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<center>
h1.  Lecture 14 
</center>

The [Clausius-Clapeyron|http://en.wikipedia.org/wiki/Clausius-Clapeyron] equation can be used to develop equations of lines on the [phase diagram|http://en.wikipedia.org/wiki/Phase_diagram], and there is a special meaning of the phase boundaries in contact with the vapor.  They tell the [vapor pressure|http://en.wikipedia.org/wiki/Vapor_pressure] above a solid or liquid.

<br>

<center>
!Phase_diagram.PNG!
</center>

<br>

The vapor pressure is a property of the [condensed phase|http://en.wikipedia.org/wiki/Condensed_matter_physics].  It is the pressure above the solid or liquid at [equilibrium|http://en.wikipedia.org/wiki/Thermodynamic_equilibrium].  A [partial pressure|http://en.wikipedia.org/wiki/Partial_pressure] of a material will be established in a vapor phase above a condensed phase.

<br>

<center>
!Reaching_equilibrium_vapor_pressure_--_diagram.PNG!
!Reaching_equilibrium_vapor_pressure.PNG!
</center> 

Imagine a pure system of a liquid in a container at zero pressure.  The vapor phase should be stable, and the vapor reaches an equilibrium pressure.  The concept of a partial pressure is useful if the vapor is not pure.  If there are other species, the partial pressure is close to the vapor pressure.

<p>
</p>

At equilibrium, the [chemical potential|http://en.wikipedia.org/wiki/Chemical_potential] of the liquid and gas will be equal.

<center>
<br>
<math> \mu_i^L = \mu_i^V </math>
</center>

<br>

The vapor pressure will change if the chemical potential changes due to other species.  In [ideal gas|http://en.wikipedia.org/wiki/Ideal_gas] mixtures, the chemical potential doesn't change.  The effect of other species is zero; the species behave as if the others are not present.  When does <math> \mu_i^L </math> change?  In a pure case, there is pressure on the liquid due to the vapor pressure.  In a mixed case, the total pressure is higher due to the presence of other species.  The molar chemical potential, <math>\underline{\mu}_i</math> is related to the molar free energy, <math>\underline{G}</math>, and below is the derivative of the free energy with respect to pressure.

<center>
<br>
<math> \frac{\partial G}{\partial P}_T  =  \underline{V} </math>
</center>

<br>

This effect is small.  The total pressure does affect the chemical potential, but the effect is not large.  There can be a good approximation by using the graph.  If [cadmium|http://en.wikipedia.org/wiki/Cadmium] were to be melted in atmosphere, one would want to know the vapor pressure, and one could use the value of vapor pressure of the pure substance even though all the other gases in the atmosphere are present.  This is covered in Ragone.

<p>
</p>



*Clausius-clapeyron with vapor*

It is useful to estimate the form of lines on the phase diagram.  The equations can take on a simple form.

<center>
<br>
<math> \frac{\Delta H}{T \Delta V}  =  \frac{dP}{dT} </math>
</center>

<br>

There are two possible approximations, and one involves <math>\Delta V</math>, which is the volume difference between two phases.

<center>
<br>
<math> \Delta \underline{V} = \underline{V}_{vapor} - \underline{V}_{l} </math>
</center>

<br>

The volume of the vapor is about a thousand times bigger than the volume of the liquid.  It is possible to neglect the volume of the condensed phase.

<p>
</p>

It is possible to get rid of a variable in the equation by considering the gas to be ideal.  Below is an ideal gas equation, and it is substituted into the Clapeyron equation.

<center>

<br>

<math> \underline{V} = \frac{RT}{P} </math>

<br>

<math> \frac{dP}{dT} = \frac{P \Delta H}{R T^2} </math>

<br>

</center>

[Separate variables|http://en.wikipedia.org/wiki/Separation_of_variables], and there is not a dependence in the final equation on the volume of transformation.

<center>

<br>

<math> \frac{dP}{P} = \frac{\Delta H}{R} \frac{dT}{T^2} </math>

<br>

<math> \frac{dT}{T^2} = -d \left ( \frac{1}{T} \right ) </math>

<br>

<math> \frac{dP}{P} = \frac{-\Delta H}{R} d \left ( \frac{1}{T} \right ) </math>

<br>
</center>

Vapor pressure is tabulated, and the differential equation can be integrated under two different assumptions.  The change in enthalpy, <math>\Delta H</math>, can be considered constant or linear.  Below is the result when <math>\Delta H</math> is considered constant.

