import math
# Given that
p = 1.013e5 # Pressure in Pa
t = 300 # Temperature in K
d = 3.5 # Effective diameter of oxygen molecule in Angstrom
r = 2 # Ratio of free path of molecules with the lambda
print "\n Example 22.1 \n"
sigma = math.pi*(d*(10**-10))**2
n = p/(t*1.38*(10**-23))
R = math.exp(-r)
print "\n Mean free path = math.exp m,\n The fraction of molecules have free path longer than 2*lambda = ",R*100, " percent"
# Answer given in the book contain round off error for mean free path.
import math
# Given that
lambda1 = (2.63e-5) # Mean free path of the molecules of the gas in m
t = 25 # Temperature in degree centigrade
r = 2.56e-10 # Radius of the molecules in m
print "\n Example 22.2 \n"
sigma = 4*math.pi*r**2
n = 0.707/(sigma*lambda1)
p = n*(t+273)*(1.38*10**-23)
N = 1.0/lambda1
print "\n Pressure of the gas = ",p," Pa,\n No of collisions made by a molecule per meter of path = math.exp",N
# Given that
import math
from scipy import integrate
lambda1 = 10.0 # Mean free path of the gas in cm
N0 = 10000.0 # No of free paths
x1 = 10.0 # In cm
x2 = 20.0 # In cm
x3 = 50.0 # In cm
x4 = 5.0 # In cm
x5 = 9.5 # In cm
x6 = 10.5 # In cm
x7 = 9.9 # In cm
x8 = 10.1 # In cm
print "\n Example 22.3 \n"
# For x>10 cm
N1 = N0*(math.exp(-1))
# For x>20 cm
N2 = N0*(math.exp(-2))
# For x>50 cm
N3 = N0*(math.exp(-5))
def f(x):
y = (-N0/lambda1)*(math.exp((-x)/lambda1)),
return y
# For 5>x>10 cm
N4,er = integrate.quad( lambda x: (-N0/lambda1)*(math.exp((-x)/lambda1)),x4,x1)
# For 9.5>x>10.5 cm
N5,e = integrate.quad( lambda x: (-N0/lambda1)*(math.exp((-x)/lambda1)),x5,x6)
# For 9.9>x>10.1 cm
N6,eor = integrate.quad( lambda x: (-N0/lambda1)*(math.exp((-x)/lambda1)),x7,x8)
# For x=10 cm
N7,eer = integrate.quad( lambda x: (-N0/lambda1)*(math.exp((-x)/lambda1)),x1,x1)
print "\n The no of free paths which are longer than, \n 10 cm = ",math. ceil(N1) ,",\n 20 cm = ",math. ceil(N2) ,",\n 50 cm = ",math. ceil(N3) ,",\n\n The no of free paths which are between,\n 5 cm and 10 cm = ",math.floor(N4) ,",\n 9.5 cm and 10.5 cm = ",math.floor(N5) ,",\n 9.9 cm and 10.1 cm = ",math.floor(N6) ,",\n\n The no of free paths which are exactly 10 cm = ",N7
import math
# Given that
p = 1.0 # Pressure in atm
t = 300.0 # Temperature in K
print "\n Example 22.4 \n"
# From previous example, we have
m = 5.31e-26 # In kg/molecule
v = 445.0 # In m/s
sigma = 3.84e-19 # In m**2
# Therefore
mu = (1.0/3.0)*(m*v/sigma)
print "\n Coefficient of viscosity = math.exp Ns/m**2",mu
import math
# Given that
p = 1.0 # Pressure in atm
t = 300.0 # Temperature in K
F = 5.0 # For oxygen gas degree of freedom
print "\n Example 22.5 \n"
v = 445.0 # In m/s as given in the book
m = 5.31e-26 # Mass of oxygen molecule in kg
sigma = 3.84e-19 # As given in the book in m**2
k = (1/6)*(v*F*(1.38*10**-23))/sigma
# If the gas has Maxwellian velocity distribution,
k_ = (1.0/3.0)*(F*(1.38*10**-23)/sigma)*((1.38*10**-23)*t/(math.pi*m))**(1/2)
print "\n Thermal conductivity = ",k ," W/mK,\n If the gas has Maxwellian velocity distribution,\n Thermal conductivity = ",k_ ," W/mK"
