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Chapter 9 Exercise
Solutions

Find V and V in 0.20 mol/kg MgSO (aq) at 20°C and 1 atm. (Answers:
MgSO 4 H 2 O 4
3
3
2.2 cm /mol, 18.04 cm /mol.)
Because of strong attractions between the solute ions and the water molecules, the
solution’s volume V in Fig. 9.3 initially decreases with increasing n MgSO 4 at fixed n H 2 O .

The negative slope means that the partial molar volume V is negative for molal-
MgSO 4
ities less than 0.07 mol/kg. The tight packing of water molecules in the solvation shells
around the ions makes the volume of a dilute MgSO solution less than the volume of

4
the pure water used to prepare the solution, and V MgSO 4 is negative.

The value of V i in the limit as the concentration of solute i goes to zero is the
q
q
infinite-dilution partial molar volume of i and is symbolized by V i . To find V i
of MgSO in water at 20°C, one draws in Fig. 9.3 the line tangent to the curve at
4
q
n 0 and takes its slope. Some V i values for solutes in aqueous solution at 25°C
MgSO 4
and 1 atm compared with the molar volumes V* of the pure solutes are:
m,i
Solute NaCl Na SO MgSO H SO CH OH n-C H OH
2 4 4 2 4 3 3 7
q 3
V i /(cm /mol) 16.6 11.6 7.0 14.1 38.7 70.7
3
V* /(cm /mol) 27.0 53.0 45.3 53.5 40.7 75.1
m,i
For a two-component solution, there is only one independent mole fraction, so

V A V A (T, P, x ) and V B V B (T, P, x ) [Eq. (9.9)], in agreement with the fact that a
A
A
two-component, one-phase system has 3 degrees of freedom. For solutions of water (W)
and methanol (M), Fig. 9.4 shows temperature, pressure, and composition dependences

of the partial molar volume V . At x 0, the solution is pure water, and the value
W
M
3
V W 18.07 cm /mol at 25°C and 1 bar in Fig. 9.4 is V* of pure H O at 25°C and 1 bar.
m
2
Other Partial Molar Quantities
The ideas just developed for the volume V apply to any extensive property of the so-
lution. For example, the solution’s internal energy U is a function of T, P, n , ..., n r
1

[Eq. (9.5)], and by analogy with V 10V>0n 2 [Eq. (9.7)], the partial molar
i

i T,P,n j i
internal energy U i of component i in the solution is defined by

U 10U>0n 2 one-phase syst. (9.18)
i
i T,P,n j i

The same arguments that gave V n V [Eq. (9.16)] give (simply replace the sym-
i i i
bol V by U in all the equations of the derivation)

U a n U one-phase syst. (9.19)
i
i
i
where U is the internal energy of the solution.

We also have partial molar enthalpies H , i partial molar entropies S , partial molar
i

Helmholtz energies A , partial molar Gibbs energies G , i and partial molar heat capac-
i

ities C :
P,i

H 10H>0n 2 , S 10S>0n 2 (9.20)
i
i
i T,P,n j i
i T,P,n j i

Figure 9.4 G 10G>0n 2 , C P,i 10C >0n 2 (9.21)
i
P
i T,P,n j i
i T,P,n j i
where H, S, G, and C are the solution’s enthalpy, entropy, Gibbs energy, and heat ca-
Partial molar volumes V W of water P
in solutions of water (W) and pacity. All partial molar quantities are defined with T, P, and n j i held constant.
methanol (M). The x 1 curves The partial molar Gibbs energy is especially important since it is identical to the
M
are infinite-dilution values. [Data chemical potential [Eq. (4.72)]:
from A. J. Easteal and L. A.
Woolf, J. Chem. Thermodyn., 17, 0G
G a b m one-phase syst. (9.22)*
i
i
49 (1985).] 0n i T,P,n j i