Note

The following notes are based on my coursework in Spring 2024 instructed by Prof. James Im. I am writing as I am preparing for the first midterm coming on March 20, 2024.

Course introduction

The course focuses on conceptual understanding rather than “crunching numbers” according to the instructor. The topics here have been chosen by the instructor based on his more than a decade working with industry companies.

Lecture 1. Basic thermodynamics

First and second laws of thermodynamics

Recall the first and second laws of thermodynamics:

\[\begin{align} dU & \equiv \delta Q + \delta W + \delta W', \\ dS & \geq \frac{\delta Q}{T}, \\ \therefore dU & = TdS - PdV, \\ \end{align}\]

where $\delta Q$ is the heat transferred, $\delta W$ is the mechanical work, and $\delta W’$ includes chemical, magnetic, and electrical work.

Single component system

Define Gibbs free energy

Gibbs free energy for a single component system is defined by

\[\begin{align} G &\equiv U + PV - TS, \\ dG &= dU + PdV + VdP - TdS - SdT, \\ \therefore dG & = -SdT + VdP. \end{align}\]

Two conditions for equlibrium

The two Conditions for equlibrium $ dG=0 $ and $ d^2G > 0 $. The curvature must be concave up to ensure the global or local minimum.

Competition between enthalpy and entropy with temperature

Recall the definition of Gibbs free energy:

\[\begin{align} G &= U + PV - TS, \\ H &\equiv U + PV, \\ \therefore G &= H - TS. \end{align}\]

Gibbs free energy is a function of $H$ and $TS$. As low temperature, $H$ dominates. At high temperature, $TS$ dominates.

Assumption in condensed systems

The internal energy of condensed systems such as liquids and solids is primarily determined by enthalpy, allowing us to approximate as $U \approx H$. Enthalpy indicates energy associated with bonding. Entropy measures the degree of randomless in a system.

4 types of entropies

  1. Configurational entropy $S_c$ is defined by the number of distinguishable arrangements of atoms within a system.
  2. Thermal entropy $S_T$ quantifies the uncertainty in atomic positions. Systems with weaker atomic bonds and heavier atoms exhibit higher levels of thermal entropy.
  3. Electornic entropy $S_E$ quantifies energy distribution among electrons.
  4. Magnetic entropy $S_M$ measures ways of arranging net magnetic moments.

Thermodynamics of single component system

1st and 2nd derivatives of Gibbs free energy

Gibbs free energy is a function of temperature and pressure. We may measure them.

\[\begin{align} dG &= \left(\frac{\partial G}{\partial T}\right)_P dT + \left(\frac{\partial G}{\partial P}\right)_T dP, \\ dG &= - SdT + VdP. \end{align}\]

First derivatives:

\[\begin{align} \left(\frac{\partial G}{\partial T}\right)_p &= -S, \\ \left(\frac{\partial G}{\partial P}\right)_T &= V. \end{align}\] \[\]

Second derivatives:

\[\begin{align} \left(\frac{\partial^2 G}{\partial T^2}\right)_P &= - \left(\frac{\partial S}{\partial T}\right)_P, \\ \left(\frac{\partial^2 G}{\partial T^2}\right)_P &= - \left(\frac{C_p}{T}\right)_P. \end{align}\]

As entropy and heat capacity must be greater than or equal to zero, the first and the second derivatives of $G$ with respect to $T$ must be zero or negative.

Order of phase transition

A first-order transition occurs in a system where the first derivative of Gibbs free energy with respect to state variables such as temperature and pressure is discontinuous.

A second-order transition occurs in a system where the first and third derivatives are continuous, but the second derivative is not continuous.

Derive Gibbs from heat capacity

Recall the definition of heat capacity $C_P$:

\[C_P \equiv \left(\frac{\delta Q}{dT}\right)_P = T\left(\frac{dS}{dT}\right)_P.\]

At constant pressure, the relationship between heat capacity and enthalpy can be expressed, leading to:

\[H = TdS,\]

and

\[C_P = \left(\frac{dH}{dT}\right).\]

To derive the full expression of enthalpy as a function of temperature, and subsequently Gibbs free energy, we integrate over temperature:

\[H(T) = H_0(T_0) + \int_{T_0}^{T} C_P dT\] \[S(T) = S_0(T_0) + \int_{T_0}^{T} \frac{C_P}{T} dT\]

Gibbs free energy then can be computed by combinting entropy and enthalpy:

\[\begin{align} G(T) &= H(T) - T S(T), \\ G(T) &= H_0 - T S_0 + \int_{T_0}^{T} \frac{C_P}{T} dT - T \int_{T_0}^{T} \frac{C_P}{T} dT. \end{align}\]

