Chapter 11 : LIquid Liquid Liquid Solid And Gas Solid Equilibrium¶
Example 11.1 Page: 272¶
In [16]:
import math
V_water = 1. #[L] volume of the water
Temp = 25. #[C]
d_benzene = 0.88 #[g/ml] density of the benzene
M_benzene = 78. #[g/mol] molecular weight of the benzene
M_water = 18. #[g/mol]
n_1drop = 1./20*(d_benzene/M_benzene) #[mol/drop]
n_water = 1000./M_water #[mol]
s_benzene = 0.000405
n_benzene_saturate = s_benzene*n_water #[mol]
N_benzene = n_benzene_saturate/n_1drop #[drops]
print "The total number of the drops of the benznene required to saturate the water is %0.0f drops"%(N_benzene)
Example 11.2 Page: 273¶
In [17]:
import math
m_benzene = 1000. #[lbm]
M_benzene = 78. #[lbm/lbmol]
n_benzene = m_benzene/M_benzene #[lbmol]
x_benzene = 0.000405
n_water = n_benzene/x_benzene #[lbmol]
m_water = n_water*18 #[lbm]
print "The amount of the ground water that will make a saturated solution will be %e lbm"%(m_water)
Example 11.3 Page: 277¶
In [23]:
import math
Temp = 25. #[C]
n_water = 3.75 #[mol]
n_ethanol = 2.5 #[mol]
n_benzene = 3.75 #[mol]
x_water = 0.375
x_ethanol = 0.250
x_benzene = 0.375
x_water_w = 64.9 #[%]
x_ethanol_w = 31.75 #[%]
x_benzene_w = 3.37 #[%]
x_water_b = 6.43 #[%]
x_ethanol_b = 18.94 #[%]
x_benzene_b = 74.62 #[%]
print "The composition of the two equilibrium phases i.e. water-reach phase and benzene reach phase is as "
print "\t\t\t\tWater-reach phase\t\tbenzene-reach phase"
print "Mol water\t\t\t%f\t\t\t\t%f"%(x_water_w,x_water_b)
print "Mol ethanol\t\t\t%f\t\t\t\t%f"%(x_ethanol_w,x_ethanol_b)
print "Mol benzene\t\t\t%f\t\t\t\t%f"%(x_benzene_w,x_benzene_b)
Example 11.4 Page: 282¶
In [20]:
import math
Temp = 25. #[C]
x_1in2 = 405. #[ppm]
x_2in1 = 3000. #[ppm]
x_water_w = (10**(6)-405.)/(10**(6))
x_benzene_w = 1-x_water_w
x_benzene_b =(10.**(6.)-3000.)/(10**(6))
x_water_b = 1-x_benzene_b
y_benzene = 1/(x_benzene_w)
y_water = 1/(x_water_b)
print " Activity coefficient of benzene in water is %f"%(y_benzene)
print " Activity coefficient of water in benzene is %f"%(y_water)
Example 11.5 Page: 283¶
In [10]:
import math
from numpy import *
from matplotlib.pyplot import *
%matplotlib inline
R = 8.314 #[J/(mol*K)] Universal gas consmath.tant
T = 298.15 #[K] Temperature
g_a_0 = 2. #[kj/mol] Gibb's free energy of the pure species 'a'
g_b_0 = 1. #[kj/mol] Gibb's free energy of the pure species 'b'
for a in range(4):
def f(x):
return x*g_a_0 + (1-x)*g_b_0 + (R*T)/1000*(x*log(x) + (1-x)*log(1-x) + x*a*(1-x)**(2) + (1-x)*a*(x)**(2))
x=linspace(0.000001,0.990001,100)
a = f(x)
plot(x,a)
xlabel(" mole fraction of species a,x_a");
ylabel(" gibb''s free energy per mole of mixture,g_mixture kJ/mol");
annotate("A = 0",xy=(.7,0))
annotate("A = 1",xy=(.6,0.5))
annotate("A = 2",xy=(.6,0.95))
annotate("A = 3",xy=(.6,1.50))
suptitle("Calculated values of the molar Gibbs energy of a binary mixture at 25 C, assuming \
the symmetrical activity coefficient equation, with various values of A.")
show()
print " The plot is shown in the graphic window."
