Page 285 - Physical Chemistry

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Chapter 9 The volume of pure substance j is V* n V* (T, P), where V* is the molar
m, j
j
j
m, j
Solutions volume of pure j. If we view a pure substance as a special case of a solution, then the

definition (9.7) of V j gives V * j 10V>0n 2 ( V*/ n ) V* . Thus
m, j
j
j T, P
j T,P,n i j

V* V* (9.10)*
j m,j
The partial molar volume of a pure substance is equal to its molar volume. However,
the partial molar volume of component j of a solution is not necessarily equal to the
molar volume of pure j.
EXAMPLE 9.2 Partial molar volumes in an ideal gas mixture
Find the partial molar volume of a component of an ideal gas mixture.
We have
V 1n n p n p n 2RT>P
2
r
i
1

V 10V>0n 2 RT>P ideal gas mixture (9.11)
i
i T,P,n j i
Of course, RT/P is the molar volume of pure gas i at the T and P of the mixture,

so V i V* for an ideal gas mixture, a result not true for nonideal gas mixtures.
m,i
Exercise
A certain two-component gas mixture obeys the equation of state

P(V n b n b ) (n n )RT, where b and b are constants. Find V 1 and
1 1 2 2 1 2 1 2
V 2 for this mixture. (Answer: RT/P b , RT/P b .)
1
2
Relation between Solution Volume and Partial Molar Volumes
We now find an expression for the volume V of a solution. V depends on temperature,
Figure 9.2 pressure, and the mole numbers. For fixed values of T, P, and solution mole fractions
x , the volume, which is an extensive property, is directly proportional to the total num-
Addition of dn moles of substance i
j
j to a solution held at constant T ber of moles n in the solution. (If we double all the mole numbers at constant T and P,
and P produces a change dV in the then V is doubled; if we triple the mole numbers, then V is tripled; etc.) Since V is pro-
solution’s volume. The partial portional to n for fixed T, P, x , x , ..., x , the equation for V must have the form
1
2
r

molar volume V j of j in the
V nf1T, P, x , x , p 2 (9.12)
2
1
solution equals dV/dn . j
where n n and where f is some function of T, P, and the mole fractions.
i i
Differentiation of (9.12) at constant T, P, x , ..., x gives
1 r
dV f1T, P, x , x , p 2 dn const. T, P, x i (9.13)
1
2
Equation (9.8) becomes for constant T and P

dV a V dn const. T, P (9.14)
i
i
i
We have x n /n or n x n. Therefore dn x dn ndx . At fixed x , we have
i i i i i i i i
dx 0, and dn x dn. Substitution into (9.14) gives
i i i

dV a x V dn const. T, P, x i (9.15)
i i
i
Comparison of the expressions (9.13) and (9.15) for dV gives (after division by dn):

f xV . Equation (9.12) becomes V nf n xV or (since x n /n)
i i i i i i i i

V a n V one-phase syst. (9.16)*
i i
i