{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 2 : Basic Chemical Calculations"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 2.1   Page 17"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "5 mol of NH4Cl =  267.5  [g]\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "# NH4Cl\n",
    "M = 14+4+35.5           # [g] (molar mass of NH4Cl)\n",
    "n=5                     # [mol]\n",
    "\n",
    "# Calculation \n",
    "m = M*n                 # [g]\n",
    "\n",
    "# Result\n",
    "print \"5 mol of NH4Cl = \",m,\" [g]\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 2.2   Page 17"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "In the formula CuSO4.5H2O, the moles of CuSO4 is one hence,  the equivalent moles of CuSO4 in the crystal is  2.0 .\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "# Variables \n",
    "\n",
    "# CuSO4.5H2O\n",
    "M1 = 159.5                  # [g] (molar mass of CuSO4)\n",
    "M2 = 159.5+5.*(2.+16)       # (molar mass of CuSO4.5H2O)\n",
    "m = 499.\n",
    "\n",
    "# Calculation \n",
    "n = m/M2                    #[mol]\n",
    "\n",
    "# Result\n",
    "print \"In the formula CuSO4.5H2O, the moles of CuSO4 is one hence,\\\n",
    "  the equivalent moles of CuSO4 in the crystal is \",n,\".\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.3   Page 17"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "  1.5  kmol of K2CO3 contains 117 kg of K.\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "# Variables \n",
    "\n",
    "# K2CO3\n",
    "m = 117.            # [kg] (wt of K)\n",
    "Mk = 39.            # [g] (at wt of K)\n",
    "\n",
    "# Calculation \n",
    "a = m/Mk            # [kg atoms] \n",
    "\n",
    "# 1 mol of K2CO3 contains 2 atoms of K\n",
    "n = a/2. # [kmol] (moles of K2CO3)\n",
    "\n",
    "# Result\n",
    "print \" \",n,\" kmol of K2CO3 contains 117 kg of K.\" \n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.4   Page 18"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Atoms present in 416.6 g BaCl2 =  1.2044e+24 \n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "# BaCl2\n",
    "M = 137.3+2*35.5                # [g] (molar mass of BaCl2)\n",
    "m = 416.6                       # [g]\n",
    "\n",
    "# Calculation \n",
    "n = m/M                         # [mol]\n",
    "N = n*6.022*10**23              # (no. of atoms)\n",
    "\n",
    "# Result\n",
    "print \"Atoms present in 416.6 g BaCl2 = \",N,\"\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.5   Page 19"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "a   eq. mass of PO4 is  31.6666666667  [g]   \n",
      "b   eq. mass of Na3PO4 is  54.6666666667  [g]  \n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "print \"a  \",\n",
    "#PO4 radical\n",
    "M = 31+4*16.                #[g] \n",
    "V = 3.                      # (valence of PO4)\n",
    "\n",
    "# Calculation and Result\n",
    "eqm = M/V \n",
    "print \"eq. mass of PO4 is \",eqm,\" [g]   \"\n",
    "print \"b  \",\n",
    "#Na3PO4\n",
    "M = 3*23+95.                #[g]\n",
    "V = 3.\n",
    "eqm = M/V\n",
    "print \"eq. mass of Na3PO4 is \",eqm,\" [g]  \"\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 2.6   Page 19"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "no. of equivalents in 3 kmol of AlCl3 is  9.0  keq.\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "# AlCl3\n",
    "v = 3.                          # valency of Al ion\n",
    "\n",
    "# Calculation \n",
    "eq = 3.*3                       # [mol]\n",
    "\n",
    "# Result\n",
    "print \"no. of equivalents in 3 kmol of AlCl3 is \",eq,\" keq.\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.7   Page 20"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "a   mass percentage of NaCl is  75.0  mass percentage of KCl is  25.0    \n",
      "b   mol percentage of NaCl is  79.2553191489  mol percentage of KCl is  20.7446808511  \n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "# (a)\n",
    "print \"a  \",\n",
    "# mass %\n",
    "m1 = 600.                   #[kg] (NaCl)\n",
    "m2 = 200.                   #[kg] (KCl)\n",
    "\n",
    "# Calculation \n",
    "m = m1+m2                   # total mass\n",
