Download Partial Molal Volume

ATENEO DE MANILA UNIVERSITY
CH 46 PHYSICAL CHEMISTRY LABORATORY I
Partial Molal Volume
Background
Thermodynamic variables fall into two types. Extensive properties of a phase are proportional to the amount of
the phase under consideration. Those representing intensive properties are independent on the amount of the
phase and includes p and T.
Partial molal volume is an intensive variable and is defined by the equation

V


V
i 


n
 i p,T,Nji
Equation 1
where Vi partial molal volume due to component i keeping pressure, temperature, and number of moles of the
component being studied constant. For a phase of one component, the partial molal volume is equal to the molal
volume of the pure component.
Partial molal volume is the increase in the volume of an infinite amount of solution (or an amount so large that
insignificant concentration change will result) when 1 mole of component i is added. This is by no means equal
to the volume of 1 mole of pure i.
Partial molal volume is connected to other partial molal quantities, one of the more interesting would be
chemical potential, also called the partial molal free energy. For any given component, chemical potential is
equal for all phases that are in equilibrium with each other. Consider a system containing a pure substance in
equilibrium with the saturated aqueous solution. The chemical potential of the solute is the same in the two
phases. For an equilibrium change, such as isothermal change in pressure, involving only expansion work,
dGVdp
Equation 2
Differentiating Equation 2 with respect to N2, the number of moles of the solute, we obtain:
dG2 V2dp
Equation 3
where the partial molal free energy (chemical potential) and partial molal volume appear. For the change in
state NaCl(s)  NaCl(aq), we can write
 

G
V

 
d
G

V
dp
2 
2
or
Equation 4
2
p T
 
2
Equation 5
Thus, if the partial molal volume of solute is greater than the molal volume of the solid solute, an increase in
temperature will cause an increase in chemical potential of solute in the solution relative to the solid phase, thus
the solute will leave the solution until equilibrium solubility is again obtained. Conversely, a lower partial molal
volume of the solute in solution relative to that of the solid will increase the solubility with an increase in
pressure.
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For an extensive homogeneous thermodynamic variable such as total volume, one can apply Euler’s theorem to
get
N
V

N
V

...

N
V
V
1
2
i
1
2
i
Equation 6
For a binary solution at constant temperature and pressure, one can derive the Gibbs-Duhem equation:
dV 1
X
 2
X1
dV 2
Equation 7
where X1 and X2 are the mole fractions of the components of the system.
Using equation 6, the total volume of a solution containing 1000 g of water (55.51 moles) and m moles of solute
is given by
V

N
V

N
V

55
.
51
V

m
V
1
2
1
2
1
2
Equation 8
where the subscripts 1 and 2 refer to the solvent and solute, respectively. If we let V 1 be the molal volume of
pure water (18.096 mL at 25.00ºC), we can define the apparent molal volume φ of the solute by the equation


V

N
V

N
55
.
51
V

m
1
1
2
1
0
0
Equation 9
Equation 9 can be rearranged to give
0
1
V

55
.51
V
1
Equation 10
m
N1V10
Since V = (1000 +mM2)/d and
= 1000/d0 where d is the density of the solution, d0 is the density of the
pure solvent, and M2 is the solute molecular weight. Substituting these equations to equation 10, we obtain
 1000

 1000

d

d
W

W
1
1






M


M





d
d
d
W

W
 m

 m

Equation 11
0
0
2
2
0
0e
In equation 11, we use the directly measured weights of the vessel, in this case, a pycnometer. We is the empty
weight of the pycnometer, W0 is the weight of the pycnometer when filled to the mark with pure water, and W
when filled to the mark with solution. This is preferable than using density values since doing so would require
greater precision to calculate the minimal difference of d – d0.
By the definition of partial molal volumes and by use of equations 8 and 9,
  



V

d
Equation 12


V



N

m
2



N

N dm
2


22
N
,
T
,
p
1


2
0 2
0 m
1

d
Equation 13


V

N
V

N

V

1
1
1 2
1


N

N
55
.
51
dm
1
2


Plotting φ versus m would produce a smooth curve with tangents at the desired concentration equal to the
measure of the slopes. However, for solutions of simple electrolytes, it has been found that many apparent molal
quantities such as φ vary linearly with m1/2, even up to moderate concentrations. This behavior is in agreement
with the prediction of the Debye-Hückel theory for dilute solutions. Since
 

d dd
m1 d


dm
dm
d
m
2
m
d
m
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Equation 14
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we obtain from equations 12 and 13:
   
Equation 15
m
d
m
3
m
d
0
V






2
2
2
2
m
d
m
d
m
m
m
d

0


V

V
1
1
2dm

55
.
51


Equation 16
where φ0 is the apparent molal volume extrapolated to zero concentration. Plotting φ versus m1/2 will give a
straight line. From the slope dφ/dm1/2 and the value of φ0, both partial molal volume values for solute and
solvent can be determined.
Procedure
Make up 250 mL of approximately 3.2m (3.0M) NaCl solution in water by weighing NaCl accurately
and by using a volumetric flask. It may be necessary to heat the solution first to dissolve all the NaCl. Never
heat volumetric equipment. Also prepare solutions of 1/2, 1/4, 1/8 and 1/16 of the initial concentration by doing
serial dilution.
The pycnometer is rinsed with distilled water and thoroughly dried before each use. Use acetone to hasten the
drying process.
Once dry, fill your pycnometer up to the cover with the solution to be measured. Place the pycnometer in the
thermostat bath at 25.0ºC. Allow at least 15 minutes for equilibration. Afterwards, remove the pycnometer from
the bath and quickly but thoroughly dry the outside surface with a towel or tissue paper, then weigh.
The empty weight of the pycnometer should be taken, as well as the weight with distilled water and the different
solutions. Do everything in duplicate.
Calculations
The volume of the pycnometer can be derived from the density of water at 25ºC (0.997044 g/mL) and
the pycnometer weight data.
The molalities m which are needed for the calculations can be obtained from the molarities M obtained from the
volumetric procedures by use of the equation

1
M 1
Equation 17
m

 

M
M
d M
2 d
2
1



d
1000
M
1000
where M2 is the molecular weight of the solute (58.45 g/mol) and d is the experimental density of the solution.
Calculate φ for each solution using equation 11 and plot φ versus m1/2. Draw the best straight line from your
graph and get the slope. Record the value of φ0, the intercept at m equal to 0.
Calculate V2 and V1 for m = 0, 0.5, 1.0,1.5, 2.0, and 2.5. Plot them against m and draw a smooth curve for each
of the two quantities.
In your report, discuss how the solubility of NaCl in water will be affected by an increase in pressure. The
density of NaCl is 2.165 g/mL at 25ºC.
Also discuss whether the curves of the partial molal volumes of both the solute and solvent behave in accord
with equation 7.
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Discuss why volumes are not additive, i. e., 10.00 mL of pure ethanol and 10.00 mL of pure water will not yield
20.00 mL of solution!
Reference
1. Sime, R.J., Physical Chemistry, Saunders College Publishing, 1990.
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