From fba055ce5aa0955e22bac2413c33493b10ae6532 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Tue, 5 May 2015 14:21:39 +0530 Subject: add books --- Physical_Chemsitry/Chapter14.ipynb | 603 ------------------------------------- 1 file changed, 603 deletions(-) delete mode 100755 Physical_Chemsitry/Chapter14.ipynb (limited to 'Physical_Chemsitry/Chapter14.ipynb') diff --git a/Physical_Chemsitry/Chapter14.ipynb b/Physical_Chemsitry/Chapter14.ipynb deleted file mode 100755 index b1948488..00000000 --- a/Physical_Chemsitry/Chapter14.ipynb +++ /dev/null @@ -1,603 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:f51aa40f01063be99829dfd06c53ed90029beb6946cceb4b92469b9c50e91012" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 14 - Development and use of activity concepts" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1 - pg 350" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the moles of Iodine present\n", - "#Initialization of variables\n", - "x1=0.0200\n", - "Kx=812.\n", - "#calculations\n", - "print \"Neglecting 2x in comparision with x1,\"\n", - "x=x1/Kx\n", - "#results\n", - "print '%s %.2e %s' %(\"Moles of Iodine present =\",x,\" mole\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Neglecting 2x in comparision with x1,\n", - "Moles of Iodine present = 2.46e-05 mole\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2 - pg 350 " - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the Concentration of H+ ions\n", - "#Initialization of variables\n", - "Kc=1.749*10**-5 #M\n", - "n1=0.1 #mole\n", - "n2=0.01 #mole\n", - "#calculations\n", - "c=n1/n2 *Kc\n", - "#results\n", - "print '%s %.1e %s' %(\"Concentration of Hplus ions =\",c,\" M\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Concentration of Hplus ions = 1.7e-04 M\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3 - pg 351" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the Concentraton of Hplus ions\n", - "#Initialization of variables\n", - "import math\n", - "c=0.01 #M\n", - "kc=1.749*10**-5 #M\n", - "#calculations\n", - "x2=c*kc\n", - "x=math.sqrt(x2)\n", - "#results\n", - "print '%s %.1e %s' %(\"Concentration of Hplus ions =\",x,\"M\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Concentration of Hplus ions = 4.2e-04 M\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4 - pg 351" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the Concentration of OH- ions\n", - "#Initialization of variables\n", - "import math\n", - "K2=1.0008*10**-14 #m^2\n", - "K1=1.754*10**-5 #m\n", - "c=0.1\n", - "#calculations\n", - "print \"Neglecting x w.r.t c,\"\n", - "x2=c*K2/K1\n", - "x=math.sqrt(x2)\n", - "#results\n", - "print '%s %.1e %s' %(\"Concentration of OH minus ions =\",x,\" m\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Neglecting x w.r.t c,\n", - "Concentration of OH minus ions = 7.6e-06 m\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5 - pg 352" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the Concentration of H plus ions\n", - "#Initialization of variables\n", - "import math\n", - "print \"from table 14.1,\"\n", - "r1=7.47*10**-5 #m\n", - "r2=4.57*10**-3 #m\n", - "mp=1.008*10**-14 #m**2\n", - "#calculations\n", - "r3=r2/r1\n", - "mH2=r3*mp\n", - "mH=math.sqrt(mH2)\n", - "#results\n", - "print '%s %.2e %s' %(\"Concentration of Hplus ions = \",mH,\" M\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "from table 14.1,\n", - "Concentration of Hplus ions = 7.85e-07 M\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6 - pg 354" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the Concentraton of H+ ions\n", - "#Initialization of variables\n", - "print \"from table 14.1,\"\n", - "import math\n", - "r1=1.75*10**-5 #m\n", - "r2=1.772*10**-4 #m\n", - "mp=1.008*10**-14 #m**2\n", - "#calculations\n", - "r3=r2/r1\n", - "mH2=r3*mp\n", - "mH=math.sqrt(mH2)\n", - "#results\n", - "print '%s %.1e %s' %(\"Concentration of Hplus ions =\",mH,\" M\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "from table 14.1,\n", - "Concentration of Hplus ions = 3.2e-07 M\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7 - pg 355" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the