Chapter 7 : Properties of Solutions¶

Example 7.2¶

In :
# Variables
V = 0.1; 			#volume of mixture required (m**3)
Ve = 0.03; 			#volume of alcohol
Vw = 0.07; 			#volume of water
de = 789.; 			#density of ethanol (kg/m**3)
dw = 997.; 			#density of water (kg/m**3)
pe = 53.6*10**-6; 			#partial molar volume of ethanol (m**3/mol)
pw = 18.*10**-6; 			#partial molar volume of water (m**3/mol)
Me = 46.; 			#molecular wt of ethanol
Mw = 18.; 			#molecular wt of water

# Calculations
#To find the volume of mixture
ne = (Ve*de*10**3)/Me; 			#number of moles of ethanol
nw = (Vw*dw*10**3)/Mw; 			#number of moles of water
xe = ne/(ne+nw) 			#mole fraction of ethanol
xw = 1-ne; 			#mole fraction of water
act_V = (ne*pe)+(nw*pw)

# Results
if (V==act_V) :
print 'It is possible to prepare the required solution'
else:
Ve_act = (Ve/act_V)*V;
Vw_act = (Vw/act_V)*V;
print ' For the given volumes of ethanol and water, it is not possible to prepare 0.1 cubic m of mixture'
print ' Required volume of ethanol is %f cubic m'%Ve_act
print ' Required volume of water is %f cubic m'%Vw_act
For the given volumes of ethanol and water, it is not possible to prepare 0.1 cubic m of mixture
Required volume of ethanol is 0.030810 cubic m
Required volume of water is 0.071890 cubic m

Example 7.3¶

In :
# Variables
V = 2.; 			#volume of desired solution (m**3)
x1 = 0.3; 			#moles fraction of methanol
x2 = 0.7; 			#moles fraction of water
V1 = 38.632*10**-6; 			#partial molar volume of methanol (m**3/mol)
V2 = 17.765*10**-6; 			#partial molar volume of water (m**3/mol)
mol_V1 = 40.727*10**-6; 		#molar volume of ethanol (m**3/mol)
mol_V2 = 18.068*10**-6; 		#molar volume of water (m**3/mol)

# Calculations
#To find the required volume of methanol and water
V_mol = (x1*V1)+(x2*V2); 			#molar volume of desired solution
n = V/V_mol; 			            #no. of moles in the desired solution
n1 = x1*n; 			#moles of methanol
n2 = x2*n; 			#moles of water
V_m = n1*mol_V1;
V_w = n2*mol_V2;

# Results
print 'Volume of methanol to be taken is %f cubic m'%V_m
print ' Volume of water to be taken is %f cubic m'%V_w
Volume of methanol to be taken is 1.017111 cubic m
Volume of water to be taken is 1.052866 cubic m

Example 7.4¶

In :
# Variables
V1_w = 0.816*10**-3; 			#partial molar volume of water in 96% alcohol solution
V1_e = 1.273*10**-3; 			#partial molar volume of ethanol in 96% alcohol solution
V2_w = 0.953*10**-3; 			#partial molar volume of water in 56% alcohol solution
V2_e = 1.243*10**-3; 			#partial molar volume of ethanol in 56% alcohol solution
d = 0.997*10**3; 			    #density of water (kg/m**3)

# Calculations
#To calculate the volume of water to be added and volume of dilute alcohol solution
#Basis:
V = 2*10**-3; 			                    #volume of alcohol solution (m**3)
V_sp = (0.96*V1_e)+(0.04*V1_w); 			#volume of 1 kg of laboratory alcohol
m_e = V/V_sp; 			                    #mass of 2*10**-3 m**3 alcohol

#(a).
#Let mass of water added be m kg
#Taking an alcohol balance
m = (m_e*0.96)/0.56 - m_e;
v = m/d;

# Results
print ' (a).'
print ' Mass of water added is %f kg'%m
print ' Volume of water added is %4.3e cubic m'%v

#(b)
m_sol = m_e + m; 			                #mass of alcohol solution obtained
sp_vol = (0.56*V2_e)+(0.44*V2_w); 			#specific volume of 56% alcohol
V_dil = sp_vol*m_sol; 			            #volume of dilute alcohol solution
print ' (b)'
print ' Volume of dilute alcohol solution is %5.4e cubic m'%V_dil
(a).
Mass of water added is 1.138558 kg
Volume of water added is 1.142e-03 cubic m
(b)
Volume of dilute alcohol solution is 3.0479e-03 cubic m

