{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 2 : Physico Chemical Calculations" ] }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.1 page number 71" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "p1=15. #in bar\n", "p2=1.013 #in bar\n", "t1=283. #in K\n", "t2=273. #in K\n", "v1=10. #in l\n", "\n", "v2=p1*v1*t2/(t1*p2);\n", "\n", "print \"volume of oxygen = %f liters\"%(v2)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "volume of oxygen = 142.842692 liters\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.2 page number 71\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "nCO2 = 2./44; #moles of CO2\n", "nO2 = 4./32; #moles of O2\n", "nCH4 = 1.5/16; #moles of CH4\n", "\n", "total_moles = nCO2+nO2+nCH4;\n", "yCO2 = nCO2/total_moles;\n", "yO2 = nO2/total_moles;\n", "yCH4 = nCH4/total_moles;\n", "\n", "print \" Composition of mixture = CH4 = %f O2 = %f CO2 = %f \"%(yCH4,yO2,yCO2)\n", "\n", "pCO2=nCO2*8.314*273/(6*10**-3);\n", "pO2=nO2*8.314*273/(6*10**-3);\n", "pCH4=nCH4*8.314*273/(6*10**-3);\n", "\n", "print \"pressure of CH4 = %f kPa pressure of O2 = %f kPa pressure of CO2 =%f kPa\"%(pCH4*10**-3,pO2*10**-3,pCO2*10**-3)\n", "\n", "total_pressure=pCO2+pCH4+pO2;\n", "print \"total pressure = %f Kpa\"%(total_pressure*10**-3)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Composition of mixture = CH4 = 0.354839 O2 = 0.473118 CO2 = 0.172043 \n", "pressure of CH4 = 35.464406 kPa pressure of O2 = 47.285875 kPa pressure of CO2 =17.194864 kPa\n", "total pressure = 99.945145 Kpa\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.3 page number 72\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "P=104.3 #total pressure in KPa\n", "pH2O=2.3 #in KPa\n", "pH2=P-pH2O; #in KPa\n", "\n", "VH2=209*pH2*273/(293*101.3)\n", "\n", "print \"volume of hydrogen obtained = %f ml\"%(VH2)\n", "\n", "\n", "m=350/196.08*11.2 #mass of metal in grams\n", "print \"mass of metal equivalent to 11.2 litre/mol of hydrogen = %f gm\"%(m)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "volume of hydrogen obtained = 196.079432 ml\n", "mass of metal equivalent to 11.2 litre/mol of hydrogen = 19.991840 gm\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.4 page number 72\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "w=2 #in gm\n", "m=0.287 #in gm\n", "\n", "mNaCl=58.5/143.4*m;\n", "\n", "print \"mass of NaCl = %f gm\"%(mNaCl )\n", "\n", "percentage_NaCl=mNaCl/w*100;\n", "print \"amount of NaCl = %f\"%(percentage_NaCl)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "mass of NaCl = 0.117082 gm\n", "amount of NaCl = 5.854079\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.5 page number 72\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "w=4.73 #in gm5\n", "VCO2=5.30 #in liters\n", "\n", "weight_CO2=44/22.4*VCO2;\n", "carbon_content=12./44*weight_CO2;\n", "percentage_content=(carbon_content/w)*100;\n", "\n", "print \"percentage amount of carbon in sample = %f\"%(percentage_content)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "percentage amount of carbon in sample = 60.027182\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.6 page number 73\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "volume_H2=0.5 #in m3\n", "volume_CH4=0.35 #in m3\n", "volume_CO=0.08 #in m3\n", "volume_C2H4=0.02 #in m3\n", "volume_oxygen=0.21 #in m3 in air\n", "\n", "H2=0.5*volume_H2;\n", "CH4=2*volume_CH4;\n", "CO=0.5*volume_CO;\n", "C2H4=3*volume_C2H4;\n", "\n", "total_O2=H2+CH4+CO+C2H4;\n", "oxygen_required=total_O2/volume_oxygen;\n", "\n", "print \"amount of oxygen required = %f cubic meter\"%(oxygen_required)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "amount of oxygen required = 5.000000 cubic