<center>

<br>

<math> \Delta H \ne f(T) </math>

<br>

<math> ln P = - \frac{\Delta H}{R}  \frac{1}{T} + C </math>

<br>
</center>

Below is a graph showing the linear relation of <math>ln P</math> and  <math>\frac{1}{T}</math>.

<center>
!Ln_P_vs_T%5E-1.PNG!
</center>

<br>

<p>
</p>

The most often tabulated form of the vapor pressure is the result of assuming a linear dependence of <math>\Delta H</math>.  Values of <math>\Delta H</math> and <math>C_p</math> can be extracted by taking appropriate derivatives.

<center>

<br>

<math> \Delta H = \Delta H(298) + \Delta C_p(T-298) </math>

<br>

<math> ln P = A \frac{1}{T} + B lnT + C </math>

<br>
</center>

*Example -- water*

<p>
</p>

Consider the vapor pressure of water at <math> 25^\circ C (298 K)</math>.

<center>

<br>

<math> P^*(298) = 0.03 \mbox{bar} </math>

<br>

!Cup_of_water_and_vapor_pressure.PNG!

</center>

There is three percent water in [saturated air|http://en.wikipedia.org/wiki/Boiling_point] at room temperature.  The vapor pressure of water goes up exponentially with temperature.  In warm air, there is much more water.

<center>

<br>

<math> P^*(313) = 0.07 \mbox{bar} </math>

<br>
</center>


This introduces the concept of [relative humidity|http://en.wikipedia.org/wiki/Relative_humidity].  Relative humidity is the water in air divided by the [saturation pressure|http://en.wikipedia.org/wiki/Boiling_point] multiplied by <math>100%</math>.

<center>

<br>

<math> RH = \frac{P_{H_2O}}{P^*} \cdot 100% </math>

<br>

</center>

A vapor pressure of water of <math>0.07 \mbox{bar}</math> at <math>25^\circ C</math> is equal to one hundred percent humidity, while the same pressure at <math>40^\circ C</math> is equal to forty-two percent humidity.  This is near the dry limit, which is arbitrarily defined at forty percent.  The wet limit is above seventy percent.

<p>
</p>

It can [rain|http://en.wikipedia.org/wiki/Rain] when saturated air drops in temperature, which drops the saturation pressure, <math> P^* </math>.  It can rain when the pressure is above one hundred percent.  Similarly, on a hot day, it feels more sticky at the end of the day.  The water content stays constant which the relative humidity increases.

*Example -- mass transport*

A difference in vapor pressure caused by a [temperature gradient|http://en.wikipedia.org/wiki/Temperature_gradient] can cause [mass transport|http://en.wikipedia.org/wiki/Mass_transport].  Imagine an ice cube in a freezer.  The ice cube tries to establish <math>P_i^*</math> in the interior.  Heat is extracted at the side of the freezer, and the temperature is lower there.  The saturation pressure is lower at the side of the refrigerator, and ice forms due to mass transport.  The ice cubes shrink, and the minerals do not evaporate.  This is a form of [distillation|http://en.wikipedia.org/wiki/Distillation]; it is done at cold temperatures.  This is the same reason that water condenses on a glass on a hot day.  There is a region of low saturation pressure near the glass.  This effect can be utilized in instrumentation.  A "[cold finger|http://en.wikipedia.org/wiki/Cold_finger]" can be used to keep an environment free of species.  Liquid nitrogen can be used to create a cold spot in the chamber, and unwanted species in a gas can condense and be kept away.

<center>
!Freezer_and_ice.PNG!
</center>

<p>
</p>

*"Strange" thermodynamic properties*

<p>
</p>

Consider the [heat capacity|http://en.wikipedia.org/wiki/Heat_capacity] of saturated air.  Calculate the heat capacity or the amount of heat needed to be extracted to cool one-hundred percent humid air from <math>25^\circ C</math> to <math>20^\circ C</math>.  The vapor at a pressure of <math>0.03 \mbox{bar}</math> is above the saturation pressure, and water vapor [condenses|http://en.wikipedia.org/wiki/Condense] when cooling.  There are several components involved when caclulating the heat capacity, including the process of the air cooling five degrees, the cooling of any liquid, and the extraction of heat of condensation, <math>\Delta H</math>.  The heat capacity is considerably larger than what would be calculated if all these processes were not included.  The heat capacity can also depend on the speed of heating or cooling.  If water doesn't evaporate during the process, then there is no need to consider the evaporation process when calculating the heat capacity.  In general, when there are two different phases coexisting in equilibrium, there is a difference in heat capacity.  Changing the temperature changes the equilibrium, and there are several process to evaluate.

