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diff --git a/Introduction_To_Chemical_Engineering/ch2.ipynb b/Introduction_To_Chemical_Engineering/ch2.ipynb new file mode 100644 index 00000000..7f49d9ae --- /dev/null +++ b/Introduction_To_Chemical_Engineering/ch2.ipynb @@ -0,0 +1,1550 @@ +{ + "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": [ + "#to find the volume of oxygen that can be obtained\n", + "\n", + "import math \n", + "\n", + "# Variables\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", + "# Calculations\n", + "v2=p1*v1*t2/(t1*p2);\n", + "\n", + "# Results\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": [ + "#to find volumetric composition,partial pressue of each gas and total pressure of mixture\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "nCO2 = 2./44; #moles of CO2\n", + "nO2 = 4./32; #moles of O2\n", + "nCH4 = 1.5/16; #moles of CH4\n", + "\n", + "# Calculations and Results\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": [ + "#to find equivalent mass of metal\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "P=104.3 #total pressure in KPa\n", + "pH2O=2.3 #in KPa\n", + "pH2=P-pH2O; #in KPa\n", + "\n", + "# Calculations and Results\n", + "VH2=209*pH2*273/(293*101.3)\n", + "\n", + "print \"volume of hydrogen obtained = %f ml\"%(VH2)\n", + "\n", + "#calculating amount of metal having 11.2l of hydrogen\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": [ + "#to find NaCl content in NaOH solution\n", + "\n", + "import math \n", + "# Variables\n", + "w=2 #in gm\n", + "m=0.287 #in gm\n", + "\n", + "# Calculations and Results\n", + "#precipitate from 58.5gm of NaCl=143.4gm\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": [ + "#to find the carbon content in sample\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "w=4.73 #in gm5\n", + "VCO2=5.30 #in liters\n", + "\n", + "# Calculations\n", + "weight_CO2=44/22.4*VCO2;\n", + "carbon_content=12./44*weight_CO2;\n", + "percentage_content=(carbon_content/w)*100;\n", + "\n", + "# Results\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": [ + "#to find the volume of air\n", + "\n", + "import math \n", + "\n", + "# Variables\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", + "# Calculations\n", + "#required oxygen for various gases\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", + "# Results\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": [ + "#to find the volume of sulphuric acid and mass of water consumed\n", + "\n", + "import math \n", + "\n", + "\n", + "# Variables\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", + "# Calculations\n", + "volume_std = mass_H2SO4/mass_std; #volume of 96%H2SO4\n", + "mass_water = mass_1 - mass_H2SO4;\n", + "\n", + "# Results\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": [ + "#to find molarity,molality and normality\n", + "\n", + "import math \n", + "\n", + "# Variables\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", + "# Calculations and Results\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": [ + "#to find normality\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "molar_mass_BaCl2=208.3; #in gm\n", + "equivalent_H2SO4=0.144;\n", + "\n", + "# Calculations\n", + "normality=equivalent_H2SO4*1000/28.8;\n", + "\n", + "# Results\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": [ + "#to find amount of KClO3 precipitated\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "solubility_70=30.2 #in gm/100gm\n", + "w_solute=solubility_70*350/130.2; #in gm\n", + "\n", + "# Calculations\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", + "# Results\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": [ + "#to find the pressure for solubility of CO2\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "absorbtion_coefficient=1.71 #in liters\n", + "molar_mass=44;\n", + "\n", + "# Calculations\n", + "solubility=absorbtion_coefficient*molar_mass/22.4; #in gm\n", + "pressure=8/solubility*101.3;\n", + "\n", + "# Results\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": [ + "#to find the vapor pressure of water\n", + "\n", + "import math \n", + "\n", + "\n", + "# Variables\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", + "# Calculations\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", + "# Results\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": [ + "#to find the boiling point of solution\n", + "\n", + "import math \n", + "\n", + "# Variables\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", + "# Calculations\n", + "delta_t=E*m;\n", + "boiling_point=100+delta_t;\n", + "\n", + "# Results\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": [ + "#to find the molar mass and osmotic pressure\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "K=1.86;\n", + "c=15 #concentration of alcohol\n", + "delta_t=10.26;\n", + "\n", + "# Calculations and Results\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": [ + "#to find u_in, M_v, k'\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "u_in = 0.575 #from the graph\n", + "u_s = 0.295 #in mPa-s\n", + "\n", + "# Calculations\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", + "# Results\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": [ + "#to find the molecular formula\n", + "\n", + "import math \n", + "\n", + "# Variables\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", + "# Calculations\n", + "ratio=molar_mass/density;\n", + "x=ratio*2;\n", + "y=ratio*1;\n", + "z=ratio*4;\n", + "\n", + "# Results\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": [ + "#to find molecular foemula \n", + "\n", + "import math \n", + "\n", + "# Variables\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", + "# Calculations\n", + "ratio=density/molar_mass;\n", + "x=round(ratio*5);\n", + "y=round(ratio*4);\n", + "\n", + "# Results\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": [ + "#to find molecular formula\n", + "\n", + "import math \n", + "\n", + "# Variables\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", + "# Calculations and Results\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": [ + "#to find the molecular formula\n", + "\n", + "import math \n", + "\n", + "# Variables\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", + "# Calculations\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", + "# Results \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": [ + "#to find sequence of deposition\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "E1=-0.25;\n", + "E2=0.80;\n", + "E3=0.34;\n", + "\n", + "# Calculations\n", + "a=[E1,E2,E3];\n", + "sorted(a)\n", + "\n", + "# Results\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": [ + "#to find the emf of cell\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "E0_Zn=-0.76;\n", + "E0_Pb=-0.13;\n", + "c_Zn=0.1;\n", + "c_Pb=0.02;\n", + "\n", + "# Calculations\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", + "# Results\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": [ + "#to find the emf of cell\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "E0_Ag=0.80;\n", + "E0_AgNO3=0.80;\n", + "c_Ag=0.001;\n", + "c_AgNO3=0.1;\n", + "\n", + "# Calculations\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", + "# Results\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": [ + "#to find emf of cell\n", + "\n", + "import math \n", + "# Variables\n", + "pH=12; #pH of solution\n", + "E_H2=0;\n", + "\n", + "# Calculations\n", + "E2=-0.059*pH;\n", + "E=E_H2-E2;\n", + "\n", + "# Results\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": [ + "#to find amount of silver deposited\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "I=3 #in Ampere\n", + "t=900 #in s\n", + "m_eq=107.9 #in gm/mol\n", + "F=96500;\n", + "\n", + "# Calculations\n", + "m=(I*t*m_eq)/F;\n", + "\n", + "# Results\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": [ + "#to find the time for electroplating\n", + "\n", + "import math \n", + "\n", + "# Variables\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", + "# Calculations\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", + "# Results\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": [ + "#to find hardness of water\n", + "\n", + "# Variables\n", + "m_MgSO4=90. #in ppm\n", + "MgSO4_parts=120.;\n", + "CaCO3_parts=100.;\n", + "\n", + "# Calculations\n", + "hardness=(CaCO3_parts/MgSO4_parts)*m_MgSO4;\n", + "\n", + "# Results\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", + "# Variables\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", + "# Calculations and Results\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", + "#part_1\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", + "#part 2\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": [ + "#to find hardness of water\n", + "\n", + "# Variables\n", + "volume_NaCl=50. #in l\n", + "c_NaCl=5000. #in mg/l\n", + "\n", + "# Calculations\n", + "m=volume_NaCl*c_NaCl;\n", + "equivalent_NaCl=50/58.5;\n", + "\n", + "hardness=equivalent_NaCl*m;\n", + "\n", + "# Results\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": [ + "#to find the total vapor pressure and molar compositions\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "m_benzene = 55. #in kg\n", + "m_toluene = 28. #in kg\n", + "m_xylene = 17. # in kg\n", + "\n", + "# Calculations and Results\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": [ + "#to find the mixture composition\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "vapor_pressure=8. #in kPa\n", + "pressure=100. #in kPa\n", + "\n", + "# Calculations and Results\n", + "#part 1\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", + "#part 2\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", + "#part 3\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", + "#part 4\n", + "w_ethanol=mass_ethanol/mass_air;\n", + "print \"weight of ethanol/kg vapor free air = %f Kg\"%(w_ethanol)\n", + "\n", + "#part 5\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": [ + "#to find relative saturation and dew point\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "vapor_pressure=8. #in kPa\n", + "volume_ethanol=0.05;\n", + "\n", + "\n", + "# Calculations and Results\n", + "#basis 1kmol of mixture\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", + "#basis 1kmol saturated gas mixture at 100kPa\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", + "#dew point\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": [ + "#to find the properties of humid air\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "p = 4.24 #in kPa\n", + "H_rel = 0.8;\n", + "\n", + "# Calculations and Results\n", + "p_partial = p*H_rel;\n", + "molal_H = p_partial/(100-p_partial);\n", + "print \"initial molal humidity = %.3f\"%(molal_H)\n", + "\n", + "#part 2\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", + "#part 3\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", + "#part 4\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": {} + } + ] +}
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