meter\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.7 page number 73\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "\n", "density_H2SO4 = 1.10 #in g/ml\n", "mass_1 = 100*density_H2SO4; #mass of 100ml of 15% solution\n", "mass_H2SO4 = 0.15*mass_1;\n", "density_std = 1.84 #density of 96% sulphuric acid\n", "mass_std = 0.96*density_std; #mass of H2SO4 in 1ml 96% H2SO4\n", "\n", "volume_std = mass_H2SO4/mass_std; #volume of 96%H2SO4\n", "mass_water = mass_1 - mass_H2SO4;\n", "\n", "print \"volume of 0.96 H2SO4 required = %f ml\"%(volume_std)\n", "print \"mass of water required = %f g\"%(mass_water)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "volume of 0.96 H2SO4 required = 9.341033 ml\n", "mass of water required = 93.500000 g\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.8 page number 73\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "w_H2SO4=0.15 #in gm/1gm solution\n", "density=1.10 #in gm/ml\n", "m=density*1000; #mass per liter\n", "weight=m*w_H2SO4; #H2SO4 per liter solution\n", "molar_mass=98;\n", "\n", "Molarity=weight/molar_mass;\n", "print \"Molarity = %f mol/l\"%(Molarity)\n", "\n", "equivalent_mass=49;\n", "normality=weight/equivalent_mass;\n", "print \"Normality = %f N\"%(normality)\n", "\n", "molality=176.5/molar_mass;\n", "print \"Molality = %f\"%(molality)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Molarity = 1.683673 mol/l\n", "Normality = 3.367347 N\n", "Molality = 1.801020\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.9 page number 74\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "molar_mass_BaCl2=208.3; #in gm\n", "equivalent_H2SO4=0.144;\n", "\n", "normality=equivalent_H2SO4*1000/28.8;\n", "\n", "print \"Normality = %f N\"%(normality)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Normality = 5.000000 N\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.10 page number 74\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "solubility_70=30.2 #in gm/100gm\n", "w_solute=solubility_70*350/130.2; #in gm\n", "\n", "w_water=350-w_solute;\n", "solubility_30=10.1 #in gm/100gm\n", "precipitate=(solubility_70-solubility_30)*w_water/100\n", "\n", "print \"amount precipitated = %f gm\"%(precipitate)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "amount precipitated = 54.032258 gm\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.11 page number 74\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "absorbtion_coefficient=1.71 #in liters\n", "molar_mass=44;\n", "\n", "solubility=absorbtion_coefficient*molar_mass/22.4; #in gm\n", "pressure=8/solubility*101.3;\n", "\n", "print \"pressure required = %f kPa\"%(pressure)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "pressure required = 241.267411 kPa\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.12 page number 74\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "\n", "w_water=540. #in gm\n", "w_glucose=36. #in gm\n", "m_water=18.; #molar mass of water\n", "m_glucose=180.; #molar mass of glucose\n", "\n", "x=(w_water/m_water)/(w_water/m_water+w_glucose/m_glucose);\n", "p=8.2*x;\n", "depression=8.2-p;\n", "\n", "print \"depression in vapor pressure = %f Pa\"%(depression*1000)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "depression in vapor pressure = 54.304636 Pa\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.13 page number 75\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "w_glucose=9. #in gm\n", "w_water=100. #in gm\n", "E=0.52;\n", "m=90/180.; #moles/1000gm water\n", "\n", "delta_t=E*m;\n", "boiling_point=100+delta_t;\n", "\n", "print \"boiling_point of water = %f degreeC\"%(boiling_point)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "boiling_point of water = 100.260000 