    "Wa = (m1/m)*100.\n",
    "Wb = (m2/m)*100.\n",
    "\n",
    "# Result\n",
    "print \"mass percentage of NaCl is \",Wa,\" mass percentage of KCl is \",Wb,\"   \"\n",
    "\n",
    "# (b)\n",
    "print \"b  \",\n",
    "#mol %\n",
    "M1 = 23+35.5                # molar mass of NaCl\n",
    "n1 = m1/M1                  # no. of moles of NaCl\n",
    "M2 = 39+35.5                # molar mass of KCl\n",
    "n2 = m2/M2                  # no. of moles of KCl\n",
    "n = n1+n2\n",
    "N1 = (n1/n)*100.\n",
    "N2 = (n2/n)*100.\n",
    "print \"mol percentage of NaCl is \",N1,\" mol percentage of KCl is \",N2,\" \"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.8   Page 21"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "mass percentage of C is  53.4941324009\n",
      " mass percentage of H is  1.57123286189\n",
      " mass percentage of O is  24.9407534679\n",
      " mass percentage of S is  19.9938812693\n",
      "molar mass =  12.2023744565  kg/kmol.\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "# CH.35O.35S.14\n",
    "# mass %\n",
    "C = 12.0107                     #[kg] \n",
    "H = 1.00794*.35                 # [kg]\n",
    "O = 15.9994*.35                 # [kg]\n",
    "S = 32.065*.14                  #[kg]\n",
    "\n",
    "# Calculation \n",
    "m = C+H+O+S\n",
    "m1 = (C/m)*100.\n",
    "m2 = (H/m)*100.\n",
    "m3 = (O/m)*100.\n",
    "m4 = (S/m)*100.\n",
    "\n",
    "# Result\n",
    "print \"mass percentage of C is \",m1\n",
    "print \" mass percentage of H is \",m2\n",
    "print \" mass percentage of O is \",m3\n",
    "print \" mass percentage of S is \",m4\n",
    "M = m/(1+.35+.35+.14)\n",
    "print \"molar mass = \",M,\" kg/kmol.\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.9   Page 22"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The sample contains  96.4285714286  percent urea.\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "# basis 100kg urea\n",
    "m1 = 45.                 #[kg] (mass of N present)\n",
    "Mu = 60.                 # (molar mass of urea)\n",
    "m2 = 14*2.               #[kg] (mass of N in 1 kmol of urea)\n",
    "\n",
    "# Calculation \n",
    "m = (Mu/m2)*m1\n",
    "\n",
    "# Result\n",
    "print \"The sample contains \",m,\" percent urea.\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.10   Page 22"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Mass percent of SiO2 is  0.006 .\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "# NaOH\n",
    "Impurity = 60.                   # [ppm] SiO2\n",
    "\n",
    "# Calculation \n",
    "m = (60/1000000.)*100.\n",
    "\n",
    "# Result\n",
    "print \"Mass percent of SiO2 is \",m,\".\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.11    Page 22"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total no. of Ca+ ions is  392.749838659  and total no. of F- ions is  2.43335898941 .\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "Ca = 40.078                         # at. wt of Ca\n",
    "F = 18.9984032                      # at wt of F\n",
    "\n",
    "# Calculation \n",
    "M1 = 3*Ca +2*(30.97762+(4*15.9994)) # molar mass of Ca3PO4\n",
    "M2 = Ca +12.0107+3*15.9994          # molar mass of CaCO3\n",
    "M3 = Ca+2*F                         # molar mass of CaF2\n",
    "m1 = 800.                           #[mg] Ca3PO4\n",
    "m2 = 200.                           #[mg] CaCO3\n",
    "m3 = 5.                             #[mg] CaF2\n",
    "n1 = ((3*Ca)/M1)*m1+(Ca/M2)*m2+(Ca/M3)*m3       # [mg] total Ca ions\n",
    "n2 = (F/M3)*2*5                     #[mg] total F ions\n",
    "\n",
    "# Result\n",
    "print \"Total no. of Ca+ ions is \",n1,\" and total no. of F- ions is \",n2,\".\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.12   Page 23"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "a   mass percent of salicylic acid is  39.0243902439  and mass percent of methanol is  60.9756097561 .   \n",
      "b   Mole percent of methanol is  87.0772337203  and Mole percent of salicylic acid is  12.9227662797 .\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# solution \n",
    "\n",
    "# Variables \n",
    "# (a)\n",
    "print \"a  \",\n",
    "# mass %\n",