Example 7.6¶

In :
# Variables
x1= .3
x2 = .7

# Calculations and Results
H = 400.*x1 + 600*x2 + x1*x2*(40*x1+20*x2)
H1 = 420.-60+40;
#Using eq. 7.28 (Page no. 264)
#H = H2_bar as x2 = 1
H2 = 600.;
print ' (b).'
print ' Pure state enthalpies are:'
print ' H1 = %i J/mol'%H1
print ' H2 = %i J/mol'%H2

#(c).
#H1_inf = H1_bar as x1 = 0, so from eq. 7.27
H1_inf = 420.;
#H2_inf = H2_bar as x2 = 0. so from eq 7.28
H2_inf = 640.;
print ' (c).'
print ' At infinite dilution:'
print ' H1 = %i J/mol'%H1_inf
print ' H2 = %i J/mol'%H2_inf
(b).
Pure state enthalpies are:
H1 = 400 J/mol
H2 = 600 J/mol
(c).
At infinite dilution:
H1 = 420 J/mol
H2 = 640 J/mol

Example 7.7¶

In :
def V(m):
y = 1.003*10**-3 + 0.1662*10**-4*m + 0.177*10**-5*m**1.5 + 0.12*10**-6*m**2
return y

# Variables
m = 0.1; 			#molality of solution (mol/kg)

# Calculations
#To calculate the partial molar volume of the components
#Differentiating Eq. 7.29 with reference to m, we get
V1_bar = 0.1662*10**-4 + 0.177*1.5*10**-5*m**0.5 + 0.12*2*10**-6*m;
V_sol = V(m) 			#volume of aqueous soluttion
n1 = m;
n2 = 1000./18;
V2_bar = (V_sol - n1*V1_bar)/n2;

# Results
print 'Partial molar volume of water = %4.3e cubic m/mol'%V2_bar
print ' Partial molar volume of NaCl = %4.3e cubic m/mol'%V1_bar
Partial molar volume of water = 1.805e-05 cubic m/mol
Partial molar volume of NaCl = 1.748e-05 cubic m/mol

Example 7.10¶

In :
# Variables
K = 4.4*10**4; 			#Henry's law constant (bar)
pp = 0.25; 			    #partial pressure of oxygen in bar
M_O2 = 32.; 			#molecular wt of oxygen
M_water = 18.; 			#molecular wt of water

# Calculations and Results
#To estimate the solubility of oxygen in water at 298 K
#Using eq. 7.72 (Page no. 275)
x_O2 = pp/K; 			#mole fraction of O2
print 'Solubility of oxygen is %5.4e moles per mole of water'%x_O2

#In mass units
sol_O2 = (x_O2*M_O2)/M_water;
print ' Solubility of oxygen in mass units is %4.3e kg oxygen per kg water'%sol_O2
Solubility of oxygen is 5.6818e-06 moles per mole of water
Solubility of oxygen in mass units is 1.010e-05 kg oxygen per kg water

Example 7.11¶

In :
%matplotlib inline

from numpy import *
from matplotlib.pyplot import *
import numpy

# Variables
xb = array([0, 0.2, 0.4, 0.6, 0.8, 1.0])
pa_bar = [0.457 ,0.355, 0.243, 0.134, 0.049, 0];
pb_bar = [0,0.046, 0.108, 0.187, 0.288, 0.386];

#To confirm mixture conforms to Raoult's Law and to determine Henry's law constant
xa = 1 - xb
plot(xa,pa_bar)
plot(xa,pb_bar)

# Calculations and Results
#For Henry's law plotting
x = [0,0.2, 0.4 ,0.6 ,0.8 ,1.0];
#Form the partial presures plot of component A and B
yh1 = [0,0,0,0,0,0]
yh1 = 0;
yh1 = 0.049; 			#For component A
for i in range(2,6):
yh1[i] = yh1[i-1]+(x[i]-x[i-1])*((yh1-yh1)/(x-x))

yh_2 = [0,0,0,0,0,0]
yh_2 = 0;
yh_2 = 0.046; 			#For component B
i = 3;
while (i>=0):
yh_2[i] = yh_2[i+1] + (x[i]-x[i+1])*((yh_2-yh_2)/(x-x))
i = i-1;

plot(x,yh1)
plot(x,yh_2)
#legend("Partial pressure "," ","Raoults law"," ","Henrys Law"
show()
#(a)
print 'From the graph it can be inferred that, in the region where Raoults law is obeyed by A, the Henrys law is obeyed by B, and vice versa'

