{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 6:The Question of Ideality"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.2,Page no:144"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from scipy.optimize import fsolve\n",
"\n",
"a=3.61 #atm L**2 mol**-2\n",
"b=4.29*10**-2 #L mol**-1\n",
"R=0.082 #L atm K**-1 mol**-1\n",
"T=500 #K\n",
"P=100 #atm\n",
"\n",
"\n",
"C1=b+(R*T/P) #L mol**-1 [aSsume]\n",
"C2=a/P #L^2 mol^-2 [assume]\n",
"C3=C2*b #L^3mol**-3\n",
"def f(x):\n",
" return(x**3-C1*x**2+C2*x-C3)\n",
"x=fsolve(f,0.3)\n",
"\n",
"\n",
"print \"x=\",round(x,3)\n",
"Vm=round(x,3)\n",
"print\"Therefore,the value of molar volume,Vm=\",Vm,\"L mol^-1\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"x= 0.366\n",
"Therefore,the value of molar volume,Vm= 0.366 L mol^-1\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.5,Page no:149"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"b=0.0391 \t\t\t#Van der waals constant[dm3/mol]\n",
"R=0.082 \t\t\t#Universal gas constant[dm3*atm/mol]\n",
"P2=1000 \t\t\t#pressure [atm]\n",
"P1=0 \t\t\t\t#pressure [atm]\n",
"T=1273\t\t \t\t#Temperature [K]\n",
"import math\n",
"\n",
"x=b*(P2-P1) \n",
"y=R*T \n",
"fc=math.exp(x/y) \t\t#fugacity coefficient\n",
"\n",
"f=P2*fc #fugacity[atm]\n",
"print\"The fugacity coefficient is\",round(fc,3) \n",
"print\"The fugacity is\",round(f),\"atm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The fugacity coefficient is 1.454\n",
"The fugacity is 1454.0 atm\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.10,Page no:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"m1=0.03 #mass of CO2(g)[gm]\n",
"w1=44.01 #molecular weight of CO2(g)[gm/mol]\n",
"m2=250 #mass of water[gm]\n",
"w2=18.02 #molecular weight of water[gm/mol]\n",
"k=1.25*10**6 #Henry's law constant[Torr]\n",
"T=298 #Temperature[K]\n",
"\n",
"n1=m1/w1 #no. of moles of CO2\n",
"n2=m2/w2 #no. of moles of water\n",
"x1=n1/(n1+n2) #mole fraction of CO2\n",
"Pco2=k*x1 #Partial pressure of CO2[Torr]\n",
"print\"The partial pressure of CO2 gas is\",round(Pco2,2),\"Torr\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The partial pressure of CO2 gas is 61.41 Torr\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.11,Page no:161"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"W=1000 \t\t\t#Total mass of a solution[gm]\n",
"x1=0.5 \t\t\t#mole fraction of Chloroform\n",
"x2=0.5 \t\t\t#mole fraction of Acetone\n",
"V1m=80.235 \t\t#Partial molar volume of chloroform[cm3/mol]\n",
"V2m=74.166 \t\t#Partial molar volume of Acetone[cm3/mol]\n",
"M1=119.59 \t\t#molecular weight of chloroform[gm/mol]\n",
"M2=58 \t\t\t#molecular weight of Acetone[gm/mol]\n",
"\n",
"nT=W/(x1*M1+x2*M2) \t#Total no. of moles\n",
"V=nT*(x1*V1m+x2*V2m) \t#Total volume[cm3]\n",
"print\"The volume of the solution is\",round(V,1),\"cm^3 (approx)\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The volume of the solution is 869.4 cm^3 (approx)\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.12,Page no:163"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"x1=0.5 #mole fraction of chloroform\n",
"x2=0.5 #mole fraction of p-xylene\n",
"T=298 #Temperature[K]\n",
"\n",
"Ve=x1*x2*(0.585+0.085*(x1-x2)-0.165*(x1-x2)**2) #Excess volume measured by using a dilatometer\n",
"print\"Ve/(cm3.mol**-1) = \",round(Ve,3) "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Ve/(cm3.mol**-1) = 0.146\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.14,Page no:169"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"m1=0.01 \t\t#molality[m]\n",
