (a) What is the molecular weight of the biological molecules? (b) What is the mass of each molecule expressed in atomic units?

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