degreeC\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.14 page number 75\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "K=1.86;\n", "c=15 #concentration of alcohol\n", "delta_t=10.26;\n", "\n", "m=delta_t/K; #molality\n", "M=c/(m*85); #molar mass\n", "print \"molar mass = %f gm\"%(M*1000)\n", "\n", "density=0.97 #g/ml\n", "cm=c*density/(M*100);\n", "print \"molar concentration of alcohol = %f moles/l\"%(cm)\n", "\n", "p=cm*8.314*293 #osmotic pressure\n", "print \"osmotic pressure = %f Mpa\"%(p/1000)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "molar mass = 31.991744 gm\n", "molar concentration of alcohol = 4.548048 moles/l\n", "osmotic pressure = 11.079055 Mpa\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.15 page number 75\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "u_in = 0.575 #from the graph\n", "u_s = 0.295 #in mPa-s\n", "\n", "M_v = (u_in/(5.80*10**-5))**(1/0.72);\n", "u_red = 0.628; #in dl/g\n", "\n", "c = 0.40 #in g/dl\n", "k = (u_red-u_in)/((u_in**2)*c);\n", "\n", "print \"k = %f Mv = %fu_in = %f dl/gm\"%(k,M_v,u_in)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "k = 0.400756 Mv = 355085.654054u_in = 0.575000 dl/gm\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.16 page number 76\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "C=54.5 #% of carbon\n", "H2=9.1 #% of hydrogen\n", "O2=36.4 #% of oxygen\n", "x=C/12.; #number of carbon molecules\n", "y=O2/16.; #number of oxygen molecules\n", "z=H2/2. #number of hydrogen molecules\n", "molar_mass=88.;\n", "density=44.;\n", "\n", "ratio=molar_mass/density;\n", "x=ratio*2;\n", "y=ratio*1;\n", "z=ratio*4;\n", "\n", "print \"x = %f y = %f z = %f\"%(x,y,z)\n", "print \"formula of butyric acid is = C4H8O2\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "x = 4.000000 y = 2.000000 z = 8.000000\n", "formula of butyric acid is = C4H8O2\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.17 page number 77\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "C=93.75 #% of carbon\n", "H2=6.25 #% of hydrogen\n", "x=C/12 #number of carbon atoms\n", "y=H2/2 #number of hydrogen atoms\n", "molar_mass=64\n", "density=4.41*29;\n", "\n", "ratio=density/molar_mass;\n", "x=round(ratio*5);\n", "y=round(ratio*4);\n", "\n", "print \"x = %f y = %f\"%(x,y)\n", "print \"formula of butyric acid is = C10H8\"\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "x = 10.000000 y = 8.000000\n", "formula of butyric acid is = C10H8\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.18 page number 77\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "C=50.69 #% of carbon\n", "H2=4.23 #% of hydrogen\n", "O2=45.08 #% of oxygen\n", "a=C/12; #number of carbon molecules\n", "c=O2/16; #number of oxygen molecules\n", "b=H2/2; #number of hydrogen molecules\n", "molar_mass=71;\n", "\n", "def f(m):\n", " return (2.09*1000)/(60*m);\n", "\n", "\n", "M=f((1.25/5.1));\n", "\n", "print \"actual molecular mass = %f\"%(M)\n", "\n", "ratio=M/molar_mass;\n", "a=round(ratio*3);\n", "b=round(ratio*3);\n", "c=round(ratio*2);\n", "\n", "print \"a = %d, b = %d, c = %d\"%(a,b,c)\n", "print \"M = %.1f g/mol\"%M\n", "print \"formula of butyric acid is = C6H6O4\"\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "actual molecular mass = 142.120000\n", "a = 6, b = 6, c = 4\n", "M = 142.1 g/mol\n", "formula of butyric acid is = C6H6O4\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.19 page number 78\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "C=64.6 #% of carbon\n", "H2=5.2 #% of hydrogen\n", "O2=12.6 #% of oxygen\n", "N2=8.8 #% of nitrogen\n", "Fe=8.8 #% of iron\n", "\n", "a=C/12; #number of carbon molecules\n", "c=8.8/14; #number of nitrogen molecules\n", "b=H2; #number of hydrogen