<p>
</p>

<center>
!Cooling_saturated_vapor_from_25_to_20_degrees.PNG!
</center>

*Example -- pressure applied to two phases*

<p>
</p>

Consider a material in which <math>\alpha</math> and <math>\beta</math> are in equilibrium.  Change the volume by changing the pressure.  If the equilibrium shifts, the response function is not just an average of the response functions of the two materials.

<p>
</p>

<center>
!Alpha_beta_hydrostatic_pressure.PNG!
</center>

*Statistical Mechanics*

<p>
</p>

[Statistical mechanics|http://en.wikipedia.org/wiki/Statistical_mechanics] involves relating microscopic and macroscopic properties.  A plan is to start statisitical mechanics, and all the material will be taken together and models will be discussed.  There will be a formalism of solution theory after statistical mechanics.

<p>
</p>

*Solution theory*

<p>
</p>

Look at the variables and functions that are relevant to [open systems|http://en.wikipedia.org/wiki/Open_system_%28system_theory%29].  [Extensive variables|http://en.wikipedia.org/wiki/Extensive_variable] are <math>S</math>, <math>V</math>, and <math>n_i</math>.  Below are differentials of the [internal energy|http://en.wikipedia.org/wiki/Internal_energy].  
<center>

<br>

<math> dU = TdS - pdV + \sum_{i} \mu_i dn_i</math>

<br>

</center>

The chemical potential is an equation of state.

<center>

<br>

<math> \mu_i = \left ( \frac{\partial U}{\partial n_i} \right )_{S, V, n_{j\ne i}}</math>

<br>

</center>

Consider three more definitions of chemical potential that are all equal to that above.  Below is a relation derived from the [Enthalpy|http://en.wikipedia.org/wiki/Enthalpy].

<center>

<br>

<math> H = U + PV</math>

<br>

<math> dH = TdS + VdP + \sum_{i} \mu_i dn_i</math>

<br>

<math> \mu_i = \left ( \frac{\partial H}{\partial n_i} \right )_{S, P, n_{j\ne i}}</math>

<br>

</center>

Consider the [Hemholtz free energy|http://en.wikipedia.org/wiki/Helmholtz_free_energy]

<center>

<br>

<math> F = U - TS</math>

<br>

<math> dF = -SdT - PdV + \sum_{i} \mu_i dn_i</math>

<br>

<math> \mu_i = \left ( \frac{\partial F}{\partial n_i} \right )_{T, V, n_{j\ne i}}</math>

<br>

</center>

Below is a derived relation from the [Gibbs free energy|http://en.wikipedia.org/wiki/Gibbs_free_energy]

<center>

<br>

<math> G = U - TS + PV</math>

<br>

<math> dG = -SdT + VdV + \sum_{i} \mu_i dn_i</math>

<br>

<math> \mu_i = \left ( \frac{\partial G}{\partial n_i} \right )_{T, P, n_{j\ne i}}</math>

<br>

</center>

Different variables are held constant in these definitions of chemical potential.  Work into others by substitution.

<p>
</p>

*Euler relation on open systems*

<p>
</p>

Use the Euler relation on open systems.  Equations of the internal energy and Gibbs free energy are below.  The sum of the chemical potential times the mole number remains


<center>

<br>

<math> U = TS - PV + \sum_{i} \mu_i n_i</math>

<br>

<math> G = U - TS + PV</math>

<br>

<math> G = \sum_{i} \mu_i n_i</math>

<br>

</center>

The chemical potential contributes to the total free energy; there is a weighted sum.

<p>
</p>

Define a potential of a system that is open with respect to all species and where there is full control of all the [intensive variables|http://en.wikipedia.org/wiki/Intensive_quantity].  The potential is equal to zero.  There is a relation between intensive quantities; they are not all independent.

<p>
</p>

<center>

<br>

<math> \Phi = G - \sum_{i} \mu_i n_i = 0</math>

</center>

<br>

*Equilibrium of two-phase system*

<p>
</p>

Consider a system of <math>\alpha</math> and <math>\beta</math> at equilibrium.  There is exchange between species but there is no transfer out.  The total system is closed at constant <math>T</math> and <math>P</math>.  The differential of the free energy is below the diagram

<center>

!Alpha_beta_constant_T%2C_P.PNG!

