Dr Cooper's comments: Great job!

**· Melting and Applied Pressure**

Since the chemical potential, µ is the same as the molar Gibbs energy,

(eqn.1)

Because V_{m} is always positive, when a pressure is applied to a substance, it affects its melting or freezing point as well.

<Figure 4-10a> <Figure 4-10b>

Generally speaking, the liquid has a higher molar volume than the solid with few exceptions including water. Therefore, when the same pressure is applied to both the solid and the liquidphases, the change in µ in the liquid is greater than the change in µ in the solid because the chemical potential is proportional to the molar volume. Consequently, it increases the freezing point as in Fig.4-10a.

On the other hand, if the molar volume of the solid is larger than that of the liquid, it results in depreciation of the freezing point when the pressure is applied as described in Fig.4-10b.

**· Vapor pressure and Applied Pressure**

Experimentally, as pressure is applied to a condensed phase which is either solid or liquid, the vapor pressure of the substance increases. In other words, the applied pressure squeezes the molecules in the solid or liquid out, and those molecules escape as a gas increasing the vapor pressure above the condensed phase. This relation explains how vapor pressure can be calculated when the vapor pressure of the condensed phase without any applied pressure and the applied pressure are given; it can be written as

(eqn.2)

where *p* is the vapor pressure, *p** is the vapor pressure without an extra pressure, and Δ*p* is the added pressure.

This can be experimentally demonstrated using a piston that is only permeable to vapor but not to liquid or solid. Another way to do it is to use inert pressurizing gas such as argon or helium without the piston like in Figure 4.11.

<Figure 4.11>

**· Location of Phase Boundaries**

In a phase diagram, each phase is separated by phase boundaries. On phase boundaries, the chemical potential of two phases is the same; chemical potential of two phases must be the same at equilibrium. It means that at certain pressure and temperature, the chemical potential of phase α is equal to the chemical potential of phase β.

(eqn.3)

Gibbs energy can be expressed as *V*d*p*-*S*d*T*. Since µ is equivalent to *G*_{m}, the above equation can be rewritten as

Therefore, at constant pressure and temperature,

By grouping the terms, it can be rearranged into

The first part of the equation is simple the entropy change of transition between phase α and phase β, and the second part is the change in molar volume of transition.

(eqn.4)

The above equation is called the **Clapeyron equation**, where d*p*/d*T* is the slope of the phase boundary.

<Figure 4.12>

Either changing temperature or pressure will break equilibrium, so there is corresponding change of pressure and temperature. As in Figure 4.12, the phase boundary is slightly curved. However, if *a* and *b* are connected, the area between the line and the phase boundary would be very small. Hence, the phase boundary can be considered linear where the slope of the curve is d*p*/d*T*.

**· Solid-liquid Boundary**

Entropy change of transition is enthalpy change of transition divided by temperature.

Substituting the above term into the Claperyron equation, for the solid-liquid phase change, the equation can be written as

For the solid to liquid transition, the enthalpy changes of fusion are usually positive. Likewise, the volume changes of fusion are generally positive and also very small, and the temperature has to be positive. Therefore, the slope of the phase boundary between the solid and the liquid is usually positive and steep like in Figure 4.13.

If both sides of the Claperyron equation is divided by d*T* and taken into integration,

Δ_{fus}*H* and Δ_{fus}*V* for solid liquid transitions are about constant. Therefore both terms can be brought out from the integral sign, and

If we solve the integral,

–*p** is subtracted from both sides, and

(eqn.5)

With this equation, any points along the phase boundary can be calculated at equilibrium between two phases when the initial pressure and are given.

If *T* and *T*^{*} are very close to each other, ln(*T* / *T*^{*}) can be replaced by *T* – *T*^{*}. Therefore,

(eqn.6)

**· Liquid-vapor Boundary**

Like in the solid-liquid transition, Δ_{vap}*S* can be substituted by Δ_{vap}*H* divided by *T* into the Clapeyron equation:

Δ_{vap}*H* is generally positive. However, Δ_{vap}*V* is very large because a gas has much larger molar volume than a liquid does. Hence the slope, d*p*/d*T* is relatively smaller than the slope of the solid-liquid boundary. It also means that d*T*/d*p* is relatively large; as the pressure is applied, the temperature would respond greater than it would for a solid-liquid transition. Therefore, the boiling point would be affected much more greatly with increasing pressure than the melting point would.