molecules\n", "d=O2/16; #number of oxygen molecules\n", "e=Fe/56 #number of iron atoms\n", "\n", "cm=243.4/(8.31*293) #concentration\n", "\n", "molar_mass=63.3/cm;\n", " \n", "print \"a = %d, b = %d, c = %d, d = %d, e = %d\"%(a*6.5,b*6.5,c*6.5,d*6.5,e*6.5)\n", "print \"formula of butyric acid is = C34H33N4O5Fe\"\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "a = 34, b = 33, c = 4, d = 5, e = 1\n", "formula of butyric acid is = C34H33N4O5Fe\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.20 page number 78\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "E1=-0.25;\n", "E2=0.80;\n", "E3=0.34;\n", "\n", "a=[E1,E2,E3];\n", "sorted(a)\n", "\n", "print \"sorted potential in volts =\"\n", "print (a)\n", "print (\"E2>E3>E1\")\n", "print (\"silver>copper>nickel\")\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "sorted potential in volts =\n", "[-0.25, 0.8, 0.34]\n", "E2>E3>E1\n", "silver>copper>nickel\n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.21 page number 79\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "E0_Zn=-0.76;\n", "E0_Pb=-0.13;\n", "c_Zn=0.1;\n", "c_Pb=0.02;\n", "\n", "E_Zn=E0_Zn+(0.059/2)*math.log10(c_Zn);\n", "E_Pb=E0_Pb+(0.059/2)*math.log10(c_Pb);\n", "E=E_Pb-E_Zn;\n", "\n", "print \"emf of cell = %f V\"%(E)\n", "print \"Since potential of lead is greater than that of zinc thus reduction will occur at\\\n", " lead electrode and oxidation will occur at zinc electrode\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "emf of cell = 0.609380 V\n", "Since potential of lead is greater than that of zinc thus reduction will occur at lead electrode and oxidation will occur at zinc electrode\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.22 page number 79\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "E0_Ag=0.80;\n", "E0_AgNO3=0.80;\n", "c_Ag=0.001;\n", "c_AgNO3=0.1;\n", "\n", "E_Ag=E0_Ag+(0.059)*math.log10(c_Ag);\n", "E_AgNO3=E0_AgNO3+(0.059)*math.log10(c_AgNO3);\n", "E=E_AgNO3-E_Ag;\n", "\n", "print \"emf of cell = %f V\" %(E)\n", "print \"since E is positive, the left hand electrode will be anode and\\\n", " the electron will travel in the external circuit from the left hand to the right hand electrode\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "emf of cell = 0.118000 V\n", "since E is positive, the left hand electrode will be anode and the electron will travel in the external circuit from the left hand to the right hand electrode\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.23 page number 79\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "pH=12; #pH of solution\n", "E_H2=0;\n", "\n", "E2=-0.059*pH;\n", "E=E_H2-E2;\n", "\n", "print \"EMF of cell = %f V\"%(E)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EMF of cell = 0.708000 V\n" ] } ], "prompt_number": 24 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.24 page number 80\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "I=3 #in Ampere\n", "t=900 #in s\n", "m_eq=107.9 #in gm/mol\n", "F=96500;\n", "\n", "m=(I*t*m_eq)/F;\n", "\n", "print \"mass = %f gm\"%(m)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "mass = 3.018964 gm\n" ] } ], "prompt_number": 25 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.25 page number 80" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "volume=10*10*0.005; #in cm3\n", "mass=volume*8.9;\n", "F=96500;\n", "atomic_mass=58.7 #in amu\n", "current=2.5 #in Ampere\n", "\n", "charge=(8.9*F*2)/atomic_mass;\n", "yield_=0.95;\n", "actual_charge=charge/(yield_*3600);\n", "t=actual_charge/current;\n", "\n", "print \"time required = %f hours\"%(t)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "time required = 3.422497 hours\n" ] } ], "prompt_number": 26 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.26 