Problems to Chapter 6
1. The densities of two aqueous NaCl solutions were determined at 25 °C by using a pycnometer. The
density of the 60 g/dm3 concentration solution was found to be 1.0315 g/cm3, and that of the 82.5 g/dm3
concentration solution 1.0463 g/cm3. Calculate the partial molar volume of NaCl and of water in the
solution.
Solution: The molar volume of the solution can be expressed in terms of the partial molar volumes and
mole fractions of the two components, according to Eqs. Error! Reference source not found. and
Error! Reference source not found.:
v
2
x V
i
i
i 1
 x H 2O VH 2O  x NaCl VNaCl
Dividing both sides of the above equation with v yields
1
x H 2O
v
VH 2 O 
x NaCl
VNaCl
v
Note that for any components i,
x i ρw i

v
Mi
By substituting this to the above equation we get to the following relation:
1
ρw H2O
M H 2O
VH2O 
ρw NaCl
VNaCl
M NaCl
Since there are only two components, NaCl and H2O in the system, w H2O  1  w NaCl and as the
weight fraction of NaCl can be expressed by dividing the mass per volume concentration of NaCl with
the density of the solution,
 c

ρ 1  NaCl 
ρ 
c
1 
VH 2O  NaCl VNaCl
M H 2O
M NaCl
By rearranging the above equation, we can express ρ as a linear function of the mass per volume
concentration of NaCl,
ρ
M H 2O
VH 2 O
 VNaCl M H2O 
c
 1 
 VH O M NaCl  NaCl
2


Applying the above equation to the experimental data yields to a system of two equations containing
two unknown quantities, VH 2O and V NaCl . Solving this system of equations gives
VH 2O  18.14 cm 3/mol,
VNaCl  20.18 cm 3/mol.
2. The two compartements of a gas container are filled at room temperature with 1 g of helium and 1 g
of hydrogen, respectively. We let the two gases mix by opening the valve separating the two
compartments. Calculate the change of entropy. (Consider the atomic masses to be 1 and 4 g / mol and
both gases as ideal.)
Solution: Taking into account the molar masses, the amounts of the gases are nH  0.5 mol and
2
nHe = 0.25 mol, summing up to a total amount of 0.75 mol, thus xH  2 / 3 and xHe = 1/3. Since the two
2
gases are ideal, thus their mixture is also an ideal mixture, and the entropy of mixing can be calculated
using Eq. Error! Reference source not found. as:
 mix S  
3 2
2 1 1
R ln  ln   0.4774 R
4 3
3 3 3
3. In a binary thermodynamic system of components A and B, x A  0.1. In the temperature range
not much different from T = 25 °C, the relative activity coefficients of the components are given by the
empirical formulae
ln γA = – 935.111 xAT –3/2
and
γA = – 0.003 .
Calculate the molar excess quantities of entropy, enthalpy and Gibbs potential of the system at
25 °C.
Solution: At the temperature of 25 °C = 298.15 K – based on the formula given – ln γA = – 0.0182, thus
the molar excess Gibbs potential of the system can be calculated usig Eq. Error! Reference source
not found.:
g E  RT xA ln  A  xB ln  B   11 .1953 J/mol .
To calculate the molar excess enthalpy using Eq.Error! Reference source not found., we need to
know the partial derivatives of the logarithms of the activity coefficients with respect to temperature:
  ln  A 
1.5
 2.5
 9.13838 10 5 K 1 ;

  140 .267 K T
 T  P ,n
  ln  B 

 0
 T  P ,n
Thus the molar excess enthalpy can be calculated as
   ln  A 
  ln  B  
h E   RT 2  xA 
  xB 
   6.7538 J/mol .
 T  P , n 
  T  P , n
The molar excess entropy can be calculated by rearranging Eq. Error! Reference source not
found.:
sE 
hE  g E
 0.0145 J/(mol K)
T