Concise Physical Chemistry

c09 JWBS043-Rogers September 13, 2010 11:26 Printer Name: Yet to Come

PARTIAL MOLAR VOLUME 129

we get

lnp=

vapH

R

(

− 1

T

)

+const.

This result can be written

pvap=ae−
vapH/RT=aeb(−^1 /T)

where a arises from the constant of integration. This is a form of the

Clausius–Clapeyron equation. If one can determinebin this exponential equation,

then one has
vapH.

If an open container of liquid under an external pressurepextis heated to its

boiling point, bubbles form at the bottom of the container, each of which forms a

tiny closed “chamber” filled with pure vapor. That is why we observe that the normal

boiling points of pure liquids observed in an open container follow the same equation

(Clausius–Clapeyron) as that derived from the coexistence curve of the liquid in a

closed container.

9.4 PARTIAL MOLAR VOLUME

Prior to Gibbs, thermodynamics was largely about the transfer of heat in the process

of driving an engine. It was an age justly called the age of steam. Gibbs’s departure

was to focus on the transfer ofmatter. That is why Gibbs’s work is so important to

chemists. Our science is largely devoted to the transfer of matter from a reactant state

to a product state. In classical physical chemistry, we ask whether the transfer occurs

(thermodynamics) and, if it does, we want to know how long it takes (kinetics).

To answer the first of these questions, Gibbs used partial molar thermodynamic

state functions. To start out, we shall consider the volume of a system and we shall

restrict mixtures to two components. The volume of a system is easy to visualize,

and restricting the system to two components simplifies the arithmetic. Later we shall

release these restrictions.

The volume of an ideal two-component liquid solution at constantpandTdepends

on how much of each component is present:

V=f(n 1 ,n 2 )

If the molar volume of pure component 1 isVm◦ 1 and the molar volume of pure

component 2 isVm◦ 2 , then an ideal solution of equal molar amounts of 1 and 2 will have

a molar volume that is the average of the two

(

Vm◦ 1 +Vm◦ 2

)

/2 (Fig. 9.4). Solutions of

other ratios of components 1 and 2 would be arrayed along a straight line connecting

Vm◦ 1 toVm◦ 2 , provided that no shrinkage or swelling takes place when 1 and 2 are mixed.

In general, for different ratios of the components, we obtain

V=Vm◦ 1 n 1 +Vm◦ 2 n 2