    "m1 = 100.                        #[kg] methanol (basis)\n",
    "m2 = 64.                         #[kg] salicylic acid\n",
    "\n",
    "# Calculation \n",
    "m = m1+m2                        # [kg] mass of solution\n",
    "w1 = m2/m*100\n",
    "w2 = 100-w1\n",
    "\n",
    "# Result\n",
    "print \"mass percent of salicylic acid is \",w1,\" and mass percent of methanol is \",w2,\".   \"\n",
    "\n",
    "#(b)\n",
    "print \"b  \",\n",
    "#mole %\n",
    "M1 = 32.                         # molar mass of methanol\n",
    "M2 = 138.                        #molar mass of salicylic acid\n",
    "n1 = m1/M1                       #[kmol] methanol\n",
    "n2 = m2/M2                       #[kmol] salicylic acid\n",
    "n = n1+n2\n",
    "N1 = n1/n*100.\n",
    "N2 = n2/n*100.\n",
    "print \"Mole percent of methanol is \",N1,\" and Mole percent of salicylic acid is \",N2,\".\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.13  Page 24"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "mass percent of HCl is  11.195554466\n",
      "mass percent of NaCl is  7.08506987007\n",
      "and mass percent of H2O is  81.719375664 .   \n",
      "Mole percent of HCl is  6.1851712101\n",
      "Mole percent of NaCl is  2.44200543258\n",
      "and Mole percent of H2O is  91.3728233573\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# solution \n",
    "\n",
    "# Variables \n",
    "#mass %\n",
    "m1 = 13.70                      # HCl\n",
    "m2 = 8.67                       # NaCl\n",
    "m3 = 100.                       # H2O\n",
    "\n",
    "# Calculation \n",
    "m = m1+m2+m3                    # mass of solution\n",
    "w1 = m1/m*100.\n",
    "w2 = m2/m*100.\n",
    "w3 = m3/m*100.\n",
    "\n",
    "# Result\n",
    "print \"mass percent of HCl is \",w1\n",
    "print \"mass percent of NaCl is \",w2\n",
    "print \"and mass percent of H2O is \",w3,\".   \"\n",
    "M1=36.4609                      #HCl\n",
    "M2=58.4428                      #NaCl\n",
    "M3=18.0153                      #H2O\n",
    "n1=m1/M1                        #HCl\n",
    "n2=m2/M2                        #NaCl\n",
    "n3=m3/M3                        #H2O\n",
    "n=n1+n2+n3\n",
    "N1=n1/n*100.\n",
    "N2=n2/n*100.\n",
    "N3=n3/n*100.\n",
    "print \"Mole percent of HCl is \",N1\n",
    "print \"Mole percent of NaCl is \",N2\n",
    "print \"and Mole percent of H2O is \",N3\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.14  Page 24"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "percentage of Na2O in the solution is  56.575 .\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# solution \n",
    "# Variables\n",
    "\n",
    "m = 100.                                     #[kg] Lye (basis)\n",
    "m1 = 73.                                     #[kg] NaOH\n",
    "M1 = 40.                                     # NaOH\n",
    "M2 = 62.                                     # Na2O\n",
    "\n",
    "# Calculation\n",
    "p = (M2*m1)/(2*M1)\n",
    "\n",
    "# Result\n",
    "print \"percentage of Na2O in the solution is \",p,\".\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.15  Page 25"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "TOC =  234.782608696  mg/l\n",
      "ThOD =  730.434782609  mg/l\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# solution \n",
    "\n",
    "# Variables \n",
    "#(CH2OH)3\n",
    "M = 92.                             # molar mass of glycerin\n",
    "C = 600.                            #[mg/l] glycerin conc.\n",
    "\n",
    "# Calculation \n",
    "TOC = (3.*12./92.)*600.             #[mg/l]\n",
    "\n",
    "# by combustion reaction we see 3.5 O2 is required for 1 mol of (CH2OH)3\n",
    "ThOD = (3.5*32.*600)/92             #[mg/l]\n",
    "\n",
    "# Result\n",
    "print \"TOC = \",TOC,\" mg/l\"\n",
    "print \"ThOD = \",ThOD,\" mg/l\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.16  Page 25"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Actual concentration of CaHCO32 in the sample water is  486.0  mg/l and of MgHCO32 is  292.6  mg/l.\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "M1 = 100.                   # CaCO3\n",
    "v1 = 2.                     # valence of CaCO3\n",
    "\n",
    "# Calculation \n",
