Applications of Statistical Mechanics

Statistical mechanics has been built from microscopic physics. Thermodynamics is done based on physical models. Get entropies of mixing, applications, and formulation in solution theory. Consider the physical origins of entropy and energy. Irreversible thermodynamics is the link to kinetics and is picked up in 3.21.

Thermodynamics of Open Systems

Why are open systems important? Systems are often closed, yet we want to deal with multi-phase systems. The two phases are open with respect to each other. Species can be exchanged. The concept of a mole number was introduced.

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</center>

Partial molar quantities have been defined, such as <math>\bar V_i</math>, <math>\bar S_i</math>, and <math>\bar M_i</math>. A general form is below. And the partial molar Gibbs free energy is particularly important.

<math>\bar Y_i = \left ( \frac

Unknown macro: {partial Y}

Unknown macro: {partial n_i}

\right ){T, P, n

Unknown macro: {j ne i}

</math>

<math>\bar G_i = \left ( \frac

Unknown macro: {partial G}

\right ){T, P, n

Unknown macro: {j ne i}

</math>

<math>\bar G_i = \mu_i \left ( T, P, n_j \right )</math>

</center>

This fit in the property matrix and is an equation of state. Most of the next lectures will be devoted to this. Equilibrium requires the chemical potential to be the same everywhere. The equilibrium condition is expressed below, where the Greek letters denote phases and the Roman letters refer to species.

<math>\mu_i^

Unknown macro: {alpha}

= \mu_i^

Unknown macro: {beta}

</math>

</center>

There is a need of an equation of state. Build two types of models: empirical or heurestic and physics based. There is a lot of nomenclature.

Empirical Descriptions

The activity is referred to as <math>a_i</math>, and there is a relation below. The temperature and pressure are generally parameters of the activity.

<math>\mu_i \left (T, P, X_j \right ) = \mu_i^

Unknown macro: {circ}

(T, P) + RT \ln a_i \left ( X_i \right )</math>

</center>

The chemical potential is defined in terms of a reference state, which is of pure <math>i</math> in the same phase. The correction factor accounts for the deviation away from <math>i</math>. Look at the chemical potential in a liquid. When the substance is pure, the activity is equal to one, and the chemical potential is equal to <math>\mu_i^

</math>. There is no information gained or lost in writing the chemical potential in the form above. There is merely a transfer of ignorance.

Consider the equilibrium relation below. This is if the reference states are the same for <math>\alpha</math> and <math>\beta</math>.

<math>a_i^

= a_i^

Unknown macro: {beta}

</math>

</center>

The parametric dependence is weak. The temperature and pressure contained in the reference activity captures compositional differences. A hope is the the activity captures composition dependence. An equation with the chemical activity written in terms of the standard state is written below.

<math>\mu_i \left (T, P, X_j \right ) = \mu_i^

Unknown macro: {circ}

(T_o, P_o) + RT \ln a_i \left ( X_i \right )</math>

</center>

The chemical potential does not show a strong pressure dependence. The order of magnitude of volume is small. Below is an expression.

<math>\frac

Unknown macro: {partial mu_i}

Unknown macro: {partial p}

= \bar V_i</math>

</center>

Find Models

Raoultian Behavior

If the reference state is one where the term <math>x_i</math> is one, the following is true.

</center>

Consider the free energy dependence of ideal gas mixtures.

</center>

The free energy can depend on <math>x_i P</math>.

Henry's Law

In the 1940s, it was seen that this doesn't work. A correction factor was introduced, and systems that display linear behavior follow Henry's law.

</center>

This cannot work over all composition ranges. When the term <math>x_i</math> is equal to one, the activity must be equal to one, which causes the term of the natural logarithm to disappear and the chemical potential to be equal to the reference state. Henry's law is true in the dilute limit. When the term <math>x_i</math> is equal to zero, the term of the natural logarithm diverges and the chemical potential approaches negative infinity.

Example

Consider a system of <math>A</math> and <math>B</math>. If component <math>B</math> follows Henry's law, component <math>A</math> is Raoultian. This can be proved from the Gibbs-Duhem relation, which states that the change in intensive variables is constant or that the variation in intensive variables is related.

</center>

Consider how the chemical potential varies with composition.

<math>d \mu_B = RT d \ln \left ( k_B x_B \right )</math>

<math> d \mu_B = RT \frac

Unknown macro: {d x_B}

Unknown macro: {x_B}

</math>

<math>x_A R T d \ln a_A = - x_B \frac

Unknown macro: {R T d x_B}

</math>

<math>x_A R T d \ln a_A = RT \frac

Unknown macro: {dx_A}

Unknown macro: {x_A}

</math>

<math>d \ln a_A = \frac

Unknown macro: {x_A}

</math>

</center>

Henry's law holds in the dilute limit. This is where <math>x_A</math> is one, which corresponds to <math>a_A</math> being equal to one. This means that <math>k_A</math> is equal to one.