Thermodynamic driving force for transformation

Here, an example of solidifcation of elemental liquids will be used to demonstrate the driving force for transformation.

\[\begin{align} ∆G= G_S - G_L < 0 \space \text{for} \space T < T_{MP} \end{align}\]

$\Delta G$ here is negative because $G_S$ is lower than $G_L$ below $T_{MP}$.

\[\begin{align} ∆G_{driving force} = | \space \Delta G \space| = G_L - G_S > 0 \space \text{for} \space T < T_{MP} \end{align}\]

We want to know $∆G_{driving force}$ any $T$ below $T_{MP}$. Use $∆G_{driving force} \equiv ∆G_{DR}$

\[\begin{align} ∆G_{DR} = G_L - G_S = (H_L - TS_L) - (H_S - TS_S) \end{align}\]

Recall

\[\begin{align} H(T) = H_0(T_0) + \int_{T_0}^{T} C_P dT \\ H_0(T_0) = H(T) - \int_{T_0}^{T} C_P dT \\ H_0(T_0) = H(T_{MP}) - \int_{T_0}^{T_{MP}} C_P dT \end{align}\]

We may determine $H_L(T)$ by subtracting the integral of heat capacity from the thermodynamic quanity at the melting point.

\[\begin{align} \therefore H_L(T) = H_L^{MP} - \int_{T_{u}}^{T_{MP}} C_{P_L} dT \\ H_S(T) = H_S^{MP} - \int_{T_{u}}^{T_{MP}} C_{P_S} dT \end{align}\]

Likewise

\[\begin{align} S_S = S_S^{MP} - \int_{T_{u}}^{T_{MP}} \frac{C_{P_{S}}}{T} dT \\ S_L = S_L^{MP} - \int_{T_{u}}^{T_{MP}} \frac{C_{P_{L}}}{T} dT \\ \end{align}\]

Find $∆H_F$ and $∆S_F$

\[\begin{align} ∆H_{Fusion} \equiv H_L^{MP} - H_S^{MP} \\ ∆S_{Fusion} \equiv S_L^{MP} - S_S^{MP} \end{align}\] \[\begin{align} G_S &= G_L \\ H_S - T_{MP}S_S &= H_L - T_{MP}S_L \\ H_L - H_S &= -T S_S + T S_L \\ ∆H_{Fusion} &= T ∆S_{Fusion} \\ \therefore ∆S_{Fusion} &= \frac{∆H_{Fuison}}{T_{MP}} \end{align}\]

Find the expression for $∆G_{DR}$

\[\begin{align} ∆G_{DR} (T_u) &= G_L - G_S \\ &= (H_L - T_u S_L) - (H_S - T_u S_S) \\ &= H_L^{MP} - \int_{T_{u}}^{T_{MP}} C_{P_L} dT - H_S^{MP} + \int_{T_{u}}^{T_{MP}} C_{P_S} dT \\ &\quad -T_u\left(S_L^{MP} - \int_{T_{u}}^{T_{MP}} \frac{C_{P_{L}}}{T} dT\right) + T_u\left(S_S^{MP} - \int_{T_{u}}^{T_{MP}} \frac{C_{P_{S}}}{T} dT\right) \\ &= ∆H_{Fusion} - \int_{T_u}^{MP}(C_{P_L} - C_{P_S}) dT - T_u\left(\frac{∆H_{Fusion}}{T_{MP}} - \int_{T_u}^{T_{MP}} \frac{(C_{P_L} - C_{P_S})}{T} dT \right) \end{align}\]

Here we assume the heat capacity of liquid and solid is the same.

\[\begin{align} ∆G_{DR}(T_u) &= ∆H_{Fusion} - T_u \frac{∆H_{Fusion}}{T_{MP}} \\ &= ∆H_{Fusion}\left(1 - \frac{T_u}{T_{MP}}\right) \\ & = ∆H_{Fusion}\left(\frac{T_{MP} - T_u}{T_{MP}} \right) \\ & = ∆H_{Fusion}\left(\frac{∆T}{T_{MP}} \right) \end{align}\]

Clausius Clapeyron equation

The following equation is used to determine the boundary slope in a P-T phase diagram for an uniary compound.

\[\begin{align} G_{\alpha} &= G_{\beta} \\ dG_{\alpha} &= dG_{\beta} \\ -S_\alpha dT + V_\alpha dP &= -S_\beta dT + V_\beta dP \\ (S_\beta - S_\alpha)dT &= (V_\beta - V_\alpha) dP \\ \frac{dT}{dP} &= \frac{∆V^{\alpha \to \beta}}{∆S^{\alpha \to \beta}} \\ \end{align}\]