    "eqm1 = M1/v1                # equivalent mass of CaCO3\n",
    "M2 = 162.                   # Ca(HCO3)2\n",
    "v2 = 2.\n",
    "eqm2 = M2/v2\n",
    "m = 500.                    # [mg/l] CaCO3\n",
    "C1 = (eqm2/eqm1)*m*.6       # [mg/l] conc. of Ca(HCO3)2\n",
    "M3 = 146.3                  # Mg(HCO3)2\n",
    "v3 = 2.\n",
    "eqm3 = M3/v3\n",
    "C2 = (eqm3/eqm1)*m*.4       #[mg/l] conc. of Mg(HCO3)2\n",
    "\n",
    "# Result\n",
    "print \"Actual concentration of CaHCO32 in the sample\\\n",
    " water is \",C1,\" mg/l and of MgHCO32 is \",C2,\" mg/l.\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.17  Page 26"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Sulphur content in LDO is  5780.0  ppm.\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "S = .68                 # sulphur content by mass\n",
    "d = .85                 # kg/l\n",
    "\n",
    "# Calculation \n",
    "s = (S*d*10**6)/100.    # [mg/l] or [ppm]\n",
    "\n",
    "# Result\n",
    "print \"Sulphur content in LDO is \",s,\" ppm.\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.18   Page 26"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Molarity =  0.385299145299 M\n",
      "Normality =  0.385299145299 N\n",
      "Molality =  0.42735042735  mol/kg.\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# solution \n",
    "\n",
    "# Variables \n",
    "m1 = 100.                    #[kg] solution (basis)\n",
    "m2 = 20.                     #[kg] NaCl\n",
    "d = 1.127                    #[kg/l]\n",
    "\n",
    "# Calculation \n",
    "V = m1/d                     # volume of 100 kg sol.\n",
    "n = (m2/58.5)*100.           # [mol] NaCl\n",
    "M = n/V                      #[M]\n",
    "v = 1.                       # valence of NaCl so,\n",
    "N = M\n",
    "m = n/(m1-m2)                #[mol/kg]\n",
    "\n",
    "# Result\n",
    "print \"Molarity = \",M,\"M\"\n",
    "print \"Normality = \",N,\"N\"\n",
    "print \"Molality = \",m,\" mol/kg.\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.19   Page 27"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Molarity of solution =  0.352348993289  M.\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "m1 = 100.                #[kg] TEA solution (basis)\n",
    "m2 = 50.                 #[kg] TEA\n",
    "M1 = 149.                # molar mass of TEA\n",
    "d = 1.05                 #[kg/l]\n",
    "\n",
    "# Calculation \n",
    "V = m1/d                 # volume of 100 kg sol.\n",
    "n = (m2/M1)*100.         # [mol] NaCl\n",
    "M = n/V                  #[M]\n",
    "\n",
    "# Result\n",
    "print \"Molarity of solution = \",M,\" M.\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###   Example 2.20   Page 27"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Mass percent of CO2 =  2.97180327869  and Mol percent =  1.44480291766 .\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "m1 = 100.                   #[kg] MEA solution (basis)\n",
    "m2 = 20.                    #[kg] MEA\n",
    "M1 = 61.                    # molar mass of MEA\n",
    "n1 = m2/M1                  # [kmol]\n",
    "C = .206 \n",
    "\n",
    "# Calculation \n",
    "n2 = C*n1                   #[kmol] dissolved CO2\n",
    "m3 = n2*44                  # [kg] mass of CO2\n",
    "n3 = (m1-m2-m3)/18          #[kmol] water\n",
    "n = (n2/(n1+n2+n3))*100.\n",
    "m = (m3/100)*100.\n",
    "\n",
    "# Result\n",
    "print \"Mass percent of CO2 = \",m,\" and Mol percent = \",n,\".\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 2.21   Page 27"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "pH of the sol is  3.50886438348 .\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "# solution \n",
    "# Variables \n",
    "#HOCl\n",
    "Ma = .1                     #molarity\n",
    "\n",
    "# Calculation \n",
    "Ka = 9.6*10**-7\n",
    "C = (Ma*Ka)**.5             # conc. of H+ ions\n",
    "pH = -math.log10(C)\n",
    "\n",
    "# Result\n",
    "print \"pH of the sol is \",pH,\".\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.22   Page 39"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "average molar mass of air is  28.96968  g.\n"
     ]
    }
   ],
   "source": [
    "\n",
    "\n",
    "# solution \n",