</center>

Real Activities

Consider the activity in a system of platinum and nickel. When the composition of the system is close to pure platinum or pure nickel, the behavior of the activity is nearly linear. The system approaches linearity there more than anywhere else.

Ideal solution

Consider a solution in which all the components are Raoultian. If one component in a solution is Raoultian, than all components are. This is a hard proof. To be Raoultian means that one atom does not interact with others. If component <math>a</math> is Raoultian and does not interact with <math>b</math> and <math>c</math>, than <math>b</math> and <math>c</math> don't interact. Placing <math>a</math> into the system breaks up interactions.

Free energy of mixing

Consider the change in free energy due to mixing. When considering the free energy of the mixture, sum over the partial molar quantities.

<math>\Delta G_

Unknown macro: {mix}

= G_

Unknown macro: {mixture}

G_
Unknown macro: {pure comp}
</math>

<math>G_

= x_A \mu_A + x_B \mu_B</math>

<math>\Delta G_

= x_A \left ( \mu_A^

+ RT \ln x_A \right ) + x_B \left ( \mu_B^

Unknown macro: {circ}

+ RT \ln x_B \right ) - G_

Unknown macro: {pure comp}

</math>

<math>\Delta G_

Unknown macro: {mix}

= RT \left [x_A \ln x_A + x_B \ln x_B \right]</math>

Unable to render embedded object: File (Delta_G_mix_with_components_A_and_B.PNG) not found.

</center>

Differentiate the expression of <math>\Delta G_

</math> and consider the slope at pure compositions. The derivative diverges. If a species behaves ideally, there is always a driving force to mix. The system mixes at every composition.

Entropy change of mixing

An expression of the entropy in terms of the free energy and temperature is below.

<math>S = \frac

Unknown macro: { - partial G}

Unknown macro: {partial T}

</math>

<math>\Delta \underline

Unknown macro: {S}

_

Unknown macro: {mix}

= -R \left [x_A \ln x_A + x_B \ln x_B \right]</math>

Unable to render embedded object: File (Delta_S_mix_with_components_A_and_B.PNG) not found.

</center>

Enthalpy change of mixing

The change in enthalpy is expressed below.

<math>\Delta H = \Delta G + T \Delta S</math>

<math>\Delta H = 0</math>

</center>

There is no enthalpy of mixing, which means there is no heat of mixing corresponding to heat being absorbed or released. But there is a <math>\Delta S</math> term. Isn't heat equal to temperature times the change in entropy? This holds only for reversible processes, and the fact that there is no enthalpy of mixing means that the process is highly irreversible.

Summary

Consider the volume of mixing. An expression of volume is written below.

<math>V = \frac

Unknown macro: {partial G}

Unknown macro: {partial P}

</math>

<math>\Delta V = \frac

Unknown macro: {partial Delta G}

</math>

</center>

When there is no pressure dependence, there is no volume of mixing. Yet the ideal solution is applied and the volume of mixing is worked with.

<math>\Delta H = \Delta U + p \Delta V_

</math>

</center>

The terminology is poor. A better description that "ideal solution" may be "simplistic solution". The model of an ideal solution is not very practical.

Applications

Example 1

Calculate solubility limit when adding bismuth to lead. When does solid bismith precipitate? Below is provided data.

Unable to render embedded object: File (Adding_Bi_to_mixture_of_Pb_and_Bi.PNG) not found.

<math>\mbox

Unknown macro: {Bismuth}

</math>

<math>\Delta \underline H_m = 11.3 \mbox

Unknown macro: {kJ/mol}

</math>

<math>\mbox

Unknown macro: {in liquid}

</math>

<math>a_

Unknown macro: {Bi}

= x_

</math>

<math>\mbox

Unknown macro: {Pb does not dissolve in solid Bi}

</math>

</center>

How is it determined whether the solid dissolves? Species move where the chemical potential is lower, which results in a lowering of the free energy. Bismuth dissolves as long as the chemical potential in the liquid is lower than in the solid. The chemical potential becomes equal at the solubility limit.

<math>\mu_

Unknown macro: {Bi}

^l = \mu_

^s</math>

</center>

Change the condition to reach the solubility limit to composition variables. Define as an ideal solution.

<math>\mu_

Unknown macro: {Bi}

^l = \mu_

^

Unknown macro: {l, circ}

+ RT \ln x_

Unknown macro: {Bi}

^l</math>

<math>\mu_

^s = \mu_

Unknown macro: {Bi}

^

Unknown macro: {s, circ}

</math>

<math>\mu_

^

Unknown macro: {l,circ}

- \mu_

Unknown macro: {Bi}

^

Unknown macro: {s,circ}

= -RT \ln x_

^l</math>

<math>\Delta G_

Unknown macro: {pure Bi}

^

Unknown macro: {m}

\approx \Delta H_m \left ( 1 - \frac

Unknown macro: {T}

Unknown macro: {T_m}

\right )</math>

<math>x_

Unknown macro: {Bi}

^l = 0.7</math>

</center>

About <math>70%</math> is dissolved before precipitating out. The prediction of the simple model is close to the actual value. The data used to calculate this value, such as the enthalpy of melting, is easily accessible. The Raoultian approximation was a first guess.