<math>dG = dG^\alpha + dG^\beta</math>

<br>

<math>dG^\alpha = -S^\alpha dT + V^\alpha dP + \sum_{i} \mu_i^\alpha n_i^\alpha</math>


</center>

Since temperature and pressure are at equilibrium, the terms <math>dT</math> and <math>dP</math> are zero.

<center>

<br>

<math>dG = \sum_{i} \mu_i^\alpha dn_i^\alpha + \sum_{i} \mu_i^\beta dn_i^\beta</math>

<br>

<math>dn_i^\alpha = -dn_i^\beta </math>

<br>

<math>dG = \sum_{i} (\mu_i^\alpha - \mu_i^\beta) dn_i^\alpha </math>

<br>

</center>

The free energy is minimal when  <math>\mu_i^\alpha = \mu_i^\beta </math>.  This is the equilibrium condition, and this is a similar derivation as seen previously. A first remark is that it is neither <math>G^\alpha</math> or <math>G^\beta</math> that is optimized by minimization.  The relevant thermodynamic potential is <math>G^{total} = G^\alpha + G^\beta</math>.  Secondly, there is no equilibrium condition for a species that is not mobile.  This is useful to remember when looking for the number of equilibrium conditions with the [phase rule|http://en.wikipedia.org/wiki/Phase_rule].  Also, it is possible to solve the problem by treating <math>\alpha</math> and <math>\beta</math> as [open systems|http://en.wikipedia.org/wiki/Open_system_%28system_theory%29] where <math>\alpha</math> is in equilibrium with <math>\beta</math>

<p>
</p>

*Equilibrium of open system with environment*

<p>
</p>

Look at the most general equation that applies to an open system in an environment.

<center>

<br>

<math>dU - T^*dS + P^*dV - \sum_{i} \mu_i^* dn_i \le 0</math>

<br>

</center>


When the system is in equilibrium <math>T^* = T</math> and <math>P^* = P</math>.  Consider one species and plug in <math>dU</math>, and divide by <math>dn_i</math>

<center>

<br>

<math>dG - \mu_i^* dn_i \le 0</math>

<br>

<math> \left ( \frac{\partial G}{\partial n_i} \right )_{T,P,n_{j \ne i}} - \mu_i^* \le 0</math>

<br>

</center>


The whole analysis is under constant temperature and pressure.  A condition of species <math>i</math> flowing into the system is <math>dn_i \ge 0</math>, and this occurs when <math>\mu_i \le \mu_i^*</math>.  Material flows to where the chemical potential is the lowest.

<p>
</p>

*Equations of state in an open system*

<p>
</p>

Perform a [Legendre transform|http://en.wikipedia.org/wiki/Legendre_transform] and write in terms of intensive variables we control.

<center>

<br>

<math>U(S,V,n_i) \rightarrow G(T,P,n_i)</math>

<br>

<math>dG = -SdT + VdP + \sum_{i} \mu_i^dn_i</math>

<br>

</center>

There are three [equations of state|http://en.wikipedia.org/wiki/Equation_of_state]

<center>

<br>

<math>S = - \left ( \frac{\partial G}{\partial T} \right )_{P, n_i} \rightarrow S(T, P, n_i)</math>

<br>

<math>U = - \left ( \frac{\partial G}{\partial P} \right )_{T, n_i} \rightarrow V(T, P, n_i)</math>

<br>

<math>\mu_i = - \left ( \frac{\partial G}{\partial n_i} \right )_{T, P, n_{j \ne i}} \rightarrow \mu_i(T, P, n_i)</math>

<br>

</center>

The chemical potential depends on all composition variables.  This is important because there is a link between equilibrium and the composition that results.  Practically, we want to know the composition of the system.  The relation between <math>\mu_i</math> and <math>n_i</math> is the essence of thermodynamics.  The chemical potential can vary, and there can be a shift in the equilibrium condition.  This last equation of state is crucial.  

<p>
</p>

Call in properties with the equations of state.  A relevant set of properties have been determined; look at the table below.