import math
# Given that
F = .90 # Fraction of electrons leaving the cathode ray reach the anode without making a collision
x = 0.2 # Distance between cathode ray and anode in m
d = 3.6e-10 # Diameter of ion in m
t = 2000.0 # Temperature of electron in K
print "\n Example 22.6 \n"
lambda1 = x/(math.log(1/F))
sigma = math.pi*(d**2)
n = 4/(sigma*lambda1)
p = n*(1.38*10**-23)*(t)
print "\n Pressure in the cathode ray tube = ",p ," Pa"
import math
# Given that
V = 1.0 # Volume of the flask in litre
p = 1.0 # Pressure in atm
t = 300.0 # Temperature in K
r = 1.8e-10 # Radius of oxygen gas molecule in m
m = 5.31e-26 # Mass of oxygen molecule in kg
print "\n Example 22.7 \n"
n = (p*(1.013e5))/((1.38e-23)*(t)*1000)
sigma = 4*math.pi*(r**2)
v = ((8*(1.38e-23)*t)/(math.pi*m))**(1/2)
z = sigma*n*v*1000
N = (1.0/4.0)*(n*0.1*v)
print "\n No of collisions per sec are made by one molecule with the other molecule =", z,"\nThe no of molecules strike the flask per sq. cm =",N,"\n No of molecules in the flask =",n
import math
# Given that
lambda1 = 2.0 # Mean free path in cm
T = 300.0 # Temperature in K
r = 0.5 # As half of the molecules did not make any collision
print "\n Example 22.8 \n"
x = lambda1*(math.log(1/r))
v = 445.58 # For oxygen at 300K in m/s
t = x/(v*100)
print "\n Time =", math.exp(t), "s"
import math
# Given that
f = 0.9 # Fraction of electrons leaving the cathode ray and reaching the anode without making any collision
x = 20.0 # Distance between cathode ray tube and anode in cm
sigma = 4.07e-19 # Collision cross section of molecules in m**2
T = 2000 # Temperature in K
print "\n Example 22.9 \n"
lambda1 = (x*0.01)/(math.log(1.0/f))
n = 1/(sigma*lambda1)
p = n*(1.38e-23)*T
print "\n Pressure =", math.exp(p), "N/m**2"
# The answer given in the book contains round off error.
import math
# Given that
l = 2.0 # Length of tube in m
a = 1e-4 # Cross section of the tube in m**2
p = 1.0 # Pressure in atm
t = 0 # Temperature in degree centigrade
r = 0.5 # Fraction of the carbon atoms which are radioactive C14
sigma = 4e-19 # Collision cross section area in m**2
print "\n Example 22.10 \n"
n = (p*1.01325e+5)/((1.38e-23)*(t+273))
C_g = -n/l
m = (46/6.023)*10**-26 # In kg/molecule
v = (2.55*(1.38e-23)*(t+273)/m)**(1/2.0)
lambda1 = (1.0/(sigma*n))
gama = (1.0/4)*(v*n) - (1/6.0)*(v*lambda1*(C_g))
gama_ = (1/4.0)*(v*n) + (1.0/6.0)*(v*lambda1*(C_g))
x = (1.0/4)*(v*n)
y = (1.0/6)*(v*lambda1*(C_g))
d = (1.0/6)*(v*lambda1*(-1*C_g))*2*(m)
a=x+y
b=x-y
print "\n Initial concentration gradient of reactive molecules =",math.exp (C_g)," molecules/m**4, \n The no of reactive molecules per sec cross a cross section at the mid point of the tube from left to right =",f , "molecules/m**2,\n The no of reactive molecules per sec cross a cross section at the mid point of the tube from right to left =",math.e ," molecule/m**2,\n Initial net rate of diffusion = ",d*1000 ,"g/m**2-s"
# The answer for lambda given in the book conatains calculation error
# The answers contains calculation error