Recall $H = TdS + VdP$ and at constnat pressure, $∆H = T∆S$.

\[\begin{align} \frac{dT}{dP} = \frac{T∆V^{\alpha \to \beta}}{∆H^{\alpha \to \beta}} \end{align}\]

Lecture 2. Thermodynamics of binary solutions

We are interested in mixing two pure components and determine the change in the overall Gibbs free energy.

\[\begin{align} G_{sol} &= G_{sep} + G_{mix} \\ G_{sol} &= X_AG_A^o + X_B G_A^o + ∆G_{mix} \\ G_{sol} &= X_A \overline{G}_A + X_B \overline{G}_B \\ \therefore ∆G_{mix} &= X_A \overline{G}_A + X_B \overline{G}_B - X_A G_A^o - X_B G_A^o \\ ∆G_{mix} &= X_A(\overline{G}_A - G_A^o) + X_B(\overline{G}_B - G_B^o) \end{align}\] \[\begin{align} ∆\overline{G}_A &\equiv (\overline{G}_A - G_A^o) \equiv RT \ln a_A = RT \ln \gamma_A X_A = RT \ln \gamma_A + RT \ln X_A \\ ∆\overline{G}_B &\equiv (\overline{G}_B - G_B^o) \equiv RT \ln a_B = RT \ln \gamma_B X_B = RT \ln \gamma_B + RT \ln X_B \\ \therefore ∆G_{mix} &= X_A ∆\overline{G}_A + X_B ∆\overline{G}_B \\ ∆G_{mix} &= X_A(RT \ln \gamma_A + RT \ln X_A) + X_B(RT \ln \gamma_B + RT \ln X_B) \\ \end{align}\]

Define the following:

\[\begin{align} ∆\overline{G}_K^{xs} &\equiv RT \ln \gamma_K \\ ∆\overline{G}_K^{id} &\equiv RT \ln X_K \\ \end{align}\] \[\begin{align} \therefore ∆G_{mix} &= X_A(∆\overline{G}_A^{xs} + ∆\overline{G}_A^{id}) + X_B(∆\overline{G}_B^{xs} + ∆\overline{G}_B^{id}) \end{align}\]

For the ideal solution model $∆\overline{G}_{K}^{xs} = 0$.

For the regular solution model $∆\overline{G}_{K}^{xs} \not ={0}$.

Recall the following equation:

\[∆G_{mix} = X_A RT \ln\gamma_A + X_A RT \ln X_A + X_B RT \ln\gamma_B + X_B RT \ln X_B\] \[\Omega X_A X_B \equiv X_A RT \ln\gamma_A + X_B RT \ln\gamma_B\]

Therefore, the Gibbs free energy of mixing for a regular solution ($∆G_{mix}^{reg}$) can be expressed as:

\[∆G_{mix}^{reg} = RT(X_A \ln X_A + X_B \ln X_B) + \Omega X_A X_B \\ ∆G_{mix}^{reg} = \Delta G_{mix}^{id} + \Delta G_{mix}^{xs}\]

Recalling the fundamental relationship $G = H - TS$, we can derive expressions for the entropy of mixing ($∆S_{mix}^{reg}$) and and the enthalpy of mixing ($∆H_{mix}^{reg}$) in a regular solution:

\[\begin{align} \Delta S_{mix}^{reg} &= \Delta S_{mix}^{id} + \Delta S_{mix}^{xs} = -R(X_A \ln X_A + X_B \ln X_B) \\ \Delta H_{mix}^{reg} & =\Delta H_{mix}^{id} + \Delta H_{mix}^{xs} = \Omega X_A X_B \end{align}\]

For an ideal solution, $\Delta H_{mix}^{id} = 0$.

\[\begin{align} \Delta S_{mix}^{xs} &= 0 \\ \Delta H_{mix}^{reg} &= \Delta H_{mix}^{xs} = \Omega X_A X_B \end{align}\]

Notice that positive $\Omega$ increases $\Delta G_{mix}^{reg}$. Higher Gibbs energy is not favored. Therefore, we can expect clustering will occur.

