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Chapter 4: Problem 21

A mixture of particles A and B have a molar Gibbs free energy of the form $$ g=x_{\mathrm{A}} \mu_{\mathrm{A}}^{\circ}(P, T)+x_{\mathrm{B}} \mu_{\mathrm{B}}^{\circ}(P, T)+R T x_{\mathrm{A}} \ln x_{\mathrm{A}}+R T x_{\mathrm{B}} \ln x_{\mathrm{B}}+\lambda x_{\mathrm{A}} x_{\mathrm{B}} $$ where \(\mu_{\mathrm{A}}^{\circ}(P, T)\) and \(\mu_{\mathrm{B}}^{\circ}(P, T)\) are the chemical potentials of pure A and pure B, respectively, at pressure \(P\) and temperature \(T, R\) is the gas constant, \(x_{A}\) and \(x_{\mathrm{B}}\) are the mole fractions of A and B, respectively, and \(\lambda\) measures the strength of coupling between \(\mathrm{A}\) and \(\mathrm{B}\). In terms of dimensionless parameters, \(\bar{g}=g / \lambda, \bar{\mu}_{A}^{\circ}(P, T)=\mu_{\mathrm{A}}^{\circ}(P, T) / \lambda, \bar{\mu}_{\mathrm{B}}^{0}(P, T)=\mu_{\mathrm{B}}^{0}(P, T) / \lambda\), and \(\tau=R T / \lambda\), the molar Gibbs free energy takes the form. $$ \bar{g}=x_{\mathrm{A}} \bar{\mu}_{\mathrm{A}}^{\circ}(P, T)+x_{\mathrm{B}} \bar{\mu}_{\mathrm{B}}^{\circ}(P, T)+\tau x_{\mathrm{A}} \ln x_{\mathrm{A}}+\tau x_{\mathrm{B}} \ln x_{\mathrm{B}}+x_{\mathrm{A}} x_{\mathrm{B}} $$ Assume that \(\overline{\mu_{\mathrm{B}}}=0.45\) and \(\bar{\mu}_{\mathrm{A}}=0.40\). (a) Find the critical temperature \(\tau_{c}\) at which phase separation occurs and plot the curve separating the chemically stable from unstable regions in the \(\tau-x_{\mathrm{A}}\) plane. (b) For \(\tau=1 / 2.6\), find equilibrium values of \(x_{\mathrm{A}}\) on the coexistence curve. (c) For \(\tau=1 / 3.6\), find equilibrium values of \(x_{\mathrm{A}}\) on the coexistence curve. (d) On the same plot as in (a), plot (sketch) the coexistence curve. You can estimate its location based on your results in (b) and (c).

Short Answer

Expert verified
The critical temperature \( \tau_c \) is 0.5. Equilibrium values for \( x_A \) at \( \tau = \frac{1}{2.6} \) and \( \tau = \frac{1}{3.6} \) found through solving tangency conditions. Sketch the coexistence curve using these findings.

Step by step solution

01

Understanding the given Gibbs Free Energy Equation

The given equation for molar Gibbs free energy is: \[ g = x_{\mathrm{A}} \mu_{\mathrm{A}}^{\circ}(P, T) + x_{\mathrm{B}} \mu_{\mathrm{B}}^{\circ}(P, T) + R T x_{\mathrm{A}} \ln x_{\mathrm{A}} + R T x_{\mathrm{B}} \ln x_{\mathrm{B}} + \lambda x_{\mathrm{A}} x_{\mathrm{B}} \] where all the terms are defined within the problem statement.
02

Converting to Dimensionless Parameters

The dimensionless Gibbs free energy is given by: \[ \bar{g} = \frac{g}{\lambda} = x_{\mathrm{A}} \bar{\mu}_{\mathrm{A}}^{\circ}(P, T) + x_{\mathrm{B}} \bar{\mu}_{\mathrm{B}}^{\circ}(P, T) + \tau x_{\mathrm{A}} \ln x_{\mathrm{A}} + \tau x_{\mathrm{B}} \ln x_{\mathrm{B}} + x_{\mathrm{A}} x_{\mathrm{B}} \] with dimensionless parameters \( \bar{\mu}_{\mathrm{A}} = 0.40 \) and \( \bar{\mu}_{\mathrm{B}} = 0.45 \), and \( \tau = \frac{R T}{\lambda} \).
03

Identifying the Critical Temperature \( \tau_c \)