page number 80\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "m_MgSO4=90. #in ppm\n", "MgSO4_parts=120.;\n", "CaCO3_parts=100.;\n", "\n", "hardness=(CaCO3_parts/MgSO4_parts)*m_MgSO4;\n", "\n", "print \"hardness of water = %f mg/l\"%(hardness)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "hardness of water = 75.000000 mg/l\n" ] } ], "prompt_number": 27 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.27 page number 81\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "calculate\n", "i) the temporary and total hardness of the sample\n", "ii) the amounts of lime and soda needed for softening of 1 l of the sample\n", "'''\n", "\n", "import math \n", "\n", "m1 = 162. #mass of calcium bi carbonate in mg\n", "m2 = 73. #mass of magnesium bi carbonate in mg\n", "m3 = 136. # mass of calsium sulfate in mg\n", "m4 = 95. # mass of magnesium cloride\n", "m5 = 500. #mass of sodium cloride in mg\n", "m6 = 50. # mass of potassium cloride in mg\n", "\n", "content_1 = m1*100/m1; #content of calcium bi carbonate in mg\n", "content_2 = m2*100/(2*m2); #content of magnesium bi carbonate in mg\n", "content_3 = m3*100/m3; # content of calsium sufate in mg\n", "content_4 = m4*100/m4; # content of magnesium cloride\n", "\n", "\n", "temp_hardness = content_1 + content_2; #depends on bicarbonate only\n", "total_hardness = content_1+content_2+content_3+content_4;\n", "print \"total hardness = %.0f mg/l temporary hardness = %.0f mg/l\"%(temp_hardness,total_hardness)\n", "\n", "wt_lime = (74./100)*(content_1+2*content_2+content_4);\n", "actual_lime = wt_lime/0.85;\n", "print \"amount of lime required = %.1f mg/l\"%(actual_lime)\n", "\n", "soda_required = (106./100)*(content_1+content_4);\n", "actual_soda = soda_required/0.98;\n", "print \"amount of soda required = %.1f mg/l\"%(actual_soda)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "total hardness = 150 mg/l temporary hardness = 350 mg/l\n", "amount of lime required = 261.2 mg/l\n", "amount of soda required = 216.3 mg/l\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.28 page number 82\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "volume_NaCl=50. #in l\n", "c_NaCl=5000. #in mg/l\n", "\n", "m=volume_NaCl*c_NaCl;\n", "equivalent_NaCl=50/58.5;\n", "\n", "hardness=equivalent_NaCl*m;\n", "\n", "print \"hardness of water = %f mg/l\"%(hardness/1000.)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "hardness of water = 213.675214 mg/l\n" ] } ], "prompt_number": 29 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.29 page number 82\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "m_benzene = 55. #in kg\n", "m_toluene = 28. #in kg\n", "m_xylene = 17. # in kg\n", "\n", "mole_benzene = m_benzene/78.;\n", "mole_toluene = m_toluene/92.;\n", "mole_xylene = m_xylene/106.;\n", "\n", "mole_total = mole_benzene+mole_toluene+mole_xylene;\n", "x_benzene = mole_benzene/mole_total;\n", "x_toluene = mole_toluene/mole_total;\n", "x_xylene = mole_xylene/mole_total;\n", "\n", "P = x_benzene*178.6+x_toluene*74.6+x_xylene*28;\n", "print \"total pressure = %f kPa\"%(P)\n", "\n", "benzene = (x_benzene*178.6*100)/P;\n", "toluene = (x_toluene*74.6*100)/P;\n", "xylene = (x_xylene*28*100)/P;\n", "\n", "print \"xylene = %f toluene = %f benzene = %f\"%(xylene,toluene,benzene)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "total pressure = 130.897438 kPa\n", "xylene = 2.932503 toluene = 14.826766 benzene = 82.240730\n" ] } ], "prompt_number": 30 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.30 page number 83\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "vapor_pressure=8. #in kPa\n", "pressure=100. #in kPa\n", "\n", "volume=1 #in