Chapter 08: Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases

Example Problem 8.2, Page Number 186

In [39]:
from math import log, exp

#Varialble Declaration
Tn = 353.24       #normal boiling point of Benzene, K
pi = 1.19e4       #Vapor pressure of benzene at 20°C, Pa
DHf = 9.95        #Latent heat of fusion, kJ/mol
pv443 = 137.      #Vapor pressure of benzene at -44.3°C, Pa
R = 8.314         #Ideal Gas Constant, J/(mol.K)
Pf = 101325       #Std. atmospheric pressure, Pa
T20 = 293.15      #Temperature in K
P0 = 1.
Pl = 10000.
Ts = -44.3        #Temperature of solid benzene, °C

#Calculations
Ts = Ts + 273.15
#Part a

DHv = -(R*log(Pf/pi))/(1./Tn-1./T20)
#Part b

DSv = DHv/Tn
DHf = DHf*1e3
#Part c

Ttp = -DHf/(R*(log(Pl/P0)-log(pv443/P0)-(DHv+DHf)/(R*Ts)+DHv/(R*T20)))
Ptp = exp(-DHv/R*(1./Ttp-1./Tn))*101325

#Results
print 'Latent heat of vaporization of benzene at 20°C %4.1f kJ/mol'%(DHv/1000)
print 'Entropy Change of vaporization of benzene at 20°C %3.1f J/mol'%DSv
print 'Triple point temperature = %4.1f K for benzene'%Ttp
print 'Triple point pressure = %4.2e Pa for benzene'%Ptp

Latent heat of vaporization of benzene at 20°C 30.7 kJ/mol
Entropy Change of vaporization of benzene at 20°C 86.9 J/mol
Triple point temperature = 267.3 K for benzene
Triple point pressure = 3.53e+03 Pa for benzene

Example Problem 8.3, Page Number 191

In [40]:
from math import cos, pi

#Varialble Declaration
gama = 71.99e-3   #Surface tension of water, N/m
r = 1.2e-4        #Radius of hemisphere, m
theta = 0.0       #Contact angle, rad

#Calculations
DP = 2*gama*cos(theta)/r
F = DP*pi*r**2

#Results
print 'Force exerted by one leg %5.3e N'%F

Force exerted by one leg 5.428e-05 N

Example Problem 8.4, Page Number 191

In [41]:
from math import cos

#Varialble Declaration
gama = 71.99e-3   #Surface tension of water, N/m
r = 2e-5          #Radius of xylem, m
theta = 0.0       #Contact angle, rad
rho = 997.0       #Density of water, kg/m3
g = 9.81          #gravitational acceleration, m/s2
H = 100           #Height at top of redwood tree, m

#Calculations
h = 2*gama/(rho*g*r*cos(theta))

#Results
print 'Height to which water can rise by capillary action is %3.2f m'%h
print 'This is very less than %4.1f n, hence water can not reach top of tree'%H

Height to which water can rise by capillary action is 0.74 m
This is very less than 100.0 n, hence water can not reach top of tree