{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 2: Physico Chemical Calculations"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.10: Precipitation_of_KClO3.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.10 page number 74\n\n')\n",
"\n",
"//to find amount of KClO3 precipitated\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",
"printf('amount precipitated = %f gm',precipitate)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.11: Solubility_of_CO2.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.11 page number 74\n\n')\n",
"\n",
"//to find the pressure for solubility of CO2\n",
"\n",
"absorbtion_coefficient=1.71    //in liters\n",
"molar_mass=44;\n",
"solubility=absorbtion_coefficient*molar_mass/22.4;   //in gm\n",
"pressure=8/solubility*101.3;\n",
"\n",
"printf('pressure required = %f kPa',pressure)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.12: Vapor_pressure_calculatio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.12 page number 74\n\n')\n",
"\n",
"//to find the vapor pressure of water\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",
"printf('depression in vapor pressure = %f Pa',depression*1000)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.13: Boiling_point_calculation.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.13 page number 75\n\n')\n",
"\n",
"//to find the boiling point of solution\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",
"printf('boiling_point of water = %f degreeC',boiling_point)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.14: Colligative_properties.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.14 page number 75\n\n')\n",
"\n",
"//to find the molar mass and osmotic pressure\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",
"printf('molar mass = %f gm\n\n',M*1000)\n",
"\n",
"density=0.97     //g/ml\n",
"cm=c*density/(M*100);\n",
"printf('molar concentration of alcohol = %f moles/l\n\n',cm)\n",
"\n",
"p=cm*8.314*293   //osmotic pressure\n",
"printf('osmotic pressure = %f Mpa\n\n',p/1000)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.15: Huggins_Equation.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.15 page number 75\n\n')\n",
"\n",
"//to find u_in, M_v, k'\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",
"printf('k = %f \nMv = %f\nu_in = %f dl/gm',k,M_v,u_in)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.16: Molecular_Formula.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.16 page number 76\n\n')\n",
"\n",
"//to find the molecular formula\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",
"printf('x = %f, y = %f, z = %f',x,y,z)\n",
"printf('\n\nformula of butyric acid is = C4H8O2')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.17: Molecular_Formula.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.17 page number 77\n\n')\n",
"\n",
"//to find molecular foemula \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",
"\n",
"x=ratio*5;\n",
"y=ratio*4;\n",
"\n",
"\n",
"printf('x = %f, y = %f',x,y)\n",
"printf('\n\nformula of butyric acid is = C10H8')\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.18: Molecular_Formula.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.18 page number 77\n\n')\n",
"\n",
"//to find molecular formula\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",
"function M=f(m)\n",
"    M=(2.09*1000)/(60*m);\n",
"\n",
"endfunction\n",
"\n",
"M=f((1.25/5.1));\n",
"\n",
"printf('actual molecular mass = %f\n\n',M)\n",
"\n",
"ratio=M/molar_mass;\n",
"a=ratio*3;\n",
"b=ratio*3;\n",
"c=ratio*2;\n",
"\n",
"\n",
"printf('a = %f, b = %f, c = %f',a,b,c)\n",
"printf('\n\nformula of butyric acid is = C6H6O4')\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.19: Molecular_Formula.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.19 page number 78\n\n')\n",
"\n",
"//to find the molecular formula\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=H2/2;    //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",
" printf('a = %f, b = %f, c = %f, d = %f, e = %f',a*6.5,b*6.5,c*6.5,d*6.5,e*6.5)\n",
"printf('\n\nformula of butyric acid is = C34H33N4O5Fe')\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.1: Ideal_gas_system.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.1 page number 71\n\n')\n",
"\n",
"//to find the volume of oxygen that can be obtained\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",
"printf('volume of oxygen = %f liters',v2)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.20: Metal_depositio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"printf('example 2.20 page number 78\n\n')\n",
"\n",
"//to find sequence of deposition\n",
"E1=-0.25;\n",
"E2=0.80;\n",
"E3=0.34;\n",
"\n",
"a=[E1;E2;E3];\n",
"b=gsort(a);\n",
"\n",
"printf('sorted potential in volts =')\n",
"disp (b)\n",
"disp ('E2>E3>E1')\n",
"disp ('silver>copper>nickel')\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.21: EMF_of_cell.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.21 page number 79\n\n')\n",