    "# Variables \n",
    "n = 100.                     # [mol] air (basis)\n",
    "n1 = 21.                     #[mol] O2\n",
    "n2 = 78.                     #[mol] N2\n",
    "n3 = 1.                      #[mol] Ar\n",
    "M1 = 31.9988                 # O2\n",
    "M2 = 28.0134                 # N2\n",
    "M3 = 39.948                  # Ar\n",
    "\n",
    "# Calculation \n",
    "m1 = n1*M1\n",
    "m2 = n2*M2\n",
    "m3 = n3*M3\n",
    "Ma = (m1+m2+m3)/n\n",
    "\n",
    "# Result\n",
    "print \"average molar mass of air is \",Ma,\" g.\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.23   Page 39"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "a   Average molar mass of gas is  26.54  g.\n",
      "b      GAS            Mass Percent  \n",
      "  Methane          27.128862095\n",
      "  Ethane           11.3036925396\n",
      "  Ethylene         26.3752825923\n",
      "  Propane          11.6051243406\n",
      "  Propylene        12.6601356443\n",
      "  n-Butane         10.9269027882\n",
      "c   Specific gravity is  0.916120124266 .\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "# Variables \n",
    "# (a)\n",
    "print \"a  \",\n",
    "n = 100.                         # [kmol] cracked gas (basis)\n",
    "n1 = 45.                         # methane\n",
    "n2 = 10.                         # ethane\n",
    "n3 = 25.                         # ethylene\n",
    "n4 = 7.                          # propane\n",
    "n5 = 8.                          # propylene\n",
    "n6 = 5.                          # n-butane\n",
    "M1 = 16.                    \n",
    "M2 = 30.\n",
    "M3 = 28.\n",
    "M4 = 44.\n",
    "M5 = 42.\n",
    "M6 = 58.\n",
    "\n",
    "# Calculation \n",
    "m1 = n1*M1\n",
    "m2 = n2*M2\n",
    "m3 = n3*M3\n",
    "m4 = n4*M4\n",
    "m5 = n5*M5\n",
    "m6 = n6*M6\n",
    "m = m1+m2+m3+m4+m5+m6\n",
    "M = m/n\n",
    "\n",
    "# Result\n",
    "print \"Average molar mass of gas is \",M,\" g.\"\n",
    "#(b)\n",
    "print \"b  \",\n",
    "\n",
    "# composition\n",
    "p1 = (m1/m)*100.\n",
    "p2 = m2*100./m\n",
    "p3 = m3*100./m\n",
    "p4 = m4*100./m\n",
    "p5 = m5*100./m\n",
    "p6 = m6*100./m\n",
    "print \"   GAS            Mass Percent  \"\n",
    "print \"  Methane         \",p1\n",
    "print \"  Ethane          \",p2\n",
    "print \"  Ethylene        \",p3\n",
    "print \"  Propane         \",p4\n",
    "print \"  Propylene       \",p5\n",
    "print \"  n-Butane        \",p6\n",
    "# (c)\n",
    "print \"c  \",\n",
    "# specific gravity\n",
    "g = M/28.97\n",
    "print \"Specific gravity is \",g,\".\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.24   Page 40"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Specific volume =  0.0287598911758  m**3/kg.\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "# Variables \n",
    "p = 100.                     #[bar]\n",
    "T = 623.15                   #[K]\n",
    "R = .083145\n",
    "\n",
    "# Calculation \n",
    "V = R*T/p                    # [l/mol] molar volume\n",
    "v = V/18.0153 \n",
    "\n",
    "# Result \n",
    "print \"Specific volume = \",v,\" m**3/kg.\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 2.25   Page 40"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Molar volume =  16.0708891875  l/mol.  \n",
      "by Vanderwall eq molar volume =  15.74  l/mol\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# solution \n",
    "# Variables \n",
    "p = 4.                   #[bar]\n",
    "T = 773.15               #[K]\n",
    "R = .083145\n",
    "V = R*T/p                # [l/mol] molar volume\n",
    "print \"Molar volume = \",V,\" l/mol.  \"\n",
    "\n",
    "# Calculation \n",
    " # using appendix III\n",
    " # calculating Tc and Pc of different gases according to their mass fractions\n",
    "Tc1 = .352*32.20            # H2\n",
    "Tc2 = .148*190.56           # methane\n",
    "Tc3 = .128*282.34           #ethylene\n",
    "Tc4 = .339*132.91           # CO\n",
    "Tc5 = .015*304.10           # CO2\n",
    "Tc6 = .018*126.09           # N2\n",
    "Tc = Tc1+Tc2+Tc3+Tc4+Tc5+Tc6            # Tc of gas\n",
    "\n",
    " # similarly finding Pc\n",
    "Pc1=.352*12.97 \n",