Activity coefficient for regular solution model

\[\begin{align} \Delta G_{mix}^{reg} &= \Omega(1 - X_A)^2 X_A + \Omega(1 - X_B)^2 X_B + X_A RT \ln X_A + X_B RT \ln X_B \\ \Delta G_{mix}^{reg} &= X_A RT \ln\gamma_A + X_B RT \ln\gamma_B + X_A RT \ln X_A + X_B RT \ln X_B \\ \end{align}\] \[\begin{align} \therefore \Omega(1 - X_A)^2 X_A &= X_A RT \ln\gamma_A \\ \therefore \Omega(1 - X_B)^2 X_B &= X_B RT \ln\gamma_B \\ \end{align}\]

Rearrange

\[\begin{align} \gamma_A^{reg} &= \left(\frac{\Omega(1 - X_A)^2}{RT}\right) \\ \gamma_B^{reg} &= \left(\frac{\Omega(1 - X_B)^2}{RT}\right) \\ a_A^{reg} &= X_A \gamma_A^{reg} \\ a_B^{reg} &= X_B \gamma_B^{reg} \end{align}\]

Derive $H$ in terms of $G$ and $T$

\[\begin{align} dG = -SdT + VdP \\ -\left(\frac{\partial G}{\partial T}\right)_P = S, \quad \left(\frac{\partial G}{\partial P}\right)_T = V \\ \end{align}\] \[\begin{align} G &= H - TS \\ \frac{G}{T} &= \frac{H}{T} - S \\ \frac{\partial}{\partial T} \left(\frac{G}{T} \right) &= \frac{\partial}{\partial T} \left(\frac{H}{T} -S \right)_P \\ &= H \frac{\partial}{\partial T}\left(\frac{1}{T}\right)_P + \frac{1}{T} \left( \frac{\partial H}{\partial T} - \frac{\partial S}{\partial T}\right)_P \\ &= H \frac{\partial}{\partial T}\left(\frac{1}{T}\right)_P + \frac{C_P}{T} - \frac{C_P}{T}\\ &= H \frac{\partial}{\partial T}\left(\frac{1}{T}\right)_P \\ &= -H T^{-2} \\ H &= -T^2 \frac{\partial}{\partial T}\left(\frac{G}{T}\right)_P \\ &= - \frac{d(G/T)}{d(1/T)} \end{align}\]

Derive partial molar Gibbs free energy

Graphically

Mathematically

Use the Gibbs Duhem equation to determine $dG_{sol}$

\[\begin{align} G'_{sol} &= \overline{G}_A n_A + \overline{G}_B n_B \\ dG'_{sol} &= \overline{G}_A dn_A + n_A d\overline{G}_A + \overline{G}_B dn_B + n_B d\overline{G}_b \\ 0 &= n_A d\overline{G}_A = n_B d\overline{G}_B \\ \therefore dG'_{sol} &= \overline{G}_A dn_A + \overline{G}_B dn_B \\ \end{align}\]

Get quantifiy per mole by dividing by $n_A + n_B$

\[\begin{align} G_{sol} &= X_A \overline{G}_A + X_B \overline{G}_B \\ G_{sol} &= X_A \left(\frac{\partial G_{sol}}{\partial X_A}\right) + X_B \left(\frac{\partial G_{sol}}{\partial X_B}\right) \end{align}\]

If we know $G_{sol}$ vs. $X_B$, we may obtain $\overline{G}_A$ and $\overline{G}_B$ graphically or via equations. IFor now, let’s focus on the mathmatically way.

First recall the definition:

\[\begin{align} X_1 + X_2 = 1 \\ dX_1 + dX_2 = 0 \\ dX_1 = -dX_2 \\ \end{align}\]

Start with the following

\[\begin{align} B &= X_1 \overline{B}_1 + X_2 \overline{B}_2 \\ B &= X_2 \overline{B}_1 + (1 - X_1) \overline{B}_2 \\ B &= -X_1 \overline{B}_1 + (1 - X_1) \overline{B}_2 \\ B &= \overline{B}_2 + X_1(\overline{B}_1 - \overline{B}_2) \\ \therefore \overline{B}_2 &= B - X_1(\overline{B}_1 - \overline{B}_2) \end{align}\] \[\begin{align} dB &= \overline{B}_1 dX_1 + \overline{B}_2 dX_2 \\ dB &= \overline{B}_1 dX_1 - \overline{B}_2 dX_1 \\ dB &= ( \overline{B}_1 - \overline{B}_2 ) dX_1 \\ \therefore \frac{dB}{dX_1} &= \overline{B}_1 - \overline{B}_2 \end{align}\]

Plug $\overline{B}_1 - \overline{B}_2$

\[\therefore \overline{B}_2 = B - X_1 \frac{dB}{dX_1}\]

The same process can be applied for $\overline{B}_1$

Lecture 3. Statistical Interpretation of Configurational Entropy

Random mixing and entropy

\[S = k_B \ln W_{conf}\]

where $W_{conf}$ is the number of distinguishable ways in which the atoms can be arranged. The value of $S$ after mixing is different.

