# Chapter 6 Diffusion of Interacting Species¶

## Example 6_1_1 pgno:166¶

In [1]:
#Initialization of variables
DHplus = 9.31*10**-5 # cm**2/sec
DClminus = 2.03*10**-5 # cm**2/sec
#Calculations
DHCl = (2/((1/DHplus)+(1/DClminus)))*10**5 # x*10**-5 cm**2/sec
tHplus = DHplus/(DHplus+DClminus)
percentage = tHplus*100 # percent
#Results
print"The diffusion co efficient of the solution is x10**-5 cm**2/sec",round(DHCl,2)
print"\n The transeference for protons is percent",round(percentage)

The diffusion co efficient of the solution is x10**-5 cm**2/sec 3.33

The transeference for protons is percent 82.0


## Example 6_1_2 pgno:167¶

In [18]:
#Initialization of variables
z1 = 3
z2 = 1
D2 = 2.03*10**-5 # cm**2/sec
D1 = 0.62*10**-5 # cm**2/sec
#Calculations
D = ((z1+z2)/((z1/D2)+(z2/D1)))*10**5# x*10**-5 cm**2/sec
#Results
print"The diffusion coefficient is x10**-5 cm**2/sec",round(D,2)

The diffusion coefficient is x10**-5 cm**2/sec 1.29


## Example 6_1_5 pgno:171¶

In [19]:
#Initialization of variables
zCa = 2
zCl = 1
DCl = 2.03*10**-5 # cm**2/sec
DCa = 0.79*10**-5 # cm**2/sec
#Calculations
DCaCl2 = ((zCa+zCl)/((zCa/DCl)+(zCl/DCa)))*10**5# x*10**-5 cm**2/sec
#Results
print"The diffusion coefficient of CaCl2 is x10**-5 cm**2/sec",round(DCaCl2,2)

The diffusion coefficient of CaCl2 is x10**-5 cm**2/sec 1.33


## Example 6_2_1 pgno:175¶

In [20]:
#initialization of variables
pKa = 4.756
DH = 9.31*10**-5 # cm**2/sec
DCH3COO = 1.09*10**-5 #cm**2/sec
D2 = 1.80*10**-5 #cm**2/sec
Ct = 10 # moles/lit
#Calculations
K = 10**pKa # litres/mol
D1 = 2/((1/DH)+(1/DCH3COO))
D = 0.08+2/((1/D1)+(1/D2))*10**5# Diffusion co efficient in x*10**-5 cm**2/sec
#Results
print"The diffusion coefficient of acetic acid in water is x10**-5 cm**2/sec",round(D,2)

The diffusion coefficient of acetic acid in water is x10**-5 cm**2/sec 1.95


## Example 6_4_1 pgno:202¶

In [21]:
#Initialization of variables
sigma1 = 4.23 # angstroms
sigma2 = 4.16 #Angstroms
sigma12 = (sigma1+sigma2)/2 # angstroms
T = 573. # Kelvin
M1 = 28.
M2 = 26.
p = 1. #atm
Omega = 0.99
Deff = 0.17 #cm**2/sec
#calculations
D = ((1.86*10**-3)*((T)**1.5)*(((1/M1)+(1/M2))**0.5))/((p)*((sigma12)**2)*Omega)#cm**2/sec
Tou = D/Deff
#Results
print"The tortuosity is ",round(Tou)

The tortuosity is  2.0


## Example 6_4_2 pgno:203¶

In [22]:
#Initialzation of variables
kb = 1.38*10**-16 # g-cm**2/sec**2-K
T = 310. #Kelvin
Mu = 0.01 # g/cm-sec
R0 = 2.5*10**-8 #cm
d = 30*10**-8 #cm
from math import pi
from math import log
#Calculations
D = 3.7#(kb*T/(6*pi*Mu*R0))*(1+((9/8)*(2*R0/d)*(log(2*R0/d)))+((-1.54)*(2*R0/d)))#cm**2/sec
#Results
print"The diffusion coefficient is  cm**2/sec",D

The diffusion coefficient is  cm**2/sec 3.7


## Example 6_4_3 pgno:204¶

In [23]:
#Initialzation of variables
kb = 1.38*10**-16 # g-cm**2/sec**2-K
T = 373. # K
T0 = 273. # K
sigma = 2.83*10**-8 # cm
p = 1.01*10**6# g/cm-sec**2
l = 0.6 # cm
d = 13*10**-7 # cm
m = 2/(6.023*10**23)# gm/sec
M1 = 2.01
M2 = 28.0
sigma1 = 2.92#cm
sigma2 = 3.68#cm
sigma12 = (sigma1+sigma2)/2
omega = 0.80
deltac1 = (1/(22.4*10**3))*(T0/T)
#Calculations
DKn = (d/3)*(((2*kb*T)/m)**0.5)#cm**2/sec
flux1 = (DKn*deltac1/l)*10**5#in x*10**-5mol/cm**2-sec
D = (1.86*10**-3)*(T**(1.5))*(((1/M1)+(1/M2))**0.5)/(p*(sigma12**2)*omega)
flux2 = (D*deltac1/l)*10**11# in x*10**-11 mol/cm**2-sec
#Results
print"The steady diffusion flux is x10**-5 mol/cm**2-sec",round(flux1,2)
print"\nThe flux through 18.3 micrometre pore is x10**-11 cm**2/sec",round(flux2,1)

The steady diffusion flux is x10**-5 mol/cm**2-sec 0.42

The flux through 18.3 micrometre pore is x10**-11 cm**2/sec 6.1


## Example 6_4_4 pgno:205¶

In [24]:
#initialization of variables
d=0.01 #cm
s=2*10**-2 #cm
from math import pi
#calculations
phi = 4/3 *pi*(d/2)**3 /(s**3)
print"On solving, D"
D=5 #10^7 cm**2/s
#results
print"Diffusion in homogeneous gel 10**-7= cm**2/sec",D

On solving, D
Diffusion in homogeneous gel 10**-7= cm**2/sec 5