# Chapter 4 : Equation of states and intermolecular forces¶

## Example 4.1 Page No : 220¶

In [41]:
import math
from numpy import *

# Variables
H2O = 1. ;
NH3 = 2. ;
CH4 = 3. ;
CH3Cl = 4. ;
CCl4 = 5. ;

M_11 = 1.85
alp_12 = 14.80
I_13 = 12.62 ;
M_12 = 1.47
alp_22 = 22.20
I_23 = 10.07 ;
M_31 = 0.00
alp_32 = 26.00
I_33 = 12.61 ;
M_41 = 1.87
alp_42 = 45.30
I_43 = 11.26 ;
M_51 = 0.00
alp_52 = 105.0
I_53 = 11.47 ;

# Calculations
k = 1.38 * 10**-16 ; 			#[ J/K]
T = 298.             			#[K]
A =[[M_11 , alp_12 , I_13, 0,0,0,0],[M_12 , alp_22 , I_23,0,0,0,0 ],[M_31 , alp_32 , I_33,0,0,0,0],[M_41 , alp_42 , I_43,0,0,0,0],[M_51 , alp_52 , I_53,0,0,0,0 ]]
print A

print (" Molecule       M       alp*10**25        I          C*10**60      Cd_d      Cind     Cdis") ;
for i in range(5):
A[i][4] = ceil( 2./3 * A[i][0]**4 / (k * T) * 10**-12) ;
A[i][5] = ceil(2 * A[i][1] * A[i][0]**2 * 10**-1) ;
A[i][6] = ceil(3./4 * A[i][1]**2 * A[i][2] * 1.6 * 10**-2) ;
A[i][3] = ceil(A[i][4] +  A[i][5] +  A[i][6]) ; 			# ....E4.1D

print "    H2O   %8.2f        %5.1f         %5.2f         %7d    %7d   %5d        %d "%(A[0][0],A[0][1],A[0][2],A[0][3],A[0][4],A[0][5],A[0][6]) ;
print "    NH3   %8.2f        %5.1f         %5.2f         %7d    %7d   %5d        %d "%(A[1][0],A[1][1],A[1][2],A[1][3],A[1][4],A[1][5],A[1][6]) ;
print "    CH4   %8.2f        %5.1f         %5.2f         %7d    %7d   %5d        %d "%(A[2][0],A[2][1],A[2][2],A[2][3],A[2][4],A[2][5],A[2][6]) ;
print "   CH3Cl  %8.2f        %5.1f         %5.2f         %7d    %7d   %5d        %d "%(A[3][0],A[3][1],A[3][2],A[3][3],A[3][4],A[3][5],A[3][6]) ;
print "   CCl4   %8.2f        %5.1f         %5.2f         %7d    %7d   %5d       %d "%(A[4][0],A[4][1],A[4][2],A[4][3],A[4][4],A[4][5],A[4][6]) ;

print "Even though it is non polar , CCl4 exhibit the largest intermolecular forces . It is due to the large polarizability accociated with the four Cl atom in CCl4 ."

# Note : Answer are slightly different because of rounding off error.

[[1.85, 14.8, 12.62, 0, 0, 0, 0], [1.47, 22.2, 10.07, 0, 0, 0, 0], [0.0, 26.0, 12.61, 0, 0, 0, 0], [1.87, 45.3, 11.26, 0, 0, 0, 0], [0.0, 105.0, 11.47, 0, 0, 0, 0]]
Molecule       M       alp*10**25        I          C*10**60      Cd_d      Cind     Cdis
H2O       1.85         14.8         12.62             235        190      11        34
NH3       1.47         22.2         10.07             146         76      10        60
CH4       0.00         26.0         12.61             103          0       0        103
CH3Cl      1.87         45.3         11.26             509        199      32        278
CCl4       0.00        105.0         11.47            1518          0       0       1518
Even though it is non polar , CCl4 exhibit the largest intermolecular forces . It is due to the large polarizability accociated with the four Cl atom in CCl4 .