\[W_{conf} = \frac{(N_A + N_B)!}{N_A! \space N_B!}\] \[S_{mixed} = k_B \ln \frac{(N_A + N_B)!}{N_A! \space N_B!}\]

Entropy is considered zero as pure materials.

\[\begin{align} S_{before} &= X_A S_A^o + X_B S_B^o \\ &= X_A \left(k_b \ln \frac{N_A^o !}{N_A^o!}\right) + X_B \left(k_b \ln \frac{N_B^o !}{N_B^o!}\right) \\ &= 0 \end{align}\]

Therefore

\[\begin{align} \Delta S_{mix}^{conf} = S_{mixed} = k_B \ln \frac{(N_A + N_B)!}{N_A! \space N_B!} \end{align}\]

Use the Stirling’s approximation.

\[\begin{align} \ln N! \approx N \ln N - N \end{align}\]

Use the approximation to simplify $\Delta S_{mix}^{conf}$

\[\begin{align} \Delta S_{mix}^{conf} &= k_B \left[\ln (N_A + N_B)! - \ln (N_A!) - \ln (N_B!)\right] \\ &= k_B \left[(N_A + N_B) \ln (N_A + N_B) - (N_A + N_B) - (N_A \ln N_A - N_A) - (N_B \ln N_B - N_B) \right] \\ &= k_B \left[(N_A + N_B) \ln (N_A + N_B) - N_A \ln N_A - N_B \ln N_B \right] \\ &= k_B \left[N_A (\ln (N_A + N_B) - \ln N_A) + N_B(\ln (N_A + N_B) - \ln N_B)\right] \\ &= k_B \left[N_A \ln \frac{N_A + N_B}{N_A} + N_B \ln \frac{N_A + N_B}{N_B}\right] \\ &= k_B \left[-N_A \ln \frac{N_A}{N_A + N_B} - N_B \ln \frac{N_B}{N_A + N_B}\right] \\ &= - k_B \left[N_A \ln X_A + N_B \ln X_B\right] \\ &= - k_B (N_A + N_B) \left[\frac{N_A}{N_A + N_B} \ln X_A + \frac{N_B}{N_A + N_B} \ln X_B\right] \\ &= - k_B N \left[X_A \ln X_A + X_B \ln X_B\right] \\ &= - R \left[X_A \ln X_A + X_B \ln X_B\right] \\ \end{align}\]

Now, the expression for $\Delta S_{mix}^{conf}$ is identical to $\Delta S_{mix}^{id}$ derived earlier. This reveals an ideal solution has an entropy contribution due to random mixing only.

Pair potential / Quasi chemical model / Nearest neighbor model

\[\begin{align} U_{initial} &= X_A U^o_A + X_B U^o_B \\ &= X_A P_{AA} \epsilon_{AA} + X_B P_{BB} \epsilon_{BB} \\ &= X_A\left[\frac{Z N_{Avo}}{2} \epsilon_{AA}\right] + X_B \left[\frac{Z N_{Avo}}{2} \epsilon_{BB}\right] \end{align}\]

$Z$ is the number of nearest neighbors.

\[\begin{align} U_{sol} = P_{AA} \epsilon_{AA} + P_{AB} \epsilon_{AB} + P_{BB} \epsilon_{BB} \end{align}\]

$P_{AB}$ is the number of bonds of each type in 1 mole of AB solution. $\epsilon_{AB}$ is the bond energy between A and B atoms.