Example 11.6 Page: 283¶
In [17]:
import math
from matplotlib.pyplot import *
from numpy import *
T = 92 + 273.15 #[K] Temperature of the system
R = 8.314 #[m**(3)*Pa/(mol*K)] universal gas consmath.tant
A = 1.2739
B = 3.9771
def f(x_a):
return (R*T/1000)*( x_a*log(x_a) + (1-x_a)*log(1-x_a) + \
x_a*(B**(2)*A*(1-x_a)**(2)/(A*x_a + B*(1-x_a))**(2)) + (1-x_a)* \
(A**(2)*B*x_a**(2)/(A*x_a + B*(1-x_a))**(2)) )
x_a = linspace(0.000001,0.99999,100)
a = f(x_a)
plot(x_a,a)
x = linspace(0.0,0.99999,100)
a = 1.2090312*x-1.251495764
plot(x,1.2090312*x-1.251495764)
xlabel(" Mol fraction of species a, x_a");
ylabel(" g_mix - sum(x_i*g_i0)");
show()
print "Thus based on the several assumptions that the Van Laar equation is an accurate representation of LLE"
print "we would conclude that at 92 deg C water-n-butanol does form two liquid phases."
Example 11.7 Page: 286¶
In [23]:
import math
R = 8.314 #[J/(ml*K)]
h_mix = -2967.7*R/1000 #[kJ/mol] Heat of mixing of water-in-benzene
h_mix_1 = 522.9*R/1000 #[kJ/mol] Heat of mixing of benzene-in-water
print " Heat of mixing of water-in-benzene is given as %f kJ/mol"%(h_mix)
print " Heat of mixing of benzene-in-water is given as %f kJ/mol"%(h_mix_1)
Example 11.8 Page: 287¶
In [24]:
import math
T_i = 50. #[F] Initial temperature of the system
T_f = 20. #[F] Final temperature of the system
M_gas = 115. #[g/mol] Molecular weight of gasoline at room temperature
M_water = 18. #[g/mol] Molecular weight of water at the room temperaature
d = 720. #[g/L] density of gasoline at the room temperature
s_50 = 0.00026 #[mol fraction]
s_20 = 0.0001 #[mol fraction]
s_rej = s_50 - s_20 # mol of water per mole of gasoline
w = (s_rej*d*M_water)/M_gas #[g water/L gasoline]
print " The amount of water that will come out of the solution in the gasoline will be %f g water/L gasoline"%(w)
print " At 20 deg F we would expect this water to become solid ice, forming a piece large enough to plug the fuel line of a parked auto."