"\n",
"//to find the emf of cell\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)*log10(c_Zn);\n",
"E_Pb=E0_Pb+(0.059/2)*log10(c_Pb);\n",
"E=E_Pb-E_Zn;\n",
"\n",
"printf('emf of cell = %f V',E)\n",
"printf('\n\nSince potential of lead is greater than that of zinc thus reduction will occur at lead electrode and oxidation will occur at zinc electrode')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.22: EMF_of_cell.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.22 page number 79\n\n')\n",
"\n",
"//to find the emf of cell\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)*log10(c_Ag);\n",
"E_AgNO3=E0_AgNO3+(0.059)*log10(c_AgNO3);\n",
"E=E_AgNO3-E_Ag;\n",
"\n",
"printf('emf of cell = %f V' ,E)\n",
"printf('\n\nsince 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')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.23: EMF_of_cell.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.23 page number 79\n\n')\n",
"\n",
"//to find emf of cell\n",
"pH=12;    //pH of solution\n",
"E_H2=0;\n",
"E2=-0.059*pH;\n",
"E=E_H2-E2;\n",
"printf('EMF of cell = %f V',E)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.24: Silver_depositio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.24 page number 80\n\n')\n",
"\n",
"//to find amount of silver deposited\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",
"printf('mass = %f gm',m)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.25: Electroplating_time.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"printf('example 2.25 page number 80\n\n')\n",
"\n",
"//to find the time for electroplating\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",
"printf('time required = %f hours',t)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.26: Water_hardness.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.26 page number 80\n\n')\n",
"\n",
"//to find hardness of water\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",
"printf('hardness of water = %f mg/l',hardness)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.27: Water_hardness.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.26 page number 80\n\n')\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",
"//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",
"printf('total hardness = %f\n temporary hardness = %f \n',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",
"printf('amount of lime required = %f \n',actual_lime)\n",
"\n",
"soda_required = (106/100)*(content_1+content_4);\n",
"actual_soda = soda_required/0.98;\n",
"printf('amount of soda required = %f \n',actual_soda)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.28: Water_hardness.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.28 page number 82\n\n')\n",
"\n",
"//to find hardness of water\n",
"\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",
"printf('hardness of water = %f mg/l',hardness/1000)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.29: Mixture_compositio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.29 page number 82\n\n')\n",
"\n",
"//to find the total vapor pressure and molar compositions\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",
"printf('total pressure = %f kPa\n',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",
"printf('xylene = %f \n toluene = %f \n benzene = %f',xylene,toluene,benzene)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.2: Mixture_properties.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.2 page number 71\n\n')\n",
"\n",
"//to find volumetric composition,partial pressue of each gas and total pressure of mixture\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",
"printf (' Composition of mixture = \nCH4 = %f \nO2 = %f \n CO2 = %f \n\n',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",
"printf ('pressure of CH4 = %f kPa \npressure of O2 = %f kPa\n pressure of CO2 =%f kPa\n\n',pCH4*10^-3,pO2*10^-3,pCO2*10^-3)\n",
"\n",
"total_pressure=pCO2+pCH4+pO2;\n",
"printf ('total pressure =  %f Kpa',total_pressure*10^-3)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.30: Mixture_compositio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.30 page number 83\n\n')\n",
"\n",
"//to find the mixture composition\n",
"\n",
"vapor_pressure=8    //in kPa\n",
"pressure=100       //in kPa\n",
"\n",
"//part 1\n",
"volume=1     //in m3\n",
"volume_ethanol=volume*(vapor_pressure/pressure);\n",
"volume_air=1-volume_ethanol;\n",
"printf('volumetric composition:- \nair composition = %f\n 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",
"printf('\n\ncomposition by weight:-\nAir = %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",
"printf('\n\nweight of ethanol/cubic meter = %f Kg',weight_ethanol)\n",
"\n",
"//part 4\n",
"w_ethanol=mass_ethanol/mass_air;\n",
"printf('\n\nweight of ethanol/kg vapor free air = %f Kg',w_ethanol)\n",
"\n",