\[\begin{align} P_{AA} &= X_A X_A \frac{N_{Avo}Z}{2} \\ P_{BB} &= X_B X_B \frac{N_{Avo}Z}{2} \\ P_{AB} &=\frac{X_A X_B N_{Avo}Z + X_A X_B N_{Avo}Z}{2} \\ P_{AB} &=X_A X_B N_{Avo}Z \end{align}\] \[\begin{align*} ∆H_{mix}^{QCM} &\approx U_{sol} - U_{ini} \\ &= X_A X_A \frac{N_{Avo}Z}{2} \epsilon_{AA} + X_B X_B \frac{N_{Avo}Z}{2} \epsilon_{BB} + X_A X_B N_{Avo}Z \epsilon_{AB} \\ &\quad - X_A\left[\frac{Z N_{Avo}}{2} \epsilon_{AA}\right] - X_B \left[\frac{Z N_{Avo}}{2} \epsilon_{BB}\right] \\ &= X_A X_A \frac{N_{Avo}Z}{2} \epsilon_{AA} - X_A\left[\frac{Z N_{Avo}}{2} \epsilon_{AA}\right] \\ &\quad + X_B X_B \frac{N_{Avo}Z}{2} \epsilon_{BB} - X_B \frac{N_{Avo} Z}{2} \epsilon_{BB} + X_A X_B N_{Avo}Z \epsilon_{AB} \\ &= N_{Avo} Z X_A X_B \left\{ \frac{X_A \epsilon_{AA}}{2X_B} - \frac{\epsilon_{AA}}{2X_B} + \frac{X_B \epsilon_{BB}}{2X_A} - \frac{\epsilon_{BB}}{2X_A} + \epsilon_{AB} \right\} \\ &= A \left\{ \frac{X_A \epsilon_{AA} - \epsilon_{AA}}{2 X_B} + \frac{X_B \epsilon_{BB} - \epsilon_{BB}}{2 X_A} + \epsilon_{AB} \right\} \\ &= A \left\{ \frac{X_A^2 \epsilon_{AA} - X_A \epsilon_{AA} + X_B^2 \epsilon_{BB} - X_B \epsilon_{BB}}{2 X_A X_B} + \epsilon_{AB} \right\} \\ &= A \left\{ \frac{X_A \epsilon_{AA} - \epsilon_{AA}}{2 X_B} + \frac{X_B \epsilon_{BB} - \epsilon_{BB}}{2 X_A} + \epsilon_{AB} \right\} \\ &= A \left\{ \frac{X_A^2 \epsilon_{AA} - X_A \epsilon_{AA} + X_B^2 \epsilon_{BB} - X_B \epsilon_{BB}}{2 X_A X_B} + \epsilon_{AB} \right\} \\ &= A \left\{ \frac{(1-X_B)^2 \epsilon_{AA} - (1-X_B)\epsilon_{AA}}{2(1-X_B)X_B} + \frac{X_B^2 \epsilon_{BB} - X_B \epsilon_{BB}}{2(1-X_B)X_B} + \epsilon_{AB} \right\} \\ &= A \left\{ \frac{\epsilon_{AA} (1-X_B) - \epsilon_{AA}}{2 X_B} + \frac{X_B \epsilon_{BB} - \epsilon_{BB}}{2(1-X_B)} + \epsilon_{AB} \right\} \\ &= A \left\{ -\frac{X_B \epsilon_{AA}}{2 X_B} + \frac{\epsilon_{BB}(X_B - 1)}{2(1-X_B)} + \epsilon_{AB} \right\} \\ &= N_{Avo} Z X_A X_B \left\{ -\frac{\epsilon_{AA}}{2} - \frac{\epsilon_{BB}}{2} + \epsilon_{AB} \right\} \end{align*}\]

Therefore, $\Omega$ using the quasi-chemical model is found from $∆H_{mix}^{reg} = \Omega X_A X_B$.

\[\begin{align} \Omega = N_{Avo} Z \left\{ -\frac{\epsilon_{AA}}{2} - \frac{\epsilon_{BB}}{2} + \epsilon_{AB} \right\} \end{align}\]

Slope of $\Delta G$ as $X_B$ approaches 0 or 1

Recall

\[\begin{align} \Delta G_{mix}^{reg} &= \Delta G_{mix}^{xs} + \Delta G_{mix}^{id} = H_{mix}^{reg} + TS_{mix}^{reg} \\ \Delta G_{mix}^{reg} &=\Omega X_B (1-X_A) + RT(X_A \ln X_A + X_B \ln X_B) \end{align}\]

Take the derivative with respect to $X_B$

\[\begin{align} \frac{d\Delta G_{mix}^{reg}}{dX_B} &= \frac{d\Delta H_{mix}^{reg}}{dX_B} - T \frac{d\Delta S_{mix}^{reg}}{dX_B} \\ &= \left[\Omega - 2\Omega X_B \right] + RT \left[ -\ln(1-X_B) + (1-X_B) \frac{-1}{(1-X_B)} + \ln X_B + 1\right] \\ &= \left[\Omega - 2\Omega X_B \right] + RT \ln \frac{X_B}{X_A} \\ \end{align}\]

As $X_B \to 0$, $X_B/X_A \to 0$, $\frac{d\Delta G_{mix}^{reg}}{dX_B} \to -\infty$

As $X_B \to 1$, $X_B/X_A \to \infty$, $\frac{d\Delta G_{mix}^{reg}}{dX_B} \to +\infty$

The above result indicates that absolute purirty is never favored. It is the $\Delta S_{mix}^{reg}$ term that is causing the slope to reach infinity. This shows that entropy is an important term.

Modification of Q.C.M for non-regular solution

Previously $∆S_{mix}^{reg} =∆S_{mix}^{id}$ and it was due to random mixing and the values of $P_{AA}$, $P_{AB}$, $P_{BB}$ were based on the assumption of random mixing.