#(b)
#Slope of Henry's law
print ' For component A, Ka = %f bar'%yh1
print ' For component B, Kb = %f bar'%yh_2
Populating the interactive namespace from numpy and matplotlib From the graph it can be inferred that, in the region where Raoults law is obeyed by A, the Henrys law is obeyed by B, and vice versa
For component A, Ka = 0.245000 bar
For component B, Kb = 0.230000 bar

Example 7.12¶

In :
from numpy import array

# Variables
xa = array([0, 0.2, 0.4, 0.6, 0.8, 1.0]);
Pa_bar = [0 ,0.049, 0.134, 0.243, 0.355, 0.457];
Pb_bar = [0.386 ,0.288, 0.187, 0.108, 0.046, 0];

# Calculations and Results
#To calculate activity and activity coeffecient of chloroform
xb = 1-xa;
Pbo = 0.386; 			#vapour pressure of pure chloroform
#(a). Based on standard state as per Lewis-Randall rule

print 'Based on Lewis Randall Rule'
print '   Activity          Activity coeffecient'
a = [0,0,0,0,0,0]
ac = [0,0,0,0,0,0]
for i in range(6):
a[i] = Pb_bar[i]/Pbo;
print '    %f'%a[i],
if(xb[i]==0):
print '         Not defined',
else:
ac[i] = a[i]/xb[i];
print '          %f'%ac[i]

#(b). Based on Henry's Law
Kb = 0.217; 			#bar (From Example 7.11 Page no. 276)

print '\n\n Based on Henrys Law'
print '   Activity         Activity coeffecient'
for i in range(6):
a[i] = Pb_bar[i]/Kb;
print '    %f'%a[i],
if(xb[i]==0):
print '         Not defined',
else:
ac[i] = a[i]/xb[i];
print '           %f'%ac[i]
Based on Lewis Randall Rule
Activity          Activity coeffecient
1.000000           1.000000
0.746114           0.932642
0.484456           0.807427
0.279793           0.699482
0.119171           0.595855
0.000000          Not defined

Based on Henrys Law
Activity         Activity coeffecient
1.778802            1.778802
1.327189            1.658986
0.861751            1.436252
0.497696            1.244240
0.211982            1.059908
0.000000          Not defined

Example 7.13¶

In :
# Variables
P = 20.; 			#pressure in bar

# Calculations and Results
def f1(x1):
y = (50*x1)-(80*x1**2)+(40*x1**3)
return y

#To determine fugacity fugacity coeffecient Henry's Law constant and activity coeffecient

#(a)
#Fugacity of component in solution becomes fugacity of pure component when mole fraction approaches 1 i.e.
x1 = 1.;
f1_pure = f1(x1);
print '(a). Fugacity f1 of pure component 1 is %i bar'%f1_pure

#(b)
phi = f1_pure/P;
print ' (b). Fugacity coeffecient is %f'%phi

#(c)
#Henry's Law constant is lim (f1/x1)and x1 tends to 0
x1 = 0;
K1 = 50 - (80*x1) + (40*x1**2);
print ' (c). Henrys Law constant is %i bar'%K1

#(d)
print ' (d). This subpart is theoretical and does not involve any numerical computation'
(a). Fugacity f1 of pure component 1 is 10 bar
(b). Fugacity coeffecient is 0.500000
(c). Henrys Law constant is 50 bar
(d). This subpart is theoretical and does not involve any numerical computation

Example 7.17¶

In :
# Variables
H1_pure = 400.; 			#enthalpy of pure liquid 1 at 298 K and 1 bar (J/mol)
H2_pure = 600.; 			#enthalpy of pure liquid 2 (J/mol)
x1 = 0.;

# Calculations
delH1_inf = 20*((1-x1)**2)*(2*x1+1);
H1_inf = H1_pure + delH1_inf; 			#(J/mol)

#For infinite dilution of 2, x1 = 1 and delH2_inf = H2_bar
x1 = 1.;
delH2_inf = 40.*x1**3;
H2_inf = delH2_inf + H2_pure; 			#(J/mol)

# Results
print 'Enthalpy at infinite dilution for component 1 is %i J/mol'%H1_inf
print ' Enthalpy at infinite dilution for component 2 is %i J/mol'%H2_inf

