</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 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

= \frac

</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

_

- \underline

</math>

<br>

<math> \frac

d \left ( \frac

\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

+ C </math>

<br>
</center>

Below is a graph showing the linear relation of <math>ln P</math> and <math>\frac

</math>.

<center>
Unable to render embedded object: File (Ln_P_vs_T%5E-1.PNG) not found.
</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

</math>

<br>

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

</center>

There is three percent water in saturated air 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

</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 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 can cause 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; 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" 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>
Unable to render embedded object: File (Freezer_and_ice.PNG) not found.
</center>

<p>
</p>

"Strange" thermodynamic properties

<p>
</p>

Consider the 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

\mu_i dn_i</math>

<br>

</center>

The chemical potential is an equation of state.

<center>

<br>

<math> \mu_i = \left ( \frac

\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.

<center>

<br>

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

<br>

<math> dH = TdS + VdP + \sum_

\right ){S, P, n{j\ne i}}</math>

<br>

</center>

Consider the Hemholtz free energy

<center>

<br>

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

<br>

<math> dF = -SdT - PdV + \sum_

\mu_i dn_i</math>

<br>

<math> \mu_i = \left ( \frac

\mu_i n_i</math>

<br>

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

<br>

<math> G = \sum_

\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>

Unable to render embedded object: File (Alpha_beta_constant_T%2C_P.PNG) not found.

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

<br>

<math>dG^\alpha = -S^\alpha dT + V^\alpha dP + \sum_

\mu_i^\alpha dn_i^\alpha + \sum_

(\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^

= 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. Also, it is possible to solve the problem by treating <math>\alpha</math> and <math>\beta</math> as open systems 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_

\right )_

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

<br>

<math>U = - \left ( \frac

\right )_

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

<br>

<math>\mu_i = - \left ( \frac

</math>

<br>

<math> \frac

= \frac

{\mbox{chemical capacitance}} </math>

<br>

</center>

There is a cross term, which is shown below. It is present but often neglected. In old metallurgy mathts the term was grouped with interaction coefficients.

<center>

<br>

<math> \left ( \frac

\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>
Unable to render embedded object: File (Alpha_beta_hydrostatic_pressure_with_A_and_B.PNG) not found.
</center>

<br>

Look at how the chemical potential changes. In the last two steps below, apply a Maxwell relation.

<center>

<br>

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

<br>

<math> \frac

</math>

<br>

<math> \frac

= \frac

\left (\frac

= \frac

_A^\alpha</math>

</center>

<br>

Consider the 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 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