# Chapter 15 : The Phase Rule¶

### Example 15.2 Page: 401¶

In [1]:
import math

print " In this system there are four identifiable chemical species%(which are C,O2,CO2 and CO. The balanced equations we can write among them are"

print "    C + 0.5O2 = CO"
print "    C + O2 = CO2"
print "    CO + 0.5O2 = CO2"
print "    CO2 + C = 2CO"

print " There are only two independent relations among these four species and"

V = 2# No of the variable
P = 2# No of the phases
C = V + P - 2
print "    C = V + P - 2"
print "    C = 4 - 2 = 2"
print " Thus, this is a two-component system"

 In this system there are four identifiable chemical species%(which are C,O2,CO2 and CO. The balanced equations we can write among them are
C + 0.5O2 = CO
C + O2 = CO2
CO + 0.5O2 = CO2
CO2 + C = 2CO
There are only two independent relations among these four species and
C = V + P - 2
C = 4 - 2 = 2
Thus, this is a two-component system


### Example 15.3 Page: 402¶

In [2]:
import math

print " The three species in this system are H2 N2 and NH3"
N = 3
print " There is only one balanced chemical reaction among these species"
Q = 1

C = N - Q
print "    C = N - Q = %0.0f"%(C)
print " Let we start with pure ammonia in the system then ammonia will dissociate in H2 and N2 in the ratio of 3:1."

print " And the relation between their mole fraction is    y_H2 = 3*y_N2"

SR = 1
c = N-Q-SR
print " We have the modified phase rule as    Components = species - independent reactions - stoichiometric restriction"
print "    C = N - Q - SR = %0.0f"%(c)

 The three species in this system are H2 N2 and NH3
There is only one balanced chemical reaction among these species
C = N - Q = 2
Let we start with pure ammonia in the system then ammonia will dissociate in H2 and N2 in the ratio of 3:1.
And the relation between their mole fraction is    y_H2 = 3*y_N2
We have the modified phase rule as    Components = species - independent reactions - stoichiometric restriction
C = N - Q - SR = 1


### Example 15.4 Page: 403¶

In [3]:
import math

N = 3# No of species
Q = 1 # no of reaction

SR = 0
C = N - Q - SR

print "Number of the components presents in the test tube are %0.0f"%(C)

Number of the components presents in the test tube are 2


### Example 15.5 Page: 403¶

In [4]:
import math

P = 3
C = 2

V = C + 2 - P

print " The no. of phases present in the system are %0.0f "%(P)
print " Total no of degrees of freedom is %0.0f "%(V)
print " Since there is only one degree of freedom so the system has a unique P-T curve"
print " which can be well represented by     logp/torr = 23.6193 - 19827/T"

 The no. of phases present in the system are 3
Total no of degrees of freedom is 1
Since there is only one degree of freedom so the system has a unique P-T curve
which can be well represented by     logp/torr = 23.6193 - 19827/T


### Example 15.6 Page: 404¶

In [5]:
import math

print " The five species present in the system are H2O%( HCl%( H+%( OH- and Cl-. "
N = 5               # Number of the species
print " Here we have two chemical relations:"
print "    H2O = H+  +  OH- "
print "    HCl = H+  +  Cl- "

Q = 2               # No of the reactions

SR = 1
C = N - Q - SR

print " Number of the components present in the system are     C = N - Q - SR = %0.0f"%(C)

 The five species present in the system are H2O%( HCl%( H+%( OH- and Cl-.
Here we have two chemical relations:
H2O = H+  +  OH-
HCl = H+  +  Cl-
Number of the components present in the system are     C = N - Q - SR = 2

In [ ]: