Chapter12-Phase equilibria

Ex1-pg346

import math
##Intitalisation of variables
k1= 0.015
k2= 0.028
p1= 0.7806 ##atm
p2= 0.21 ##atm
##CALCULATIONS
P1= k1*p1*100./(k1*p1+k2*p2)
P2= 100.-P1
##RESULTS
print'%s %.2f %s'% ('moles of Nitrogen = ',P1,' moles')
print'%s %.2f %s'% ('\n moles of Oxygen = ',P2,' moles')

moles of Nitrogen =  66.57  moles

 moles of Oxygen =  33.43  moles

Ex2-pg349

import math
##Intitalisation of variables
Ma= 153.8 ##gms
Mb= 169.9 ##gms
pa= 114.9 ##mm
pb= 238.3 ##mm
##CALCULATIONS
xa= (1./Ma)/((1./Ma)+(1./Mb))
xb= 1.-xa
Pa= pa*xa
Pb= pb*xb
Pt= Pa+Pb
##RESULTS
print'%s %.2f %s'% ('PA = ',Pa,' mm')
print'%s %.2f %s'% ('\n PB = ',Pb,' mm')
print'%s %.2f %s'% ('\n Total vapour pressure = ',Pt,' mm')

PA =  60.31  mm

 PB =  113.22  mm

 Total vapour pressure =  173.53  mm

Ex3-pg352

import math
##Intitalisation of variable
pa= 114.9 ##mm
pb= 238.3 ##mm
xa= 0.525
xb= 0.475
##CALCULATIONS
xa1= xa*pa/((xa*pa)+(xb*pb))
xb1= 1.-xa1
##RESULTS
print'%s %.2f %s'% ('Mole fraction CCl4 =  ',xa1,'')
print'%s %.2f %s'% ('\n Mole fraction of SiCl4 =  ',xb1,'')

Mole fraction CCl4 =   0.35 

 Mole fraction of SiCl4 =   0.65 

Ex4-pg363

import math
##Intitalisation of variables
p1= 55. ##per cent
P1= 744. ##mm
P2= 634. ##mm
MB= 18. ##gms
##CALCULATIONS
MA= p1*P2*MB/((P1-P2)*(100.-p1))
##RESULTS
print'%s %.2f %s'% ('Molecular weight of terpinene = ',MA,' gms') 

Molecular weight of terpinene =  126.80  gms

Ex5-pg367

import math
##Intitalisation of variables
ci= 0.1896 ##mole per liter
cKI= 0.02832 ##mole per liter
r= 625.
##CALCULATIONS
CI2= ci/r
dc= cKI-CI2
##RESULTS
print'%s %.2f %s'% ('Conc of I2 in KI layer = ',CI2,' mole per litre') 
print'%s %.2f %s'% ('\n Conc of I3- ions = ',dc,' mole per litre') 

Conc of I2 in KI layer =  0.00  mole per litre

 Conc of I3- ions =  0.03  mole per litre