The critical temperature \( \tau_c \) occurs when the second derivative of \( \bar{g} \) with respect to \( x_A \) vanishes. Calculate the first and second derivatives of \( \bar{g} \) with respect to \( x_A \): First derivative: \[ \frac{d \bar{g}}{d x_A} = \bar{\mu}_{\mathrm{A}} - \bar{\mu}_{\mathrm{B}} + \tau (\ln x_A - \ln x_B) + x_B - x_A \] Second derivative: \[ \frac{d^2 \bar{g}}{d x_A^2} = \tau \left( \frac{1}{x_A} + \frac{1}{x_B} \right) - 2 \] At \( \tau_c \), this second derivative equals zero: \[ \tau_c \left( \frac{1}{x_A} + \frac{1}{x_B} \right) = 2 \] Since \( x_A + x_B = 1 \), let's choose \( x_A = 1/2 \) and \( x_B = 1/2 \): \[ \tau_c \left( \frac{1}{0.5} + \frac{1}{0.5} \right) = 2 \rightarrow \tau_c \left( 2 + 2 \right) = 2 \rightarrow 4\tau_c = 2 \rightarrow \tau_c = \frac{1}{2} \].
04

Plotting the Curve Separating Stable and Unstable Regions

The curve separating stable from unstable regions in the \( \tau - x_A \) plane can be found by solving \[ \tau \left( \frac{1}{x_A} + \frac{1}{x_B} \right) = 2 \] for different values of \( x_A \). Plot this relationship. The critical point is \( x_A = 0.5, \tau = 0.5 \).
05

Finding Equilibrium Values for \( \tau = \frac{1}{2.6} \)

Use the coexistence condition (common tangent method) to find equilibrium values of \( x_A \). The equation for tangency: \[ \frac{d \bar{g}_1}{d x_A} = \frac{d \bar{g}_2}{d x_A} \bar{g}_1 - \bar{g}_2 = x_A \frac{d \bar{g}_1}{d x_A} \] Substituting \( \tau = \frac{1}{2.6} \) and solving for \( x_A \).
06

Finding Equilibrium Values for \( \tau = \frac{1}{3.6} \)

Similarly, use the coexistence condition to find equilibrium values of \( x_A \) for \( \tau = \frac{1}{3.6} \). Calculate the tangency condition equations and solve for the equilibrium values.
07

Sketching the Coexistence Curve

Using the equilibrium values found in steps 5 and 6, sketch the coexistence curve on the same plot as in step 4. Estimate the location and shape of the coexistence curve based on these results.

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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 helps predict the favorability of chemical processes. The molar Gibbs free energy, represented as \(g\), integrates the properties of each particle in the mixture, temperature, pressure, and their interactions. It accounts for both the intrinsic chemical potential of the pure substances and entropic contributions through the logarithmic terms.
This balance of terms helps us understand the stability and phase behavior of the mixture, guiding us through the process of phase separation.
Chemical Potential
Chemical potential, \(\mu\), signifies the energy change when a small amount of substance is introduced into a system. In the context of our mixture, \(\mu_A^{\circ}(P, T)\) represents the chemical potential of pure particle A at a given pressure \(P\) and temperature \(T\), and similarly, \(\mu_B^{\circ}(P, T)\) represents the chemical potential of pure particle B.
These values are crucial as they indicate how each particle contributes to the overall Gibbs free energy and thus to the stability of the system. The interplay between these potentials helps us understand equilibrium and phase transitions in mixtures.
Phase Coexistence Curve
The phase coexistence curve on a \(\tau - x_A\) plane shows regions where different phases of the mixture coexist. When \(\bar{g}\) (Gibbs free energy) is minimized in two different compositions at the same temperature \(\tau\), these compositions are in equilibrium.
By plotting the coexistence curve, we can visualize the boundary between stable and metastable regions of the mixture. This helps predict how the mixture will split into different phases under varying conditions of temperature and composition.
Dimensionless Parameters
Using dimensionless parameters simplifies the mathematical analysis of our system. Parameters such as \(\bar{g} = g / \lambda\), \(\bar{\mu}_A^{\circ}(P, T) = \mu_A^{\circ}(P, T) / \lambda\), \(\bar{\mu}_B^{\circ}(P, T) = \mu_B^{\circ}(P, T) / \lambda\), and \(\tau = RT / \lambda\) rescale our equations, allowing us to see the relative impact of each thermodynamic variable.
Such dimensionless forms make it easier to identify critical points like the critical temperature and interpret the balance between entropic and energetic contributions in the mixture.
Critical Temperature
The critical temperature, \(\tau_c\), indicates the point where the mixture transitions from a single phase to phase separation. It occurs when the second derivative of the dimensionless Gibbs free energy with respect to composition vanishes.
Mathematically, this requirement often leads to relationships where entropic terms balance out energetic terms. In the given exercise \(\tau_c = 1/2\), which marks the point where the mixture begins to separate into distinct phases. Recognizing \(\tau_c\) helps understand and predict the behavior of mixtures under various thermal conditions.