Example 11.9 Page: 290¶
In [19]:
from numpy import *
from matplotlib.pyplot import *
import math
Temp = 25. #[C]
x_water = 5. #[mo]
x_benzene = 0.1 #[mol]
x_ethanol_water_rich = 3.817 #[%]
x_ethanol_benzene_rich = 1.010 #[%]
K = x_ethanol_water_rich/x_ethanol_benzene_rich
X = array([3.817,7.968,12.977,18.134,23.540,24.069,27.892,31.725,35.510,39.382,41.062,41.771])
Y = array([1.010,3.323,5.860,9.121,12.939,13.340,16.090,18.943,22.444,26.216,29.341,33.093])
Z = X/Y
plot(Y,Z)
xlabel("Mol% ethanol in benzene-rich phase ");
ylabel("Distribution coefficient of ethanol, K_ethanol");
show()
K_1 = 4
m_water_rich = 100*K_1/(K_1+1)
m_benzene_rich = 100/(K_1+1)
print " Ethanol''s 0.1 mol distributed in the water rich phase will be %f mol%% of the total mol"%(m_water_rich)
print " Ethanol''s 0.1 mol distributed in the benzene rich phase will be %f mol%% of the total mol"%(m_benzene_rich)
Example 11.10 Page: 293¶
In [26]:
import math
Temp = 20. #[C]
s = 36.0 #[g per 100 g of water]
M_NaCl = 58.5 #[g/mol] molecular weight of NaCl
M_water = 18. #[g/mol] molecular weight of water
w = s/(s+100)
w_percent = w*100 #[wt %]
x = (s/M_NaCl)/((s/M_NaCl)+(100/M_water))
x_percent = x*100 #[mol %]
print " Weight fraction of the NaCl in the saturated solution is %0.1f wt %%"%(w_percent)
print " Mol fraction of the NaCl in the saturated solution is %0.0f mol %%"%(x_percent)
Example 11.11 Page: 293¶
In [2]:
import math
T_inlet = 68. #[F]
T_outlet = 110. #[F]
s_inlet_carbonate = 60. #[ppm]
s_inlet_sulphate = 2020. #[ppm]
s_outlet_carbonate = 30. #[ppm]
s_outlet_sulphate = 2000. #[ppm]
# This is close enough to the solubility of the gypsum at 68F
# so we conclude that we would not expect either form of CaSO4 to prdcipitate
w = s_inlet_carbonate - s_outlet_carbonate #[ppm]
print " Total amount of the solid left behind in the heater will be %0.1f ppm"%(w)
# Now if a typical houshold water heater heats 100 gallons/per day , we would expect to deposite
w_per_day = w*10**(-6)*100*8.33 #[lb/day]
print " Total amount of the solid left behind in the heater per day will be %.3f lb/day"%(w_per_day)
Example 11.12 Page: 298¶
In [28]:
import math
x_2 = 0.1
x_ideal_benzene = 0.114
x_ideal_CCl4 = 0.152
y_i_1 = x_ideal_benzene/x_2# corresponding to practically ideal solution
y_i_2 = x_ideal_CCl4/x_2# corresponding to mild type II behavior
print " Activity coefficient in benzene corresponding to practically ideal solution is %0.2f"%(y_i_1)
print " Activity coefficient in CCl4 corresponding to mild type II behavior is %0.2f"%(y_i_2)
Example 11.13 Page: 299¶
In [29]:
import math
T = 273.15+20 #[K]
T_m = 273.15+800 #[K]
delta_h_fusion = 30219. #[J/g]
R = 8.314 #[J/(mol*K)]
a = delta_h_fusion/(R*T)*(T_m/T-1)
x_NaCl_into_y_i_1 = 1/math.exp(a)
x_NaCl = 1/math.exp(a)*1
print " The solubility of the NaCl in water at 20 deg C is %e "%( x_NaCl)
print " But the experimental value is 0.1 so Similar to the results in book our results are very far wrong"
Example 11.14 Page: 301¶
In [30]:
import math
P = 1*14.7 #[psia]
T = 30. #[F]
p_ice = 0.0808 #[psia]
x_water_in_ice = 1.00
y_water_vapour = x_water_in_ice*p_ice/P
print " Equilibrium concentration of water vapour in the air is %0.4f"%(y_water_vapour)
Example 11.15 Page: 302¶
In [31]:
import math
T = 273.15+35 #[K]
p_d = 100. #[atm]
R = 82.06 #[(cm**(3)*atm)/(mol*K)]
p_naphthalene = 0.00065 #[atm]
x_naphthalene = 1.00
P = p_d
y_naphthalene = x_naphthalene*p_naphthalene/P
v = 132. #[cm**(3)/mol]
p_c = 1. #[atm]
f_d_by_f_c = math.exp(v/(R*T)*(p_d-p_c))
y_naphthalene = f_d_by_f_c*y_naphthalene
print "Estimated solubility of naphthalene in CO2 is %e"%(y_naphthalene)
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