Enthalpy at infinite dilution for component 1 is 420 J/mol
Enthalpy at infinite dilution for component 2 is 640 J/mol

Example 7.19¶

In :
# Variables
R = 8.314; 			#ideal gas constant
n1 = 100.; 			#moles of nitrogen
n2 = 100.; 			#moles of oxygen

# Calculations
#To determine the change in entropy of the contents of the vessel
x1 = n1/(n1+n2);
x2 = n2/(n1+n2);
import math

#Using eq. 7.122 (Page no. 292)
S = -R*(x1*math.log (x1) + x2*math.log (x2));
S_tot = S*(n1+n2);

# Results
print 'Change in entropy of components are %f J/K'%S_tot
Change in entropy of components are 1152.565132 J/K

Example 7.20¶

In :
# Variables
Hf = -408.610; 			#heat of formation (kJ)
#For reaction 2
#LiCl + 12H2O --> LiCl(12H2O)
H_sol = -33.614; 			#heat of solution (kJ)

#To determine heat of formation of LiCl in 12 moles of water
#Adding reaction 1 and 2% we get

# Calculations
#Li + 1/2Cl2 + 12H2O --> LiCl(12H2O)
H_form = Hf+H_sol;

# Results
print 'Heat of formation of LiCl in 12 moles of water is %f kJ'%H_form
Heat of formation of LiCl in 12 moles of water is -442.224000 kJ

Example 7.21¶

In :
# Variables
R = 8.314; 			#ideal gas constant
n1 = 3.; 			#moles of hydrogen
n2 = 1.; 			#moles of nitrogen
T = 298.; 			#temperature in K
P1 = 1.; 			#pressure of hydrogen in bar
P2 = 3.; 			#pressure of nitrogen in bar
import math

# Calculations
#To calculate the free energy of mixing
V1 = (n1*R*T)/(P1*10**5); 			#volume occupied by hydrogen
V2 = (n2*R*T)/(P2*10**5); 			#volume occupied by nitrogen
V = V1+V2; 			#total volume occupied
P = ((n1+n2)*R*T)/(V*10**5); 			#final pressure attained by mixture (bar)
G1 = R*T*(n1*math.log(P/P1) + n2*math.log(P/P2));

#For step 2, using eq. 7.121 (Page no. 292)
x1 = n1/(n1+n2);
x2 = n2/(n1+n2);
G2 = (n1+n2)*R*T*(x1*math.log (x1) + x2*math.log (x2));
G = G1+G2; 			#free energy in J

# Results
print 'The free energy of mixing when partition is removed is %f kJ'%(G/1000)
The free energy of mixing when partition is removed is -6.487935 kJ

Example 7.22¶

In :
# Variables
C_water = 4.18*10**3; 			#heat capacity of water (J/kg K)
C_ethanol = 2.58*10**3; 		#heat capacity of ethanol (J/kg K)
G1 = -758.; 			#heat of mixing 20 mol percent ethanol water at 298 K(J/mol)
G2 = -415.; 			#heat of mixing 20 mol percent ethanol water at 323 K (J/mol)
n_wat = 0.8; 			#moles of water
n_eth = 0.2; 			#moles of ethanol
T1 = 323.; 			#initial temperature in K
T2 = 298.; 			#final temperature in K

# Calculations
#Step 1: Water is cooled from 323 K t0 298 K
H1 = n_wat*18*C_water*(T2-T1)/1000; 			#(J)

#Step 2: Ethanol is cooled from 323 to 298 K
H2 = n_eth*46*C_ethanol*(T2-T1)/1000; 			#(J)

#Step 3: 0.8 mol water and 0.2 mol ethanol are mixed at 298 K
H3 = G1; 			#(J)

#Step 4:
H = G2;
Cpm = (H-H1-H2-H3)/(T1-T2);

# Results
print 'Mean heat capacity of solution is %f J/mol K'%Cpm
Mean heat capacity of solution is 97.648000 J/mol K

Example 7.23¶

In :
# Variables
To = 298.; 			#initial temperature (K)
Cpm = 97.65; 			#Mean heat capacity of solution (J/mol K)
Hs = -758.; 			#heat of mixing (J/mol)
H = 0.;

# Calculations
T = (H-Hs)/Cpm + To;

# Results
print 'The final temperature attained by the mixing is %f K'%T
The final temperature attained by the mixing is 305.762417 K