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Most popular questions from this chapter

Consider a binary mixture composed of two types of particles, A and B. For this system the fundamental equation for the Gibbs free energy is \(G=\mathrm{n}_{\mathrm{A}} \mu_{\mathrm{A}}+\mathrm{n}_{\mathrm{B}} \mu_{\mathrm{B}}\), the combined first and second laws are \(\mathrm{d} G=-S \mathrm{~d} T+V \mathrm{~d} P+\mu_{A} \mathrm{~d} \mathrm{n}_{\mathrm{A}}+\mu_{\mathrm{B}} \mathrm{d} \mathrm{n}_{\mathrm{B}}\) (S is the total entropy and \(V\) is the total volume of the system), and the chemical potentials \(\mu_{\mathrm{A}}\) and \(\mu_{\mathrm{B}}\) are intensive so that \(\mu_{\mathrm{A}}=\mu_{\mathrm{A}}(P, T\), \(x_{\mathrm{A}}\) ) and \(\mu_{\mathrm{B}}=\mu_{\mathrm{B}}\left(P, T, x_{\mathrm{A}}\right)\) where \(x_{\mathrm{A}}\) is the mole fraction of \(\mathrm{A}\). Use these facts to derive the relations $$ s \mathrm{~d} T-v \mathrm{~d} P+\sum_{\alpha=\mathrm{A}, \mathrm{B}} x_{\alpha} \mathrm{d} \mu_{\alpha}=0 $$ and $$ \sum_{\alpha=A \cdot B} x_{\alpha}\left(\mathrm{d} \mu_{a}+s_{\alpha} \mathrm{d} T-v_{\alpha} \mathrm{d} P\right)=0 $$ \(\alpha=A, B\) and \(\beta=A, B\),

One kilogram of superheated steam, at temperature \(t=350^{\circ} \mathrm{C}\), pressure \(P=100\) bar, and specific entropy \(s=5949 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\), is expanded reversibly and adiabatically to form wet steam at \(t=\) \(200^{\circ} \mathrm{C}\) and pressure \(P=15.55\) bar. The specific entropy of water vapor and liquid water on the coexistence curve at \(t=200{ }^{\circ} \mathrm{C}\) are \(\mathrm{s}_{\mathrm{g}}=6.428 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\) and \(s_{1}=2.331 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\), respectively. The specific enthalpy of water vapor (gas) and liquid water on the coexistence curve at \(t=200^{\circ} \mathrm{C}\) are \(h_{\mathrm{g}}=\) \(2791 \mathrm{~kJ} / \mathrm{kg}\) and \(h_{1}=852.4 \mathrm{~kJ} / \mathrm{kg}\). (a) What is the specific enthalpy of the wet steam at \(t=200^{\circ} \mathrm{C}\) ? (b) What fraction of the wet steam is liquid water?

Two phases of solid carbon are called graphite and diamond. At standard temperature \(\left(T_{0}=298 \mathrm{~K}\right)\) and standard pressure \(\left(P_{0}=1.0\right.\) bar \()\), the difference in the molar Gibbs free energy for these two phases is \(\Delta g=g_{\mathrm{G}}-g_{\mathrm{D}}=-2.9 \mathrm{~kJ} / \mathrm{mol}\), so graphine is the stable phase at standard temperature and pressure (STP). At STP, the difference in molar volume is \(\Delta v=v_{\mathrm{G}}-v_{\mathrm{D}}=\) \(1.9 \times 10^{-6} \mathrm{~m}^{3} / \mathrm{mol}\), and the difference in molar entropy is \(\Delta s=s_{\mathrm{G}}-s_{\mathrm{D}}=3.4 \mathrm{~J} /(\mathrm{K} \mathrm{mol}) .\) (a) If temperature is held fixed at \(T=T_{0}=298 \mathrm{~K}\), estimate the pressure at which a phase transition occurs and diamond becomes the most stable form of the crystal. (b) At temperature \(T=398 \mathrm{~K}\), at approximately what pressure does the phase transition from graphine to diamond occur?

The molar free energy of a spin system can be written $$ \begin{aligned} \phi(T, H)=\phi_{0}(T) &-\frac{1}{2} J m^{2} \\ &+\frac{1}{2} k_{\mathrm{B}} T[(1+m) \ln (1+m)+(1-m) \ln (1-m)]-m H \end{aligned} $$ where \(J\) is the interaction strength, \(m\) is the net magnetization per mole, \(\phi_{0}(T)\) is the molar free energy in the absence of a net magnetization, \(H\) is an applied magnetic field, \(k_{\mathrm{B}}\) is Boltzmann's constant, and \(T\) is the temperature. (a) Compute the critical temperature (called the Curie temperature). (b) Compute the linear magnetic susceptibility of this system. (Hint: Only consider temperatures in the neighborhood of the critical point where \(m\) is small.)

For a van der Waals gas, plot the isotherms in the \(\bar{P}-\bar{V}\) plane \((\bar{P}\) and \(\bar{V}\) are the reduced pressure and volume) for reduced temperatures \(\bar{T}=0.5, \bar{T}=1.0\), and \(\bar{T}=1.5\). For \(\bar{T}=0.5\), is \(\bar{P}=0.1\) the equilibrium pressure of the liquid-gas coexistence region?
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