## Example 4.2 Page No : 222¶

In [2]:
# Variables
import math
C6_Ar_HCl_tab = 76 * 10**-60 ;			#From table E4.2
C6_Ar_Ar_tab = 52 * 10**-60 ;			#From table E4.2
C6_HCl_HCl_tab = 134 * 10**-60 ;			#From table E4.2

# Calculations
C6_Ar_HCl_gmean = math.sqrt(C6_Ar_Ar_tab * C6_HCl_HCl_tab) ; 			#[erg/cm**6]
x = (C6_Ar_HCl_gmean - C6_Ar_HCl_tab) / C6_Ar_HCl_tab * 100 ;

# Results
print "The geometric mean is different from that in table E4.2 by %d %%"%(x)

The geometric mean is different from that in table E4.2 by 9 %


## Example 4.4 Page No : 230¶

In [27]:
# Variables
import math
Psat_wat_25 = 3.169 * 10**3 ;			# From steam table
Psat_wat_50 = 1.235 * 10**4 ;			# From steam table
Psat_wat_100 = 1.014 * 10**5 ;			# From steam table
A =11.9673 ;
B = 3626.55 ;
C = -34.29 ;
T1 = 25 ; 			#[*C]
T2 = 50 ; 			#[*C]
T3 = 100 ; 			#[*C]

# Calculations
M = [[T1, Psat_wat_25 , 0],[T2 , Psat_wat_50, 0],[T3 , Psat_wat_100, 0]]
#M = array(M)

print ("  T(*C)   Water(Pa)   Methanol(Pa)")
for i in range(3):
M[i][2] = math.exp(A - B / (M[i][0] + 273 + C)) * 10**5 ;
print "%5d     %7.3e    %7.2e"%(M[i][0],M[i][1],M[i][2])
#t.append(math.exp(A - B / (M[i][0] + 273 + C)) * 10**5)
#M.append(t)

# Results

print "1)     Water can form two hydrogen bonds . While CH4Oh can form only one . Thus at\
a given temperature, water has stronger attractive forces in the liquid and a lower vapour pressure ."
print "2)      Since the Maxwell-Boltzmann distribution depends exponentially on temperature,\
Psat also increses exponentially with temperature ."

# Note: Answers may vary because of rounding off error.

  T(*C)   Water(Pa)   Methanol(Pa)
25     3.169e+03    1.68e+04
50     1.235e+04    5.52e+04
100     1.014e+05    3.53e+05
1)     Water can form two hydrogen bonds . While CH4Oh can form only one . Thus at a given temperature, water has stronger attractive forces in the liquid and a lower vapour pressure .
2)      Since the Maxwell-Boltzmann distribution depends exponentially on temperature, Psat also increses exponentially with temperature .


## Example 4.6 Page No : 236¶

In [32]:
# Variables
Pc_B = 49.1 ; 			# [bar] , From table
Pc_T = 42.0 ; 			# [bar] , From table
Pc_C = 40.4 ; 			# [bar] , From table
Tc_B = 562 ; 			# [K] , From table
Tc_T = 594 ; 			# [K] , From table
Tc_C = 553 ; 			# [K] , From table
R = 8.314 ;

# Calculations
A = [[Pc_B , Tc_B, 0,0],[Pc_T , Tc_T,0,0],[Pc_C , Tc_C,0,0]]

# Results
print "   P_c     T_c        a      b "
for i in range(3):
A[i][2] = 27./64 * (R * A[i][1])**2 /( A[i][0] * 10**5) ;
A[i][3] = R * A[i][1] / (8 * A[i][0] * 10**5) ;
print "  %5.1f %5d   %7.2f  %7.2e"%(A[i][0],A[i][1],A[i][2],A[i][3])

print "The attractive interactions of all three compounds are dominated by\
print ersion interactions ( parameter a) , while size affects parameter b ."

   P_c     T_c        a      b
49.1   562      1.88  1.19e-04
42.0   594      2.45  1.47e-04
40.4   553      2.21  1.42e-04
The attractive interactions of all three compounds are dominated by print ersion interactions ( parameter a) , while size affects parameter b .