m3\n", "volume_ethanol=volume*(vapor_pressure/pressure);\n", "volume_air=1-volume_ethanol;\n", "print \"volumetric composition:- air composition = %f ethanol compostion = %f\"%(volume_air*100,volume_ethanol*100)\n", "\n", "molar_mass_ethanol=46;\n", "molar_mass_air=28.9;\n", "mass_ethanol=0.08*molar_mass_ethanol; #in kg\n", "mass_air=0.92*molar_mass_air; #in kg\n", "fraction_ethanol=(mass_ethanol*100)/(mass_air+mass_ethanol);\n", "fraction_air=(mass_air*100)/(mass_air+mass_ethanol);\n", "print \"composition by weight:-Air = %f Ethanol vapor = %f\"%(fraction_air,fraction_ethanol)\n", "\n", "mixture_volume=22.3*(101.3/100)*(299./273); #in m3\n", "weight_ethanol=mass_ethanol/mixture_volume;\n", "print \"weight of ethanol/cubic meter = %f Kg\"%(weight_ethanol)\n", "\n", "w_ethanol=mass_ethanol/mass_air;\n", "print \"weight of ethanol/kg vapor free air = %f Kg\"%(w_ethanol)\n", "\n", "moles_ethanol=0.08/0.92;\n", "print \"kmol of ethanol per kmol of vapor free air = %f\"%(moles_ethanol)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "volumetric composition:- air composition = 92.000000 ethanol compostion = 8.000000\n", "composition by weight:-Air = 87.841945 Ethanol vapor = 12.158055\n", "weight of ethanol/cubic meter = 0.148739 Kg\n", "weight of ethanol/kg vapor free air = 0.138408 Kg\n", "kmol of ethanol per kmol of vapor free air = 0.086957\n" ] } ], "prompt_number": 31 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.31 page number 84\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "vapor_pressure=8. #in kPa\n", "volume_ethanol=0.05;\n", "\n", "\n", "partial_pressure=volume_ethanol*100;\n", "relative_saturation=partial_pressure/vapor_pressure;\n", "mole_ratio=volume_ethanol/(1-volume_ethanol);\n", "print \"mole ratio = %f \\nrelative saturation = %f %%\"%(mole_ratio,relative_saturation*100)\n", "\n", "volume_vapor=(8./100)*100;\n", "ethanol_vapor=volume_vapor/100.;\n", "air_vapor=1-ethanol_vapor;\n", "saturation_ratio=ethanol_vapor/air_vapor;\n", "percentage_saturation=mole_ratio/saturation_ratio;\n", "\n", "print \"percentage saturation = %f %%\"%(percentage_saturation*100)\n", "\n", "print \"corresponding to partial pressure of 5kPa we get a dew point of 17.3 degree celcius\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "mole ratio = 0.052632 \n", "relative saturation = 62.500000 %\n", "percentage saturation = 60.526316 %\n", "corresponding to partial pressure of 5kPa we get a dew point of 17.3 degree celcius\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "example 2.32 page number 84\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "p = 4.24 #in kPa\n", "H_rel = 0.8;\n", "\n", "p_partial = p*H_rel;\n", "molal_H = p_partial/(100-p_partial);\n", "print \"initial molal humidity = %.3f\"%(molal_H)\n", "\n", "P = 200. #in kPa\n", "p_partial = 1.70 #in kPa\n", "final_H = p_partial/(P-p_partial);\n", "print \"final molal humidity = %.4f\"%(final_H)\n", "\n", "p_dryair = 100 - 3.39;\n", "v = 100*(p_dryair/101.3)*(273./303);\n", "moles_dryair = v/22.4;\n", "vapor_initial = molal_H*moles_dryair;\n", "vapor_final = final_H*moles_dryair;\n", "water_condensed = (vapor_initial-vapor_final)*18;\n", "print \"amount of water condensed = %f kg\"%(water_condensed)\n", "\n", "total_air = moles_dryair+vapor_final;\n", "final_v = 22.4*(101.3/200)*(288./273)*total_air;\n", "print \"final volume of wety air = %f m**3\"%(final_v)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "initial molal humidity = 0.035\n", "final molal humidity = 0.0086\n", "amount of water condensed = 1.832428 kg\n", "final volume of wety air = 46.307275 m**3\n" ] } ], "prompt_number": 6 }, { "cell_type": "code", "collapsed": false, "input": [], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }