Chapter 20 : Liquid and Gases in Equilibrium with Solids¶
Example 20.1 Page no. 594¶
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%matplotlib inline
from matplotlib import pyplot as plt
# Variables
p_CO2 = [0,25,50,100,200,400,760] ; # Values of partial pressure of CO2 - [mm Hg]
y = [0,6.69*10**-2,9.24*10**-2,0.108,0.114,0.127,0.137] ; # adsorption of CO2 -[g adorbed / g seives]
# Results
plt.plot(p_CO2,y);
plt.title('Figure E20.1 The Freundlich and Langmuir iotherms coincide for the adsorption of CO2 on 5A molecular seives');
plt.show()
Example 20.2 page no. 596¶
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# Variables
G = 1000.0 ; # Volume of solution - [L]
S_ad = 1.56 ; # amount of Steptomycin adsorbed per gram resin-[g strep./g resin]
cn_S = 6. ; # Concentration of streptomycin solution-[g/L]
# Calculations
# Assume equilibrium occurs so that total(max) amount of streptomycin is adsorbed
max_S = cn_S*G ; # Maximum streptomycin adsorbed-[g]
#Use streptomycin balance to get amount of resin required
R = max_S/S_ad ; #Amount of resin required to adsorb required amount of streptomycin
# Results
print 'Amount of resin required to adsorb required amount of streptomycin is %.0f g . '%R
Example 20.3 page no. 596¶
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# Variables
G = 1000. ; # Volume of solution - [L]
x = [19.2,17.2,12.6,8.6,3.4,1.4] ; # concentration of solute- [g/L]
ac = [0,0.01,0.04,0.08,0.20,0.40] ; # Activated charcoal added-[g/1000g sol]
# Assume all concentration can be treated as g solute/1000 g sol.
# Calculations
y2 = (x[0]-x[1])/ac[1] ; # -[ g solute/g carbon]
y3 = (x[0]-x[2])/ac[2] ; # -[ g solute/g carbon]
y4 = (x[0]-x[3])/ac[3] ; # -[ g solute/g carbon]
y5 = (x[0]-x[4])/ac[4] ; # -[ g solute/g carbon]
y6 = (x[0]-x[5])/ac[5] ; # -[ g solute/g carbon]
# Use polymath to get Freundlich isotherm to bo y= 37.919*x**(0.583)
y = 37.919*x[5]**(0.583) ; #From Freundlich isotherm
A_by_G = (x[0]-x[5])/y ; #Minimum mss of activated carbon required- [g carbon/1000 g sol.]
# Results
print 'Minimum mass of activated carbon required is %.2f g carbon/1000 g sol. '%A_by_G
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