What happens to the solubility limit if lead does not dissociate in bismuth? Consider what happens to the chemical potential. The chemcial potential goes down. An upper bound was found. Allowing lead to go into bismuth results in less of a driving force. The solubility limit goes down. (compositional equilibrium)

Example 2

Consider the general equilibrium between two phases. There are four composition variables. Write an equilibrium condition. Both species are mobile.

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</center>

Write these two expressions in terms of activities. Use model of activity and find a relation between compositions. The procedure to find an expression for species <math>A</math> is the same.

<math>\mu_B^s = \mu_B^

Unknown macro: {s, circ}

+ RT \ln a_B^s</math>

<math>\mu_B^l = \mu_B^

+ RT \ln a_B^l</math>

<math>\mu_B^

Unknown macro: {l, circ}

\mu_B^
Unknown macro: {s, circ}
= RT \ln \frac
Unknown macro: {a_B^s}

Unknown macro: {a_B^l}
</math>

<math>\Delta G_

Unknown macro: {mu, B}

= \mu_B^

- \mu_B^

Unknown macro: {s, circ}

</math>

<math>\frac{\Delta G_{m, B}}

Unknown macro: {RT}

= \ln \frac

Unknown macro: {a_B^s}

Unknown macro: {a_B^l}

</math>

<math>\frac{\Delta G_

Unknown macro: {m, A}

}

= \ln \frac

Unknown macro: {a_A^s}

Unknown macro: {a_A^l}

</math>

</center>

There are two equations that define equilibrium to two others. There are certain assumptions about ideality. They are ideal solutions.

Case 1

Consider the melting temperature as a function of composition. When the temperature is below the melting point of pure <math>A</math> and pure <math>B</math>, the free energies of melting are positive.

<math>T < T_

Unknown macro: {m(A)}

< T_

Unknown macro: {m(B)}

</math>

</center>

No solution is possible. The physical condition imposed is not achievable. There is no equilibrium between the liquid and solid above the melting point of <math>B</math> and below the melting point of <math>A</math>.

Case 2

Consider when the temperature is between the melting temperature of pure <math>A</math> and pure <math>B</math>. There is an allowable solution between the two melting points.

<math>T_

< T < T_

Unknown macro: {m(B)}

</math>

<math>\Delta G_

Unknown macro: {m(A)}

< 0</math>

<math>\frac

Unknown macro: {x_A^s}

Unknown macro: {x_A^l}

< 1</math>

<math>\Delta G_

> 0</math>

</center>

The composition of <math>B</math> in the solid is larger. Assume that both the liquid and solid are ideal solutions. If species mix very well, the two phases behave very well. The interpolation is what is expected, but it is unusual to see this. It is rare that solids are miscible in the solid state.

Unable to render embedded object: File (Phase_diagram_of_two_components.PNG) not found.

</center>

Consider an mixed liquid of <math>A</math> and <math>B</math>. Write the equilibrium condition for <math>A</math>.

<math>\frac{\Delta G_{m, A}}

Unknown macro: {RT}

= \ln \frac

Unknown macro: {a_A^l}

</math>

<math>\frac{\Delta G_{m, A}}

Unknown macro: {RT}

= \Delta H_m \left ( 1 - \frac

Unknown macro: {T}

Unknown macro: {T_m}

\right )</math>

<math>\frac{\Delta H_{m(A)}}

Unknown macro: {R}

\left ( \frac

Unknown macro: {T_m - T}

Unknown macro: {T T_m}

\right ) \approx x_B^l</math>

<math>T_m - T = - \Delta T</math>

<math>T \cdot T_m \approx T_m^2</math>

<math>x_B^l = - \frac{\Delta H_{m(A)}}

Unknown macro: {R T_m^2}

\Delta T_m</math>

</center>

All the terms are positive, and the change in temperature must be negative. There are no properties of <math>B</math> in the expression. Liquid and solid remain in equilibrium by decreasing temperature. Under the conditions above, the melting point is always depressed. Lower the melting point by adding <math>B</math>. Most species lower the melting point when added. Dirt lowers the melting temperature of water; salt dissolves easily and the magnitude of the decrease of the melting temperature is greater.

In liquid, there are more degrees of freedom. Consider a more realistic assumption.

Child pages

Lecture 15

Applications of Statistical Mechanics

Thermodynamics of Open Systems

Empirical Descriptions

Find Models

Raoultian Behavior

Henry's Law

Example

Real Activities

Ideal solution

Free energy of mixing

Entropy change of mixing

Enthalpy change of mixing

Summary

Applications

Example 1

Example 2

Case 1

Case 2