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Chemistry 351

Partial Molar Quantities

Why Do We Need Partial Molar Quantities?

For a pure substance, the volume is easy to describe. If one mole of a liquid occupies \(18.0\ \mathrm{cm^3}\), then adding another mole increases the volume by another \(18.0\ \mathrm{cm^3}\).

Mixtures are more complicated.

Consider mixing ethanol and water. If volumes were perfectly additive, the total volume of the mixture would simply be the sum of the volumes of the pure liquids. Experimentally, however, the volume of the mixture is smaller than expected because the molecules pack together more efficiently in the mixture than they do in the pure liquids.

This observation raises an important question:

How much does the volume of a mixture change when a small amount of one component is added?

The answer is provided by a partial molar quantity.

Defining the Partial Molar Volume

Consider a binary mixture containing components A and B.

The volume of the mixture may depend on temperature, pressure, and the amounts of both components:

\[ V = V(T,p,n_A,n_B) \]

The total differential of the volume is therefore

\[ dV = \left( \frac{\partial V}{\partial T} \right)_{p,n_A,n_B}dT + \left( \frac{\partial V}{\partial p} \right)_{T,n_A,n_B}dp + \left( \frac{\partial V}{\partial n_A} \right)_{T,p,n_B}dn_A + \left( \frac{\partial V}{\partial n_B} \right)_{T,p,n_A}dn_B \]

The quantities

\[ \boxed{ \bar V_A = \left( \frac{\partial V} {\partial n_A} \right)_{T,p,n_B} } \]

and

\[ \boxed{ \bar V_B = \left( \frac{\partial V} {\partial n_B} \right)_{T,p,n_A} } \]

are called the partial molar volumes of components A and B.

They describe how the volume of the mixture changes when a small amount of one component is added while holding temperature, pressure, and the amount of the other component constant.

A Simpler Expression

Most measurements on mixtures are performed at constant temperature and pressure.

Under these conditions,

\[ dT = 0 \qquad dp = 0 \]

and the differential simplifies to

\[ \boxed{ dV = \bar V_A\,dn_A + \bar V_B\,dn_B } \]

This expression states that the change in the volume of a mixture is the sum of the contributions from each component.

Total Volume of a Mixture

Integrating the previous expression yields

\[ \boxed{ V = n_A\bar V_A + n_B\bar V_B } \]

for a binary mixture.

More generally, if \(\bar X_i\) represents the partial molar value of an extensive property \(X\), then

\[ \boxed{ X = \sum_i n_i\bar X_i } \]

This relationship applies not only to volume, but also to enthalpy, entropy, Gibbs free energy, and many other thermodynamic properties.

Physical Interpretation

A partial molar quantity is not necessarily equal to the molar property of the pure substance.

In fact, partial molar quantities often depend strongly on composition. Adding one mole of ethanol to pure water may change the volume by a different amount than adding one mole of ethanol to a solution that is already rich in ethanol.

Partial molar quantities can even be negative. For example, some ionic solutes cause water molecules to pack more closely together, reducing the total volume of the solution when the solute is added.

Big picture: Partial molar quantities describe the contribution of a component to a thermodynamic property of a mixture. They provide the bridge between the properties of individual components and the properties of the mixture as a whole. The most important partial molar quantity is the partial molar Gibbs function, known as the chemical potential, which will be developed in the next module.

Worked examples

Worked example: Calculating the volume of a mixture from partial molar volumes

A solution contains \(55.5\ \mathrm{mol}\) of water and \(1.00\ \mathrm{mol}\) of sodium chloride. At this composition, the partial molar volumes are

\[ \bar V_{\mathrm{H_2O}} = 17.8\ \mathrm{cm^3\,mol^{-1}} \]

and

\[ \bar V_{\mathrm{NaCl}} = 15.6\ \mathrm{cm^3\,mol^{-1}}. \]

Calculate the total volume of the solution.

The total volume of a mixture is found from

\[ V = \sum_i n_i\bar V_i \]

For this two-component solution,

\[ V = n_{\mathrm{H_2O}}\bar V_{\mathrm{H_2O}} + n_{\mathrm{NaCl}}\bar V_{\mathrm{NaCl}} \]

Substitute the values:

\[ V = (55.5\ \mathrm{mol}) (17.8\ \mathrm{cm^3\,mol^{-1}}) + (1.00\ \mathrm{mol}) (15.6\ \mathrm{cm^3\,mol^{-1}}) \]


\[ V = 988\ \mathrm{cm^3} + 15.6\ \mathrm{cm^3} \]


\[ V = 1.00\times10^3\ \mathrm{cm^3} \]

\[ \boxed{ V = 1.00\ \mathrm{L} } \]

Interpretation: The sodium chloride does not contribute its pure solid molar volume to the solution. Instead, it contributes its partial molar volume, which describes how the volume of this particular solution changes when NaCl is added at this composition.

Practice

Partial Molar Volume Practice

A two-component mixture will be generated at random. Use the partial molar volumes to calculate the total volume of the mixture.

Key points (one glance)

Big picture: Partial molar quantities describe the contribution of each component to a thermodynamic property of a mixture. They are particularly important because the most useful partial molar quantity—the partial molar Gibbs function—is known as the chemical potential, the quantity that governs how systems respond to changes in composition.