Chapter 5 - The phase rule and solutions

Example 1 - pg 261

In [1]:
#calculate the molality of the mixture
#initialisation of variables
m= 98.08 #gms
d= 1.102 #g ml^-1
m1= 165.3 #gm
v= 1000 #ml
wt=.15
#CALCULATIONS
form=d*v*wt/m
M= d*v-m1
norm=2*form
m2= m1*v/(m*M)
#RESULTS
print '%s %.3f %s' % (' molality = ',m2,'molal')
print '%s %.3f %s' %('Formality = ',form,'gm formula wt/l')
print '%s %.3f %s' %('Normality = ',norm,'N')

 molality =  1.799 molal
Formality =  1.685 gm formula wt/l
Normality =  3.371 N

Example 2 - pg 272

In [2]:
#calculate the Increase in enthalpy
#initialisation of variables
T= -40 #C
v= 217.4 #cm^3
r= 8.8 # atm deg^-1
m= 18 #gms
#CALCULATIONS
H= (273+T)*(-v*m/1000)*r*(1.987/82.05)
#RESULTS
print '%s %.1f %s' % (' Increase in enthalpy =',H,'cal mole^-1')

 Increase in enthalpy = -194.3 cal mole^-1

Example 3 - pg 279

In [3]:
#calculate the density
#initialisation of variables
T= 27 #C
R= 0.08206 #cal/mol T
W= 28.6 #gms
#CALCULATIONS
d= W/((273.2+T)*R)
#RESULTS
print '%s %.3f %s' % (' density =',d,' g l^-1')

 density = 1.161  g l^-1

Example 4 - pg 289

In [4]:
#calculate the mole fraction and total pressure
#initialisation of variables
P= 408. #mm of Hg
P1= 141. # mm of Hg
p= 60.
#CALCULATIONS
P2= P*(100-p)/100.
P3= P1*p/100.
N= P2/(P2+P3)
P4= P2+P3
#RESULTS
print '%s %.3f' % (' mole fraction = ',N)
print '%s %.1f %s' % (' \n total pressure =',P4,' mm of Hg')

 mole fraction =  0.659
 
 total pressure = 247.8  mm of Hg

Example 5 - pg 289

In [5]:
#calculate the molality
#initialisation of variables
P2= 760. #mm of Hg
m2= 2.18*10**-3
v= 23.5 #ml
p= 21.
p1= 79.
#CALCULATIONS
K= P2*55.5/m2
K1= 760*55.5*22.4*10**3/v
m= 55.5*(p*760/(100*K))+55.5*(p1*760/(100*K1))
#RESULTS
print '%s %.2e %s' % (' molality =',m,'molal')

 molality = 1.29e-03 molal

Example 7 - pg 297

In [6]:
#calculate the percentage of Br in the vapors in steam distillation
#initialisation of variables
Ph= 643. #mm of Hg
Mh= 18. #gms
Po= 117. #mm of Hg
Mo= 157. #gms
#CALCULATIONS
r= Ph*Mh/(Po*Mo)
P= 100*(1/(1+r))
#RESULTS
print '%s %.1f %s' % (' percentage =',P,'percent')

 percentage = 61.3 percent

Example 8 - pg 306

In [7]:
#calculate the amounts of phases present at 375 and 370 C
#initialisation of variables
n= 1 
n1= 0.5
n3= 0.36
n4= 0.67
n5= 0.34
r= 3
#CALCULATIONS
A= (n-n1)/(n1-n3)
A1= r*(n4-n1)/(n1-n5)
#RESULTS
print '%s %.1f' % (' amount of phase at 375 C = ',A)
print '%s %.1f' % (' \n amount of phase at 370 C = ',A1)

 amount of phase at 375 C =  3.6
 
 amount of phase at 370 C =  3.2

Example 9 - pg 311

In [8]:
#calculate the vapour pressure
#initialisation of variables
m= 100 #gms
m1= 1 #gms
m2= 2 #gms
P= 23.756 #mm of Hg
n= 18.02 
n1= 60.06
n2= 342.3 
#CALCULATIONS
r= ((m1/n1)+(m2/n2))/((m1/n1)+(m2/n2)+(m/n))
dp= P*r
P1= P-dp
#RESULTS
print '%s %.3f %s' % (' vapour pressure =',P1,' mm of Hg')

 vapour pressure = 23.660  mm of Hg

Example 11 - pg 315

In [9]:
#calculate the boiling point of solution
#initialisation of variables
kf= 0.514 #K/molal
m= 0.225 #molal
#CALCULATIONS
dT= kf*m
T2=dT+100.
#RESULTS
print '%s %.3f %s' % (' boiling point =',T2,' C')

 boiling point = 100.116  C

Example 12 - pg 315

In [10]:
#calculate the molecular weight of the solute
#initialisation of variables
kb= 2.64 #C gm
dT= 0.083 #C
m= 120 #gms
W2= 0.764 #gms
#CALCULATIONS
m2= dT/kb
M2= W2*1000/(m2*m)
#RESULTS
print '%s %.1f %s' % (' molecular weight of solute =',M2,'gms')