However, this assumption is only true when $\Omega =0$. If $\Omega \not ={0}$, then the random mixing assumption is no longer valid. As mentioned, if $\Omega > 0$, clustering occurs where $P_{AB}^{NR} <P_{AB}^{random}$. If $\Omega < 0$, ordering occurs where $P_{AB}^{NR} > P_{AB}^{random}$.

The following is used to determine the ratio of $P_{AB}^{NR}$ to $P_{AB}^{random}$

\[\begin{align} \theta & \equiv \frac{P_{AB}^{NR}}{P_{AB}^{random}} \\ &= \frac{P_{AB}^{NR}}{X_A X_B N_{Avo} Z} \\ &\approx \left[1 - \frac{2 X_A X_B \Omega}{ZRT} \right] \end{align}\]

From the above equation, when $\Omega > 0$, clustering occurs. Fewer A-B bonds form. We expect $\theta < 1$ The sign flips when $\Omega < 0$.

Lecture 4. Equlibrium in heterogeneous systems

Equlibrium condition

Previously, a homogeneous system of two components was studied. Here, we introduce a heterogeneous system of two components. Hence, there are a total of four distinct entities: component A and component B in the $\alpha$ phase and component A and component B in the $\beta$ phase.

The folloiwng is the partial Gibbs free energy of component B in the $\alpha$ and $\beta$ phase.

\[\begin{align} \overline{G}_B^\alpha &= G_B^{\alpha, o} + RT \ln a_B^\alpha \\ \overline{G}_B^\beta &= G_B^{\beta, o} + RT \ln a_B^\beta \\ ∆\overline{G}_B^{\alpha \to \beta} &= \overline{G}_{B}^\beta - \overline{G}_B^\alpha \\ &= \left[G_B^{\beta, o} - G_B^{\alpha, o}\right] + RT \ln \frac{a_B^\beta}{a_B^\alpha} \end{align}\]

At equlibrium $∆\overline{G}B^{\alpha \to \beta} =0$ where $\overline{G}{B}^\beta = \overline{G}_B^\alpha$ or $\mu_B^\alpha = \mu_B^\beta$

\[\begin{align} \therefore G_B^{\alpha, o} - G_B^{\beta, o} = ∆G_B^{\beta \to \alpha, \space o} = RT \ln \frac{a_B^\beta}{a_B^\alpha} \end{align}\]

Gibbs phase rule

Derive the Gibbs phase rule of $P+F = C+2$.

Recall

\[\begin{align} \text {DoF} = \text{number of variables (V)} - \text{number of independent relations (R)} \end{align}\]

$P(C-1) + 2$ variables exist where P is the number of phases, C is the number of components, 2 comes from pressure and temperature. $C-1$ occurs is derived instead of $C$ is used. As an example, to find the composition of the last component in a 3 component system, the mole fraction sum of 1 allows us to to know the mole fraction of the thrid component from the first two.

$C(P-1)$ relations exist. An example below clarifies.

\[\begin{align} \overline{G}_1^\alpha = \overline{G}_1^\beta = \overline{G}_1^\gamma \end{align}\]

Although there are 3 phases, only two relations are indepednent where

\[\begin{align} \overline{G}_1^\alpha &= \overline{G}_1^\beta \\ \overline{G}_1^\beta &= \overline{G}_1^\gamma \\ \overline{G}_1^\alpha &= \overline{G}_1^\gamma \space \text{is dependent} \end{align}\]

Therefore,

\[\begin{align} DoF &= [P(C-2) + 2] - [C(P-1)] \\ & = PC - P + 2 - CP + C \\ & = C - P + 2 \end{align}\]

The above result is based on the condition that temperature and pressure are not fixed. If we fix pressure for example, then

\[\begin{align} DoF &= [P(C-2) + 1] - [C(P-1)] \\ & = PC - P + 1 - CP + C \\ & = C - P + 1 \end{align}\]

In the case of the eutetic point in a binary phase diagram where we can only control temperature $DoF = C - P + 1$. $P=3$ at the eutectic point.

\[\begin{align} DoF &= C - 3 + 1 \\ & = 2 - 3 + 1 \\ &= 0 \end{align}\]

Zero degrees of freedom at the eutetic point at fixed pressure for a binary solution.