"//part 5\n",
"moles_ethanol=0.08/0.92;\n",
"printf('\n\nkmol of ethanol per kmol of vapor free air = %f',moles_ethanol)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.31: Mixture_properties.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.31 page number 84\n\n')\n",
"\n",
"//to find relative saturation and dew point\n",
"\n",
"vapor_pressure=8   //in kPa\n",
"volume_ethanol=0.05;\n",
"\n",
"//basis 1kmol of mixture\n",
"\n",
"partial_pressure=volume_ethanol*100;\n",
"relative_saturation=partial_pressure/vapor_pressure;\n",
"mole_ratio=volume_ethanol/(1-volume_ethanol);\n",
"printf('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",
"printf('\n\npercentage saturation = %f',percentage_saturation)\n",
"\n",
"//dew point\n",
"printf('\n\ncorresponding to partial pressure of 5kPa we get a dew point of 17.3 degree celcius')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.32: Humidity.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.32 page number 84\n\n')\n",
"\n",
"//to find the properties of humid air\n",
"\n",
"p = 4.24   //in kPa\n",
"H_rel = 0.8;\n",
"p_partial = p*H_rel;\n",
"molal_H = p_partial/(100-p_partial);\n",
"printf('initial molal humidity = %f\n\n',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",
"printf('final molal humidity = %f\n\n',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",
"printf('amount of water condensed = %f \n\n',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",
"printf('final volume of wety air = %f \n\n',final_v)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.3: Equivalent_metal_mass.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.3 page number 72\n\n')\n",
"\n",
"//to find equivalent mass of metal\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",
"printf('volume of hydrogen obtained = %f ml\n\n',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",
"printf('mass of metal equivalent to 11.2 litre/mol of hydrogen = %f gm',m)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.4: Purity_of_Sodium_Hydroxide.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.4 page number 72\n\n')\n",
"\n",
"//to find NaCl content in NaOH solution\n",
"\n",
"w=2   //in gm\n",
"m=0.287  //in gm\n",
"\n",
"//precipitate from 58.5gm of NaCl=143.4gm\n",
"\n",
"mNaCl=58.5/143.4*m;\n",
"\n",
"printf('mass of NaCl = %f gm\n',mNaCl )\n",
"\n",
"percentage_NaCl=mNaCl/w*100;\n",
"printf('amount of NaCl = %f',percentage_NaCl)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.5: Carbon_content_formulation.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.5 page number 72\n\n')\n",
"\n",
"//to find the carbon content in sample\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",
"\n",
"percentage_content=(carbon_content/w)*100;\n",
"\n",
"printf('percentage amount of carbon in sample = %f',percentage_content)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.6: Combustion_of_gas.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.6 page number 73\n\n')\n",
"//to find the volume of air\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",
"//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",
"printf('amount of oxygen required = %f cubic meter',oxygen_required)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.7: Sulphuric_acid_preparatio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.7 page number 73\n\n')\n",
"\n",
"//to find the volume of sulphuric acid and mass of water consumed\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",
"printf('volume of 0.96 H2SO4 required = %f ml',volume_std)\n",
"printf('\nmass of water required = %f g',mass_water)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.8: Molarity_Molality_Normality_Calculatio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.8 page number 73\n\n')\n",
"\n",
"//to find molarity,molality and normality\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",
"printf('Molarity = %f mol/l\n\n',Molarity)\n",
"\n",
"equivalent_mass=49;\n",
"normality=weight/equivalent_mass;\n",
"printf('Normality = %f N\n\n',normality)\n",
"\n",
"molality=176.5/molar_mass;\n",
"printf('Molality = %f',molality)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.9: Normality_calculatio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear \n",
"printf('example 2.9 page number 74\n\n')\n",
"\n",
"molar_mass_BaCl2=208.3;      //in gm\n",
"equivalent_H2SO4=0.144;\n",
"normality=equivalent_H2SO4*1000/28.8;\n",
"\n",
"printf('Normality = %f N',normality)"
   ]
   }
],
"metadata": {
		  "kernelspec": {
		   "display_name": "Scilab",
		   "language": "scilab",
		   "name": "scilab"
		  },
		  "language_info": {
		   "file_extension": ".sce",
		   "help_links": [
			{
			 "text": "MetaKernel Magics",
			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
			}
		   ],
		   "mimetype": "text/x-octave",
		   "name": "scilab",
		   "version": "0.7.1"
		  }
		 },
		 "nbformat": 4,
		 "nbformat_minor": 0
}