"v11=1.0 \n",
"v12=2.0 \n",
"Y1=0.71 \n",
"m2=0.005 \t\t#molality[m]\n",
"v21=1.0 \n",
"v22=1.0 \n",
"Y2=0.53 \n",
"\n",
"\n",
"v1=(v11)+(v12) \n",
"v2=(v21)+(v22) \n",
"a1=(m1**v1)*(v11**v11)*(v12**v12)*(Y1**v1) \n",
"a2=(m2**v2)*(v21**v21)*(v22**v22)*(Y2**v2) \n",
"x=1.0/v1 \n",
"a1m=a1**x \n",
"m1m=m1*(v11**v11*v12**v12)**x #molality[m]\n",
"y=1.0/v2 \n",
"m2m=m2*(v21*v21*v22**v22)**y #molality[m]\n",
"a2m=a2**y \n",
"print\"The activity of the electrolyte ZnCl2 is\",round(a1,8)\n",
"print\"The activity of the electrolyte CuSO4 is\",round(a2,8)\n",
"print\"The mean molality of ZnCl2 in [m]\",round(m1m,4)\n",
"print\"The mean molality of CuSO4 in [m]\",m2m \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The activity of the electrolyte ZnCl2 is 1.43e-06\n",
"The activity of the electrolyte CuSO4 is 7.02e-06\n",
"The mean molality of ZnCl2 in [m] 0.0159\n",
"The mean molality of CuSO4 in [m] 0.005\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.15,Page no:172"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"m2=3 \t\t\t#mass of the sucrose[gm]\n",
"m1=0.1 \t\t\t#mass of water [Kg]\n",
"Kf=1.86 \t\t#cryoscopic constant of water[K*Kg/mol]\n",
"dTf=0.16 \t\t#Lowering in freezing point[K]\n",
"\t\n",
"a=m1*dTf \n",
"b=Kf*m2 \n",
"M2=b/a \t\t\t#molecular weight\n",
"print\"M2=molecular weight , then M2=\",M2 "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"M2=molecular weight , then M2= 348.75\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.16,Page no:173"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"dTf=0.088 \t\t\t#Lowering in freezing point[K]\n",
"m2=0.45 \t\t\t#mass of sulphur[gm]\n",
"m1=0.09955 \t\t\t#mass of benzene[gm]\n",
"Kf=5.07 \t\t\t#cryoscopic constant for benzene[K*Kg/mol]\n",
"\n",
"a=m1*dTf \n",
"b=Kf*m2 \n",
"M2=b/a \t\t\t\t#molecular weight of sulphur\n",
"print\"The molecular weight of sulphur is\",round(M2,1) \n",
"x=M2/32 \t\t\t#no. of sulphur atoms\n",
"print\"\\n The molecular formula of sulphur is S\",round(x) "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The molecular weight of sulphur is 260.4\n",
"\n",
" The molecular formula of sulphur is S 8.0\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.17,Page no:174"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"m2=1.35 \t\t\t#mass of a macromolecule[gm]\n",
"V=100\t \t\t\t#volume of solution[cm^3]\n",
"R=82 \t\t\t\t#Universal gas constant[atm.cm^3.K^-1]\n",
"T=300 \t\t\t\t#Temperature[K]\n",
"II=9.9 \t\t\t\t#osmotic pressure of the solution[cm]\n",
"d=1 \t\t\t\t#density\n",
"p=1013250 \t\t\t#Atmospheric pressure\n",
"g=980.67 \t\t\t#gravitational field\n",
"\n",
"\n",
"a=m2*R*T*p \n",
"b=V*9.9*d*g \n",
"M2=a/b #molar mass of macromolecule\n",
"print\" M2 = molar mass of macromolecule , therefore M2 = \",round(M2),\"g.mol^-1\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" M2 = molar mass of macromolecule , therefore M2 = 34660.0 g.mol^-1\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.18,Page no:175"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"R=82 \t\t\t#Universal gas constant[atm.ml.K^-1.mol^-1]\n",
"T=298 \t\t\t#Temperature[K]\n",
"V=250 \t\t\t#volume of water[ml]\n",
"m2=2.6 \t\t\t#mass of the protein\n",
"M2=85000 \t\t#molar mass of protein[g.mol^-1]\n",
"\n",
"\t\n",
"n2=m2/M2 \t\t\t#no. of moles of protein\n",
"II=(n2*R*T)/V \t\t\t#Osmotic pressure of a solution[atm]\n",
