# Chapter 11 : Phase Equilibria¶

## Example 11.1 Page No : 225¶

In [1]:
# Variables
T = 0.       			#C
sv = 1.0001 			#cc g**-1
sv1 = 1.0907 			#cc g**-1
R = 0.0242 	    		#atm**-1 cc**-1 cal
p = 79.8 		    	#atm

# Calculations
r = (273.2+T)*(sv-sv1)*R/p

# Results
print  'rate of change of melting point  = %.4f deg atm**-1'%(r)

rate of change of melting point  = -0.0075 deg atm**-1


## Example 11.2 Page No : 225¶

In [2]:
# Variables
T = 95.5 			#C
p = 1.   			#atm
v = 0.0126 			#cc g**-1
a = 0.0242 			#cal cc**-1 atm**-1
r = 0.035 			#K atm**-1

# Calculations
dH = (273.2+T)*v*a/r

# Results
print  'Heat of transition  = %.1f cal g**-1'%(dH)

Heat of transition  = 3.2 cal g**-1


## Example 11.3 Page No : 226¶

In [3]:
# Variables
T = 100. 			#C
j = 0.0242 			#cal cc**-1 atm6-1
k = 539. 			#cal g**-1
p = 1664. 			#cc g**-1

# Calculations
r = (273.2+T)*(p-1)*j/k

# Results
print  'Rise in temperature per unit of pressure = %.1f deg atm**-1'%(r)

Rise in temperature per unit of pressure = 27.9 deg atm**-1


## Example 11.4 Page No : 228¶

In [4]:
# Variables
T1 = 100. 			#C
T2 = 90. 			#C
p = 76. 			#cm of hg
H = 542*18.02 			#cal mole**-1

# Calculations
p1 = p/10**((H/4.576)*((T1-T2)/((273.2+T1)*(273.2+T2))))

# Results
print  'Final vapour pressure of water  = %.1f cm'%(p1)

Final vapour pressure of water  = 52.9 cm


## Example 11.5 Page No : 230¶

In [5]:
# Variables
T = 239.05 			#K
r = 0.0242 			#cal cc6-1 atm**-1
Vv = 269.1 			#cc g**-1
Vl = 0.7 			#cc g**-1
r1 = 3.343 			#cm of mercury deg6-1
p = 76. 			#cm

# Calculations
tbyp = r1/p
dH = T*(Vv-Vl)*tbyp*r

# Results
print  'heat of vapourisation of liquid chlorine  = %.1f cal g**-1'%(dH)

heat of vapourisation of liquid chlorine  = 68.3 cal g**-1


## Example 11.6 Page No : 231¶

In [6]:
# Variables
Ta = 441. 		    	#C
Tb = 882. 	    		#C
Tb1 = 1218. 			#C

# Calculations
Ta1 = (273+Tb1)*(Tb+273)/(273+Ta)
Tb = Ta1-273

# Results
print  'Normal boiling point of silver  = %.f K'%(Ta1)
print  ' Normal boiling point of silver in degrees  = %.f degrees'%(Tb)

Normal boiling point of silver  = 2412 K
Normal boiling point of silver in degrees  = 2139 degrees


## Example 11.7 Page No : 233¶

In [7]:
# Variables
import math
T = 40. 			#C
T1 = 80.1 			#C

# Calculations
H = 2*(273.2+T1)
p = math.e**(-(H/(4.576*(273.2+T)))+4.59)/3.07

# Results
print  'vapour pressure  = %.1f cm'%(p)

vapour pressure  = 19.6 cm


## Example 11.8 Page No : 237¶

In [8]:
# Variables
p = 23.76 			#mm
R = 0.082 			#atm-lit deg**-1 mol**-1
T = 25. 			#C
vl = 18 			#ml
p1 = 1. 			#atm

# Calculations
dP = 0.001*vl*p*p1/(R*(273+T))
p2 = p+dP

# Results
print  'vapour pressure  = %.2f mm'%(p2)

# Note : ANSWER GIVEN IN THE TEXTBOOK IS WRONG

vapour pressure  = 23.78 mm


## Example 11.9 Page No : 246¶

In [9]:
# Variables
T = 25. 			    #C
R = 8.314*10**7 		#ergs /mol K
st = 72. 		    	#dynes cm**-1
mv = 18. 	    		#cc mole**-1
r = 10.**-5 			#cm
p = 23.76    			#cm

# Calculations
p1 = p*10**(2*st*mv/(r*R*2.303*(273.2+T)))

# Results
print  'vapour pressure  = %.2f mm'%(p1)

vapour pressure  = 24.01 mm

In [ ]: