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
```

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
```

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
```