"print\"The osmotic pressure is\",round(II,5),\"atm \"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The osmotic pressure is 0.00299 atm \n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.19,Page no:175"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"R=8.314 \t\t\t#Universal gas constant[J.K**-1.mol**-1]\n",
"Tb=373.15 \t\t\t#Boiling point temperature[K]\n",
"M1=0.018 \t\t\t# mass of water[kg]\n",
"Hvap=40.7 \t\t\t#Enthalpy of vaporization[KJ.mol**-1]\n",
"\n",
"a=R*0.001*Tb**2*M1 \n",
"b=Hvap \n",
"Kb=a/b \t\t\t\t#Ebullioscopic constant of water[K.Kg.mol**-1]\n",
"print\"The Ebullioscopic constant of water is\",round(Kb,2),\"K.Kg.mol-1\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Ebullioscopic constant of water is 0.51 K.Kg.mol-1\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.20,Page no:176"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"print\"CaF2(s)<=>CaF2(aq)<=>Ca+2(aq) + 2F-(aq)\"\n",
"\n",
"Ksp=4.0*(10**-11) \t#Solubility product of sparingly soluble salt CaF2\n",
"\n",
"x=Ksp/4.0 \n",
"Cs=x**(1.0/3.0) \t\t#Solubility \n",
"y=Cs**2 \n",
"Y=(x/y)**(1.0/3.0) \t\t#activity coefficient\n",
"print\"The activity coefficient is\",Y \n",
"print\"NOTE:please note that the value of Cs is wrongly calculated as 4.64*10^-11 in book\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"CaF2(s)<=>CaF2(aq)<=>Ca+2(aq) + 2F-(aq)\n",
"The activity coefficient is 0.0599484250319\n",
"NOTE:please note that the value of Cs is wrongly calculated as 4.64*10^-11 in book\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.21,Page no:177"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"R=8.314 \t\t\t#Universal gas constant[J/K/mol]\n",
"T=298 \t\t\t\t#Temperature[K]\n",
"F=96500 \t\t\t#Faraday's constant\n",
"Eo=0.98 \t\t\t#Standard e.m.f of the cell[Volts]\n",
"E=1.16 \t\t\t\t#e.m.f of the cell[Volts]\n",
"m=0.01 \n",
"import math\n",
"\n",
"a=R*T \n",
"b=2*F \n",
"x=a/b \n",
"Y=math.exp((Eo-E-(x*math.log(4*m*m*m)))/(3*x)) #mean activity coefficient\n",
"print\"The mean activity coefficient is\",round(Y,2) "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The mean activity coefficient is 0.59\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.22,Page no:184"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"M1=0.01 \t\t\t#no. of moles of KCl\n",
"M2=0.005 \t\t\t#no. of moles of MgCl2\n",
"M3=0.002 \t\t\t#no. of moles of MgSO4\n",
"M=0.1 \t\t\t\t#mass of water[Kg]\n",
"z11=1 \n",
"z12=1 \n",
"z21=2 \n",
"z22=1 \n",
"z31=2 \n",
"z32=2 \n",
"\t\n",
"m1=M1/M \t\t\t#molality of KCL[m]\n",
"m2=M2/M \t\t\t#molality of MgCl2[m]\n",
"m3=M3/M \t\t\t#molality of MgSO4[m]\n",
"\n",
"I=0.5*((m1*z11**2+m1*z12**2+m2*z21**2+2*m2*z22**2+m3*z31**2+m3*z32**2)) #[mol/Kg]\n",
"print\"The Ionic strength of a solution is\",I,\"mol/Kg\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Ionic strength of a solution is 0.33 mol/Kg\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.23,Page no:185"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"T=298 \t\t\t\t#Temperature[K]\n",
"P=1 \t\t\t\t#pressure [atm]\n",
"m=0.02\t \t\t\t#Ionic strength of HCl solution in CH3OH[mol/Kg]\n",
"E=32.6 \t\t\t\t#Di-electric constant\n",
"d=0.787 \t\t\t#Density[gm/cm3]\n",
"\t\n",
"I=0.5*(0.02*1*1+0.02*1*1) \t#Ionic strength of HCl solution[mol/Kg]\n",
"a=I*d \n",
"b=(E**3)*(298**3) \n",
"x=(a/b)**0.5 \n",
"Y=10**(-1.825*1000000*1*1*x) \t#mean activity coefficient\n",
"print\"The mean activity coefficient is\",round(Y,2) "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The mean activity coefficient is 0.58\n"
]
}
],
"prompt_number": 14
}
],
"metadata": {}
}
]
}