## Example 4.11 Page No : 246¶

In [5]:
# Variables
import math
B = 0.0486 * 10**-3 ;
T1 = 20 + 273 ; 			#[K]
T2 = 500 + 273 ; 			#[K]
v1 = 7.11 ; 			# [cm**3/mol]

# Calculations
v2 = v1 * math.exp( B * (T2 - T1)) ;

# Results
print "     Molar volume of solid state 2 = %.2f cm**3/mol"%( v2);

     Molar volume of solid state 2 = 7.28 cm**3/mol


## Example 4.12 Page No : 248¶

In [8]:
# Variables
P_c = 37.9 * 10**5 ;			#[N/m**2] , From compressibility chart
T_c = 425.2         			# [K , From compressibility chart
P = 50. * 10**5 ; 	    		#N/m**2]
T = 333.2 ;			            #[K]
R = 8.314 ;
z_0 = 0.2148 ;
z_1 = -0.0855 ;
w = 0.199 ;
m = 10. ;
MW = 0.05812 ;

# Calculations
a = (0.42748 * R**2 * T_c**2.5) / P_c ;
b = 0.08664 * R * T_c / P_c ;
A = P * T**(1./2) ;
B = -R * T**(3./2) ;
C = (a - P * T**(1./2) * b**2 - R * T**(3./2)*b) ;
D = - a * b;

#mycoeff = [ D , C , B , A] ;
mycoeff = [ A,B,C,D]
#p = poly1d(mycoeff , "v" , "coeff" );
M = roots(mycoeff);

# Results
for i in range(3):
ans = sign(M[i])
if ans == 1:
V = m / MW *(M[i]) ;
#print "Using Redlich Kwong equation the volume is = %.3f m**3"%(V)

z = z_0 + w * z_1 ;
v = z * R * T / P ;
V = m / MW * v ;
print "Using compressibility chart the volume is = %.3f m**3"%(V)

Using compressibility chart the volume is = 0.019 m**3


## Example 4.13 Page No : 253¶

In [4]:
# Variables
import math
T = 100. + 273 ; 			#[K]
P = 70. * 10**5 ; 			#[N/m**2]
P_c = 42.2 * 10 ** 5 ;
T_c = 370. ;     			#[K]
w = 0.153   	    		# Interpolating from table C.1 and C.2
z_0 = 0.2822 ;		    	# Interpolating from table C.1 and C.2
z_1 = - 0.0670 ;			# Interpolating from table C.1 and C.2
m = 20. * 10**3 ;			#[g]
MW = 44. ; 		        	#[g/mol]
R = 8.314 ;

# Calculations and Results
P_r = P / P_c ;
T_r = T / T_c ;
z = z_0 + w * z_1 ;
V = m / MW *z * R * T / P ;

print "1)          Volume = %.4f m**3 "%( V )

T = 295. ;			#[K]
n = 50. ; 			# [mol]
a = 0.42748 * R**2 * T_c**2.5 / P_c ;
b = 0.08664 * R * T_c / P_c ;
v = 0.1 ;
P = R * T / (v - b) - a / (T**0.5 * v * (v + b)) ;
x = P * n * 10**-6 ;

print "2)          Pressure = %.1f MPa "%( x )

y1 = 0.4 ;
y2 = 1 - y1 ;
n = 50. ;
P_c = 48.7 * 10**5 ;			#[N/m**2]
T_c = 305.5 ; 			#[K]
a1 = a ;
b1 = b ;
a2 = 0.42748 * R**2 * T_c**2.5 / P_c ;
b2 = 0.08664 * R * T_c / P_c ;

a_mix = y1**2 * a1 + 2 * y1 * y2 * math.sqrt(a1 * a2) + y2**2 * a2 ;
b_mix = y1 * b1 + y2 * b2 ;
P = R * T / (v - b_mix) - a_mix /(T**0.5 * v * (v + b_mix));
x = P * n * 10**-6 ;

print "3)          Pressure = %.2f MPa "%( x )

# Note : Answers are slightly different because of rounding off error.

1)          Volume = 0.0548 m**3
2)          Pressure = 1.2 MPa
3)          Pressure = 1.22 MPa