    "Pc2=.148*45.99\n",
    "Pc3=.128*50.41\n",
    "Pc4=.339*34.99\n",
    "Pc5=.015*73.75\n",
    "Pc6=.018*33.94\n",
    "Pc=Pc1+Pc2+Pc3+Pc4+Pc5+Pc6              # Pc of gas\n",
    "a = (27*R**2*Tc**2)/(64*Pc)             # [bar/mol**2]\n",
    "b = (R*Tc)/(8*Pc)                       # l/mol\n",
    "\n",
    "# substituting these values in vanderwall eq and solving by Newton Rhapson method we get\n",
    "V = 15.74                               # [l/mol]\n",
    "\n",
    "# Result\n",
    "print \"by Vanderwall eq molar volume = \",V,\" l/mol\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.26   Page 43"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "component        mol percent        mass percent\n",
      " n-Butane        43.1          570378.642948\n",
      " 1-Butene        56.9          727039.966353\n",
      " Furfural       -26.0350       570378.642948\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# solution \n",
    "# Variables \n",
    "m = 6.5065                      #[g] mixture (basis)\n",
    "Pv = 2.175                      #[kPa] V.P. of water over KOH\n",
    "Pa = 102.5-2.175                #[kPa] Partial P of n-butane and 1butene\n",
    "V = 415.1*10**-3                #[l]\n",
    "R = 8.314472\n",
    "T = 296.4                       #[K]\n",
    "\n",
    "# Calculation \n",
    "n = (Pa*V)/R*T                  # moles of butene and butane\n",
    "n1 = n*.431                     # n-butane\n",
    "m1 = n1*58                      # [g]\n",
    "n2 = n-n1                       # 1 butene\n",
    "m2 = n2*56                      #[g]\n",
    "m3 = m-m1                       # [g] furfural\n",
    "n3 = m3/96.\n",
    "\n",
    "# Result\n",
    "print \"component        mol percent        mass percent\"\n",
    "print \" n-Butane       \",n1/n*100,\"        \",m1/m*100\n",
    "print \" 1-Butene       \",n2/n*100,\"        \",m2/m*100\n",
    "print \" Furfural       %.4f\"%(n3/n*100),\"     \",m1/m*100\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 2.27   Page 44"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 27,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "mol fraction of Furfural is  0.00379701058122 mol fraction of 1-Butene is  0.996202989419 .\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "P = 5.7+1.01                    #[bar] absolute total P\n",
    "\n",
    "# Calculation \n",
    "# using Roult's law\n",
    "vp = 3.293*.7737                #[kPa] vap P of furfural\n",
    "# using Dalton's law of partial P\n",
    "n1 = vp/(P*100)                 # mol fraction of furfural\n",
    "n2 = 1-n1                       # mol fraction of 1 -butene\n",
    "\n",
    "# Result\n",
    "print \"mol fraction of Furfural is \",n1,\"mol fraction of 1-Butene is \",n2,\".\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.28   Page 44"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "absolute humidity =  0.0161586369044 .\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "P = 100.                     #[kPa] total P\n",
    "Pw = 2.5326                  #[kPa] V.P> of water at dew point\n",
    "\n",
    "# Calculation \n",
    "#absolute humity = mass of water vapour/ mass of dry air\n",
    "H = (Pw/(P-Pw))*(18.0153/28.9697)       # absolute humidity\n",
    "\n",
    "# Result\n",
    "print \"absolute humidity = \",H,\".\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###  Example 2.29   Page 45"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Outlet temperature is  335.758  K\n"
     ]
    }
   ],
   "source": [
    "# solution \n",
    "\n",
    "# Variables \n",
    "#Ti-Tf = mu*(Pi-Pf)\n",
    "Pi = 20.7                   #[bar]\n",
    "Pf = 8.7                    # [bar]\n",
    "mu = 1.616                  #[K/bar]\n",
    "Ti = 355.15                 #[K]\n",
    "\n",
    "# Calculation \n",
    "Tf = Ti-mu*(Pi-Pf)\n",
    "\n",
    "# Result\n",
    "print \"Outlet temperature is \",Tf,\" K\"\n"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 2",
   "language": "python",
   "name": "python2"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 2
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython2",
   "version": "2.7.6"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 0
}