 molecular weight of solute = 202.5 gms

Example 13 - pg 318

In [11]:
#calculate the value of n
#initialisation of variables
T= 176.5 #C
T1= 158.8 #C
Kf= 37.7
W1= 0.522 #gms
W2= 0.0386 #gms
m= 12 #gms
m1= 1 #gm
#CALCULATIONS
m3= (T-T1)/Kf
M2= W2*1000/(m3*W1)
r= M2/(m+m1)
#RESULTS
print '%s %d' % ('value of n = ',r)
print '%s %d %s' %('Molecular weight = ',M2,'gm')

value of n =  12
Molecular weight =  157 gm

Example 14 - pg 319

In [12]:
#calculate the triple point of the system
#initialisation of variables
T= 273.2 #K
P= 0.0060 #atm
P1= 1 #atm
H= 3290 #cal
dV= -0.0907 #cc
#CALCULATIONS
dT= T*dV*(P-P1)/H
#RESULTS
print '%s %.4f %s' % (' triple point =',dT,'C') 

 triple point = 0.0075 C

Example 16 - pg 323

In [13]:
#calculate the fraction of impurity in both cases
#initialisation of variables
n= 100.
K= 2.
V= 100. #ml
V2= 1000. #ml
n= 10.
n1= 100.
#CALCULATIONS
x= (K*V/(K*V+(V2/n)))**n
y= (K*V/(K*V+(V2/n1)))**n1
#RESULTS
print '%s %.4f' % (' fraction of impurity = ',x)
print '%s %.4f' % (' \n fraction of impurity = ',y)

 fraction of impurity =  0.0173
 
 fraction of impurity =  0.0076

Example 17 - pg 328

In [14]:
#calculate the molecular weight of the protein
#initialisation of variables
T= 27 #C
m= 0.635 #gms
V= 100 #ml
R= 0.08205 #cal/mol K
p= 2.35 #cm
#CALCULATIONS
M= 13.6*76*m*R*(T+273)*1000/(p*V)
#RESULTS
print '%s %d %s' % (' molecular weight =',M,'gms')
print 'The answer is a bit different due to rounding off error in textbook'

 molecular weight = 68747 gms
The answer is a bit different due to rounding off error in textbook

Example 18 - pg 328

In [15]:
#calculate the osmotic pressure
#initialisation of variables
import math
R= 0.08205 #cal/mol K
v1= 0.0180#cc
N= 0.9820
T= 273.2
#CALCULATIONS
P= -R*T*math.log(N)/v1
#RESULTS
print '%s %.1f %s' % (' osmotic pressure =',P,'atm')

 osmotic pressure = 22.6 atm

Example 19 - pg 331

In [16]:
#calculate the osmotic pressure
#initialisation of variables
kf= 1.86
dT= 0.402 #K
T= 310 #K
R= 0.08205 #cal/mol K
#CALCULATIONS
P= dT*T*R/kf
#RESULTS
print '%s %.2f %s' % (' osmotic pressure =',P,'atm')

 osmotic pressure = 5.50 atm

Example 20 - pg 333

In [17]:
#calculate the Degrees of ionisation
#initialisation of variables
m= 0.100 #gms
kf= 1.86 #K/gms
dT= 0.300 #K
v= 2
#CALCULATIONS
T= kf*m
i= dT/T
a= (i-1)/(v-1)
#RESULTS
print '%s %.2f' % (' Degrees of ionisation = ',a)

 Degrees of ionisation =  0.61

Example 21 - pg 335

In [18]:
#calculate the lowering of the freezing point
#initialisation of variables
W= 0.0020 #M
W1= 0.0010 #M
W2= 0.0040 #M
T= 1.86 #C
n= 1 #moles
n1= 1 #moles
n2= 2 #moles
a= 1.122
#CALCULATIONS
dT= T*(W+W1+W2)
I= 0.5*(n**2*W+n1**2*W2+n2**2*W1)
g= 1-(2*a*I**1.5/(3*(W+W1+W2)))
dT1= g*dT
#RESULTS
print '%s %.4f %s' % (' lowering the freezing point =',dT1,'C ')

 lowering the freezing point = 0.0125 C

Example 22 - pg 338

In [19]:
#calculate the activity coefficient of acetone and water
#initialisation of variables
p= 1820 #mm
n= 2.5 #mole percent
f= 0.470
P= 420 #mm
n1= 97.5 #percent
#CALCULATIONS
P1= p*n/(100*760)
F= f/P1
F1= (1-f)*760.*100/(P*n1)
#RESULTS
print '%s %.2f' % (' activity coefficient of acetone =  ',F)
print '%s %.2f' % (' \n activity coefficient of water =  ',F1)

 activity coefficient of acetone =   7.85
 
 activity coefficient of water =   0.98