The molar volume of the gas is much bigger than the molar volume of the liquid. In addition, according to Ideal Gas Law, the molar volume of a gas is *RT*/*p*. If we approximate that Δ_{vap}*V* is about the same as V_{m}(*g*),

which rearranges into the **Clausius-Clapeyron** equation:

This equation is used to calculate the enthalpy of vaporization, and it can be integrated as

where p^{*} and p are the initial and final pressures, and T^{*} and T are the initial and final temperatures. When Δ_{vap}*H* is considered as a constant,

It is valid as long as the temperature is not too high. If we evaluate the integral,

which then is rearranged as

Because it is an exponential function, the liquid-gas phase boundary is a curved line as in Figure 4.14.

<Figure 4.14>

**· Solid-vapor Boundary**

For the solid-gas phase boundary, the reasoning is exactly the same as the liquid-gas boundary, but the only difference is that instead of Δ_{vap}*V*, Δ_{sub}*V* is used:

(Adam)

__Begin Chapter 5__

In Chemistry, we encounter mixtures of substances that can react together. In this chapter we shall mainly consider binary mixtures involving two components,

where x = mole fraction

The partial molar volume of a substance, X, in a mixture is the change in the mixture's volume per mole of X added to the mixture

Imagine a large volume of pure water (such as a pool) with a temperature of 25 degrees celcius. If 1 additional mole of pure water was added to the initial large volume of pure water, the total volume increases by 18mL. So 18mL per mole is the molar volume of pure water.

Now imagine a large volume of ethanol. Add 1 mole of pure water to that large volume. The total volume of this binary mixture will only increase by 14mL instead of 18mL. This change is due to the H-bonding interactions between the water and the ethanol. The molecules of water are capable of being packed closer together to molecules of ethanol in a water-ethanol mixture as opposed to the molecular packing of water molecules to water molecules in a water-water mixture. The quantity of 14mL per mole is the **partial molar volume** of water in pure ethanol. The partial molar volume of water and ethanol across the full composition range at 25 degrees celcius are shown in fig 5-1.

The partial molar volumes of the components in a mixture depend on the mixture's composition as the molecular interactions vary as the composition changes from purely component A to purely component B.

The partial molar volume, VJ,of a substance J at some general composition is defined as:

(eqn 5.1)

where n' signifies that the number of moles of all components present are constant

The partial molar volume is applicable only to very specific conditions. The partial molar volume is the slope of the plot of the total volume as the amount of substance (A) is changed while the pressure, temperature, and amount of the other components are all constant (fig 5-2).

When the composition of the mixture is changed by the addition of dnA of composition A and dnB of composition B then the total volume of the mixture changes by:

(eqn 5.2)

The final volume can be calculated by integration because the partial molar volumes are constant, also the composition is held constant as the amounts of components A and B are increased. To calculate the final volume:

(eqn 5.3)

**Symbols**

we know that V(NaCl, aq) is the volume of sodium chloride in aqueous solution

V (with a horizontal line directly over it) (NaCl,aq) = the partial molar volume of sodium chloride in water

**Measuring partial molar volumes**

Partial molar volumes can be measured in several ways. One method is to measure the dependence of the volume on the composition and fit the observed volume to a function of the amount of substance. Once the function has been graphed, its slope can be determined at any composition by differentiation and thus the partial molar volume (the slope) can be determined.

Molar volumes are always positive but partial molar quantities are not always positive. An example is the limiting partial molar volume of MgSO4 in water is

-1.4mL per mole. Which means that the addition of 1 mole of MgSO4 to a large volume of water results in a decrease in volume of 1.4mL. The mixture contracts because the salt breaks up the open structure of water as the ions become hydrated and the structure collapses slightly.

Example problem:

ethanol is added to 1.000Kg of pure water

the total volume of the mixture, as measured by experiment, fits the following equation:

where , , and is the amount of ethanol present. The partial molar volume of ethanol, , is therefore:

then because

we can conclude that

(fig 5-3 is a graph of this function)

## Comments (2)

## Obed said

at 12:08 am on Nov 6, 2009

It never gets old that i can't see some pictures sometimes.

## Obed said

at 12:09 am on Nov 6, 2009

By "some" I mean every picture except for the last one.

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