Construction and analysis of ideal solution phase diagram

Here,

\[\begin{align} G_{sol}^S &= G_{sep}+ ∆G_{mix}^S \\ G_{sol}^S &= [X_A G_A^o + X_B G_B^o] + ∆G_{mix}^S \\ X_A \overline{G}_A^S + X_B \overline{G}_B^S &= [X_A G_A^o + X_B G_B^o] + ∆G_{mix}^S \\ \therefore ∆G_{mix}^S &= X_A∆\overline{G}_A^S + X_B ∆\overline{G}_B^S \end{align}\]

Similarly

\[\begin{align} ∆G_{mix}^L &= X_A∆\overline{G}_A^L + X_B ∆\overline{G}_B^L \end{align}\] \[\begin{align} ∆G_{mix}^i &= G_{mix}^i - G_{Standard \space Unmixed}^o \\ &= G_{mix}^i - (X_A G_A^o + X_B G_B^o) \end{align}\]

This section is inadequate.

Calculation of solidus and liquidus lines in ideal solution

Recall from Lecture 2. and $\gamma_K = 1$ for an ideal solution.

\[\begin{align} ∆\overline{G}_K = \overline{G}_K - G_K^o &= RT \ln a_K = RT \ln \gamma_K X_K \\ ∆\overline{G}_K = \overline{G}_K - G_K^o &= RT \ln X_K \\ \therefore \overline{G}_K &= RT \ln X_K + G_K^o \end{align}\]

If the phase is added,

Recall from Lecture 2. and $\gamma_K = 1$ for an ideal solution.

\[\begin{align} \overline{G}_B^{L} &= RT \ln X_B^{L} + G_B^{L,o} \\ \overline{G}_B^{S} &= RT \ln X_B^{S} + G_B^{S,o} \\ \end{align}\]

At equlibrium,

\[\begin{align} \overline{G}_B^L &= \overline{G}_B^S \\ G_B^{L,o} + RT \ln X_B^L &= G_B^{S,o} + RT ln X_B^S \\ G_B^{S, o} - G_B^{L, o} &= RT \ln \frac{X_B^L}{X_B^S} \end{align}\]

Recall we derived $∆G_{Driving \space Force}$ in Lecture 1.

\[\begin{align} ∆G_{DR}(T_u) &= G_L- G_S > 0 \\ ∆G_{DR}(T_u) &= ∆H_{Fusion}\left(1 - \frac{T_u}{T_{MP}}\right) \\ \therefore G_B^{S, o} - G_B^{L, o} &= -∆H_{Fusion}\left(1 - \frac{T_u}{T_{MP}}\right) \\ \end{align}\] \[\begin{align} RT \ln \frac{X_B^L}{X_B^S} &= - ∆H_{Fusion}\left(1 - \frac{T_u}{T_{MP}^B}\right) \\ \frac{X_B^L}{X_B^S} &= exp\left[- \frac{∆H_{Fusion}^B[1 - \frac{T_u}{T_{MP}^B}]}{RT}\right] \end{align}\]

The goal is to determine $X_B^L$ and $X_B^S$. Only 1 equation is currently available. The same process can be appliedt to A where $\mu_A^L = \mu_A^S$. We will be able to find the ratio of $(1-X_B^L)/(1-X_B^S)$. We have two equations, two unknowns. We can solve for the compositions where the two phase region occurs.

\[\begin{align} RT \ln \frac{X_A^L}{X_A^S} &= - ∆H_{Fusion}\left(1 - \frac{T_u}{T_{MP}^A}\right) \\ \frac{X_A^L}{X_A^S} &= exp\left[- \frac{∆H_{Fusion}^A[1 - \frac{T_u}{T_{MP}^A}]}{RT}\right] \\ \frac{1-X^L_B}{1-X^S_B} & = exp\left[- \frac{∆H_{Fusion}^A[1 - \frac{T_u}{T_{MP}^A}]}{RT}\right] \end{align}\]

Now, by knowing $∆H_{Fusion}, T_{MP}^A, T_{MP}^B$ we may determine $X_B^L, X_B^S$ at any temperature in theory. The main assumption used was the $C_{P_S} \approx C_{P_L}$ when we derived the $\Delta G_{Driving Force}$.

Calculation of solidus and liqudus lines in regular solution

In the previous section we considered $\gamma_K=0$. Here, $\gamma \neq0$.

Recall

\[\begin{align} ∆\overline{G}_K & = \overline{G}_K - G_K^o \equiv RT \ln a_K = RT \ln \gamma_K X_K\\ &= RT \ln \gamma_K + RT \ln X_K \\ &= ∆\overline{G}_K^{XS} + ∆\overline{G}_K^{Id} \end{align}\]

Similary to what we did in the previous section, let us use the equlibrium condition of $\mu_B^L =\mu_B^S$.

\[\begin{align} \overline{G}_B^ \end{align}\]