<center>

<table>

<tr>
<td></td>
<td><center><math>T</math></center></td>
<td><center><math>P</math></center></td>
<td><center><math>n_i</math></center></td>
</tr>

<tr>
<td><math>T</math></td>
<td><math>\mbox{heat capacity}</math></td>
<td><math>\mbox{thermal expansion}</math></td>
<td><math>- \left ( \frac{\partial S}{\partial n_i} \right )_{T,P,n_{j \ne i}}</math></td>
</tr>

<tr>
<td><math>P</math></td>
<td><math>\mbox{thermal expansion}</math></td>
<td><math>\mbox{compressibility}</math></td>
<td><math>\left ( \frac{\partial U}{\partial n_i} \right )_{T,P,n_{j \ne i}}</math></td>
</tr>

<tr>
<td><math>n_i</math></td>
<td><math>- \left ( \frac{\partial S}{\partial n_i} \right )_{T,P,n_{j \ne i}}</math></td>
<td><math>\left ( \frac{\partial U}{\partial n_i} \right )_{T,P,n_{j \ne i}}</math></td>
<td><math>- \left ( \frac{\partial \mu_i}{\partial n_i} \right )_{T,P,n_{j \ne i}}</math></td>
</tr>

</table>

</center>

<br>

*Partial molar quantities (PMQ)*

<p>
</p>

Partial molar quantities are derivatives of extensive variables with respect to the number of moles.

<center>

<br>

<math>X(T, P, n_i)</math>

<br>

<math>U = \left ( \frac{\partial X}{\partial n_i} \right )_{T, P, n_{j \ne i}} = \overline{X}_j</math>

<br>

<table>

<tr>
<td><math>\underline{X}_i</math></td>
<td><math>\mbox{molar property of pure} i</math></td>
</tr>

<tr>
<td><math>\underline{V}_i</math></td>
<td><math>\mbox{molar volume of pure} i</math></td>
</tr>

<tr>
<td><math>\overline{X}_i</math></td>
<td><math>\mbox{partial molar quantity} X_i</math></td>
</tr>

</table>

</center>

*"Chemical susceptibility"*

<p>
</p>

The chemical susceptibility describes the change in chemical potential resulting from moles of a species added to a system.  This is analogous to the free energy increase that results from adding charge to a system.  The change of free energy that results from adding a charge is equal to the inverse of the [capacitance|http://en.wikipedia.org/wiki/Capacitance], and the change in chemical potential resulting from the addition of a particular species is the inverse of the chemical capacitance.  If the potential increases dramatically with <math>n_i</math>, the system is stiff with respect to <math>i</math>.  If the change is small, much can be added with little change in potential.

<center>

<br>

<math> \frac{\partial G}{\partial q} = \Phi </math>

<br>

<math> \frac{\partial^2 G}{\partial q^2} =  \frac{\partial \Phi}{\partial q} </math>

<br>

<math> \frac{\partial \Phi}{\partial q} =  \frac{1}{C} </math>

<br>

<math> \frac{\partial \mu_i}{\partial n_i} =  \frac{1}{\mbox{chemical capacitance}} </math>

<br>

</center>

There is a cross term, which is shown below.  It is present but often neglected.  In old [metallurgy|http://en.wikipedia.org/wiki/Metallurgy] mathts the term was grouped with interaction coefficients.

<center>

<br>

<math> \left ( \frac{\partial u_i}{\partial n_j} \right )_{T, P, n_{i \ne j}} </math>

<br>

</center>

*Partial molar quantities*

<p>
</p>

It is possible to identify partial molar quantities in solution and tell about physics though macroscopic properties.  Consider an equilibrium concentration of <math>A</math> and <math>B</math> in <math>\alpha</math> and <math>\beta</math>.  Apply hydrostatic pressure.  One may lose more <math>A</math> in <math>\alpha</math> to <math>\beta</math> and vice versa.  This is analogous to applying pressure to salt water and determining the change in distribution as the salt passes through a membrane.

<center>
!Alpha_beta_hydrostatic_pressure_with_A_and_B.PNG!
</center>

<br>

Look at how the chemical potential changes.  In the last two steps below, apply a [Maxwell relation|http://en.wikipedia.org/wiki/Maxwell_relation].

