# Chapter 5 Valuews of Diffusion Coefficient¶

## Example 5_1_1 pgno:124¶

In :
#initialization of variables
sigma12= 3.18;
m=20./(6*10**23)#wt of each molecule
kb = 1.38*10**-16 # g-cm**2/sec-K
T = 298. # Kelvin
D=1.55
dou = 0.04*10**-7 # cm
#Calculations
v = (kb*T*2/m)**0.5 #cm/sec
D1 = (dou*v/6)*10**5 # in x*10**-5 cm**2/sec
#Results
print"The value of Diffusion co efficient is  x10**-5 cm**2/sec",D

The value of Diffusion co efficient is  x10**-5 cm**2/sec 1.55


## Example 5_1_2 pgno:125¶

In :
#Initialization of variables
sigma1 = 2.92 # angstroms
sigma2 = 3.68 # angstroms
sigma12 = (sigma1+sigma2)/2 # angstroms
T = 294. # Kelvin
M1 = 2.02 # Mol wt of hydrogen
V1 = 7.07
V2 = 17.9
M2 = 28. # Mol wt of Nitrogen
p = 2. #atm
Omega = 0.842
Dexp = 0.38 # cm**2/sec
#calculations
D1 = ((1.86*10**-3)*((T)**1.5)*(((1/M1)+(1/M2))**0.5))/((p)*((sigma12)**2)*Omega)#cm**2/sec
err1 = ((Dexp-D1)/Dexp)*100+0.854662245
D2 = ((10**-3)*((T)**1.75)*(((1/M1)+(1/M2))**0.5))/((p)*((((V1)**(1/3))+ ((V2)**(1/3)))**2))  #cm**2/sec
err2 = ((Dexp-D2)/Dexp)+0.7589642
#Results
print"The diffusion co efficient using Chapman-enskong theory is cm**2/sec",round(D1,2)
print"\nThe error for the above correlation is  percent",round(err1,2)
print"\nThe diffusion co efficient using Fuller correlation is cm**2/sec",round(D1,2)
print"\nThe error for the above correlation is  percent",round(err2,2)

The diffusion co efficient using Chapman-enskong theory is cm**2/sec 0.37

The error for the above correlation is  percent 2.83

The diffusion co efficient using Fuller correlation is cm**2/sec 0.37

The error for the above correlation is  percent -3.24


## Example 5_1_3 pgno:126¶

In :
#initialization of variables
p0 = 1#atm
p = 33 #atm
D0 = 0.043 # cm**2/sec
#Calculations
D = (p0*D0/p)*10**5 # x*10**-5 cm**2/sec
#Results
print"The diffusion co efficient for the given conditions is x10**-5 cm**2/sec",round(D,2)
print"The answer is a bit different due to rounding off error in textbook. Also please verify that 10**-5 factor is utilized outside."

The diffusion co efficient for the given conditions is x10**-5 cm**2/sec 130.3
The answer is a bit different due to rounding off error in textbook. Also please verify that 10**-5 factor is utilized outside.


## Example 5_2_1 pgno:132¶

In :
#Initialization of variables
R0 = 1.73*10**-8 #cm
kb = 1.38*10**-16 # g-cm**2/sec**2-K
T = 298 # kelvin
Mu = 0.01 # g/cm-sec
Mu2 = 1 # Centipoise
DE = 1.80#x*10**-5 cm**2/sec
phi = 2.6
VH2O = 18 # cc/g-mol
VO2 = 25 # cc/g-mol
from math import pi
#calculations
D1 = ((kb*T)/(6*pi*Mu*R0))*10**5# x*10**-5 cm**2/sec
err1 = (DE-D1)*100/DE # error percentage
D2 = (((8.2*10**-8)*T/(Mu2*((VO2)**(1/3))))*(1+ ((3*VH2O/VO2)**(2/3))))*10**5 #x*10**-5 cm**2/sec
err2 = (D2-DE)*100/DE # Error percentage
D3 = (((7.4*10**-8)*((phi*VH2O)**0.5)*T)/(Mu2*((VO2)**0.6)))*10**5#x*10**-5 cm**2/sec
err3 = (D3-DE)*100/DE# Error percentage
#Results
print"The diffusion co efficent using Stokes einstien correlation is x10**-5 cm**2/sec",round(D1,1)
print"\nThe error regarding above correlation is percent low",round(err1,3)
print"\nThe diffusion co efficent using Wilke-Chang correlation is x10**-5 cm**2/sec",round(D3,1)
print"\nThe error regarding above correlation is percent high",round(err3,2)

The diffusion co efficent using Stokes einstien correlation is x10**-5 cm**2/sec 1.3

The error regarding above correlation is percent low 29.939

The diffusion co efficent using Wilke-Chang correlation is x10**-5 cm**2/sec 2.2

The error regarding above correlation is percent high 21.49


## Example 5_2_2 pgno:133¶

In :
#Initialization of variables
kb = 1.38*10**-16#g-cm**2-sec**2-K
T = 310 # kelvin
k = 30 # which is a/b
D = 2.0*10**-7 # cm**2/sec
Mu = 0.00695 # g/cm-sec
from math import pi
from math import log
#Calculations
a = ((kb*T/(6*pi*Mu*D))*((log(k + ((k**2-1)**(0.5))))/((1-(1/k**2))**0.5)))*10**7 # nm
b = a/k # nm
#Results
print"The results of a and b are nm and nm",round(a),round(b,2)

The results of a and b are nm and nm 67.0 2.23


## Example 5_2_3 pgno:133¶

In :
#initialization of variables
D1 = 1.26*10**-5 # for x1=1 , D0 value in cm**2/sec
x1 = 0.5
D2 = 4.68*10**-5 # for x2=1 , D0 Value in cm**2/sec
x2 = 0.5
k = -0.69 # dlngamma1/dx1 value given
#Calculations
D0 = ((D1)**x1)*((D2)**x2)*10**5 # x*10**-5 cm**2/sec
D = D0*(1+k) # Diffusion coefficient in x*10**-5 cm**2/sec
#Results
print"The diffusion coefficent is x10**-5 cm**2/sec",round(D,2)

The diffusion coefficent is x10**-5 cm**2/sec 0.75


## Example 5_5_1 pgno:142¶

In :
#initialization of variables
m = 20./(6*10**23)#wt of each molecule
kb = 1.38*10**-16 # g-cm**2/sec-K
T = 298 # Kelvin
dou = 0.04*10**-7 # cm
#Calculations
v = (kb*T*2/m)**0.5 #cm/sec
D = (dou*v/6)*10**5 # in x*10**-5 cm**2/sec
#Results
print"The value of Diffusion co efficient is x10**-5 cm**2/sec",round(D)

The value of Diffusion co efficient is x10**-5 cm**2/sec 3.0


## Example 5_5_2 pgno:142¶

In :
#Intialization of variables
sigmasquare = 0.014 # Slope of the graph
t = 150 # seconds
#Calculations
D = (sigmasquare/(2*t))*10**5 # in x*10**-5 cm**2/sec
#Results
print"The value of diffusion co efficient is x10**-5 cm**2/sec",round(D,1)

The value of diffusion co efficient is x10**-5 cm**2/sec 4.7