<center>

<br>

<math> \mu_A^\alpha = \mu_A^\beta </math>

<br>

<math> \frac{\partial \mu_A^\alpha}{\partial P} = \frac{\partial \mu_A^\beta}{\partial P} </math>

<br>

<math> \frac{\partial \mu_A^\alpha}{\partial P} = \frac{\partial}{\partial P} \left (\frac{\partial G^\alpha}{\partial n_A} \right )</math>

<br>

<math> \frac{\partial \mu_A^\alpha}{\partial P} = \frac{\partial}{\partial n_A} \left (\frac{\partial G^\alpha}{\partial P} \right )</math>

<br>

<math> \frac{\partial G^\alpha}{\partial P} = \frac{\partial V^\alpha}{\partial n_A}</math>

<br>

<math> \frac{\partial V^\alpha}{\partial n_A} = \overline{V}_A^\alpha</math>

</center>

<br>

Consider the [partial molar volume|http://en.wikipedia.org/wiki/Partial_molar_volume] of <math>A</math>.  If it is less in <math>\alpha</math> than in <math>\beta</math>, the chemical potential will go up less on one side than the other with the application of pressure.  Look at the partial molar quantity of [entropy|http://en.wikipedia.org/wiki/Entropy] to evaluate the concentration shift with temperature.

<p>
</p>

*Physical interpretation of PMQ*

Below is a definition of the partial molar volume.

<center>

<br>

<math> \overline{V}_i = \left ( \frac{\partial V}{\partial n_i} \right )_{T, P, n_{j \ne i}} </math>

<br>

</center>

The total volume is a weighted sum.

<center>

<br>

<math> V = \sum_{i} \overline{V}_i n_i </math>

<br>

</center>

Consider adding <math>A</math> to interstitial sites of <math>B</math>.

<center>
!A_adding_interstitially_to_B.PNG!
</center>

<br>

The total volume doesn't change, and the value of the partial molar volume gives information about the substitution effect.


<center>

<br>

<math> \overline{V}_A = \left ( \frac{\partial V}{\partial n_A} \right )_{T, P, n_B} </math>

<br>

<math>\overline{V}_A = 0</math>

</center>

<br>

Consider a system entirely of <math>A</math>.

<br>

<center>

<math>X_A</math> = <math>1</math>, <math>\overline{V}_A = \underline{V}_A</math>

</center>

<br>

It is possible for the partial molar volume to be negative.  In this case, the volume of the system goes down with the addition of a species.  Stuff is pulled closer together.  Consider adding <math>MgSO_4</math> to a system.  Ions dissociate, polarizes water, and pulls the molecules closer together.  A negative partial molar volume is common to ionic materials.  Oxides commonly pull the environment closer.  Another example is <math>Li_xCOO_2</math>.  This is a most famous material; it is used in batteries.  Reversibly taking <math>Li</math> is a way to store charge.  Adding <math>Li</math> results in a decrease in volume.  This never happens in metals.  Oxides behave very differently than metals.

<center>

!Adding_MgSO4_to_water_-_diagram.PNG!
!Adding_MgSO4_to_water_-_graph.PNG!

</center>

<br>

*Partial molar magnetization*

<p>
</p>

Consider an open system and magnetization.  Below is the partial molar magnetization; it is a measure of how much <math>i</math> contributes to the [magnetic moment|http://en.wikipedia.org/wiki/Magnetic_moment].

<br>

<center>

<math>\overline{M}_i = \left ( \frac{\partial M}{\partial n_i} \right )_{T, P, n_{j \ne i}}</math>

</center>

<br>

Consider adding <math>Fe^{3+}</math> ions to a [paramagnetic material|http://en.wikipedia.org/wiki/Paramagnetic].  The large moment of <math>Fe^{3+}</math> polarizes the surrounding environment, and the total moment added is much larger than the moment of the ion alone.

<center>
!Adding_Fe3%2B_to_paramagnetic_material.PNG!
</center>
<br>

*Magnetic Semiconductors*

<p>
</p>

Consider adding <math>Mn</math>, a ferromagnetic material, in <math>GaAs</math>, a semiconductor.  Ions of <math>Mn</math> are acceptors; they create [holes|http://en.wikipedia.org/wiki/Electron_hole] in the [valence band|http://en.wikipedia.org/wiki/Valence_band] and polarize the holes created.  When the unpaired electrons in the valence band are not polarized, the up and down spins balance.  The [spins|http://en.wikipedia.org/wiki/Spin_%28physics%29] can be polarized up or down in the valence band by <math>Mn</math>, and the result is a magnetic semiconductor.  Magnetic semiconductors are used in [spintronics|http://en.wikipedia.org/wiki/Spintronics] applications.

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!Conduction_and_valence_band_undoped.PNG!
!Conduction_and_valence_band_doped.PNG!
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