Concentration of H+ ions\n", - "#Initialization of variables\n", - "import math\n", - "c=1*10**-6 #m\n", - "K=1.754*10**-5 #m\n", - "Kp=1.008*10**-14 #m**2\n", - "#calculations\n", - "mH=c\n", - "#Iteration 1\n", - "mOH=Kp/mH\n", - "mA=mH-mOH\n", - "mHA=mH*mA/K\n", - "mH2=mH-mHA+mOH\n", - "#Iteration 2\n", - "mOH2=Kp/mH2\n", - "mA2=mH2-mOH2\n", - "mHA2=mH2*mA2/K\n", - "mH3=mH2-mHA2+mOH2\n", - "#From x2\n", - "x2=math.sqrt(Kp)\n", - "x1=c\n", - "mOH3=Kp/x2\n", - "y2=x1\n", - "#From x1\n", - "mOH4=Kp/c\n", - "mA4=mH-mOH4\n", - "mHA4=mH*mA4/K\n", - "y1=c-mHA4-mA4\n", - "#upon further iterations, we get\n", - "mHplus=mH3\n", - "#results\n", - "print '%s %.2e %s' %(\"Concentration of H plus ions =\",mHplus,\"m\")\n", - "print 'The answer is a bit different due to rounding off error.'\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Concentration of H plus ions = 9.13e-07 m\n", - "The answer is a bit different due to rounding off error.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8 - pg 358" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the values of dS0, dH0, Krm\n", - "#Initialization of variableH\n", - "print \"From table 14-3,\"\n", - "HH=0\n", - "HHcoo=-98.\n", - "HHcooh=-98.\n", - "SH=0\n", - "SHcoo=21.9\n", - "SHcooh=39.1\n", - "KH=0\n", - "KHcoo=58.64\n", - "KHcooh=62.38\n", - "#calculationH\n", - "dH=HH+HHcoo-HHcooh\n", - "dS=SH+SHcoo-SHcooh\n", - "dK=KH+KHcoo-KHcooh\n", - "K=10**dK\n", - "#results\n", - "print '%s %.1f %s' %(\" dS0 =\",dS,\"eu\")\n", - "print '%s %.1f %s' %(\"\\n dH0 =\",dH,\"kcal\")\n", - "print '%s %.2f' %(\"\\n log Krm =\",dK)\n", - "print '%s %.1e %s' %(\"\\n Krm =\",K,\"m\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "From table 14-3,\n", - " dS0 = -17.2 eu\n", - "\n", - " dH0 = 0.0 kcal\n", - "\n", - " log Krm = -3.74\n", - "\n", - " Krm = 1.8e-04 m\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 9 - pg 369" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the Activity of cl and ca\n", - "#Initialization of variables\n", - "mca=0.01 #m\n", - "mcl=0.02 #m\n", - "#calculations\n", - "Mu=0.5*(mca*4 + mcl*1)\n", - "print \"From table 14-5,\"\n", - "aca=6 #A\n", - "acl=3 #A\n", - "print \"From table 14-6,\"\n", - "gaca=0.555 \n", - "gacl=0.843\n", - "Aca=gaca*mca\n", - "Acl=gacl*mcl\n", - "#results\n", - "print '%s %.4f' %(\"Activity of cl = \",Acl)\n", - "print '%s %.4f' %(\"\\n Activity of ca = \",Aca)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "From table 14-5,\n", - "From table 14-6,\n", - "Activity of cl = 0.0169\n", - "\n", - " Activity of ca = 0.0056\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 10 - pg 369" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the Concentration of H+ ions\n", - "#Initialization of variables\n", - "import math\n", - "m1=0.1 #m\n", - "m2=0.1 #m\n", - "K=1.754*10**-5 #m\n", - "#calculations\n", - "mu=0.5*(m1*1**2 + m2*1**2)\n", - "print(\"From table 14.5,\")\n", - "aH=9 #A\n", - "aA=4.5 #A\n", - "print(\"From table 14.6\")\n", - "gH=0.825\n", - "gA=0.775\n", - "gHA=1\n", - "x1=gHA*K/(gH*gA)\n", - "print(\"Assuming x to be small w.r.t m1,\")\n", - "x=math.sqrt(x1*m1)\n", - "#results\n", - "print '%s %.2e %s' %(\"Concentration of H plus ions =\",x,\" m\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "From table 14.5,\n", - "From table 14.6\n", - "Assuming x to be small w.r.t m1,\n", - "Concentration of H plus ions = 1.66e-03 m\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11 - pg 372" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the concentration of H+ ions\n", - "#Initialization of variables\n", - "import math\n", - "import numpy\n", - "K=1.754*10**-5 #m\n", - "c=0.1\n", - "#calculations\n", - "print(\"Neglecting x w.r.t c,\")\n", - "x2=K\n", - "x=math.sqrt(K)\n", - "mu=x\n", - "print(\"From tables 14-5 and 14-6,\")\n", - "gH=0.963\n", - "gA=0.960\n", - "x22=K/(gH*gA)\n", - "p=([1,x22, -c*x22])\n", - "z=numpy.roots(p)\n", - "alpha=z[1]\n", - "#results\n", - "print '%s %.2e %s' %(\"concentration of H plus ions =\",alpha,\" m\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Neglecting x w.r.t c,\n", - "From tables 14-5 and 14-6,\n", - "concentration of H plus ions = 1.37e-03 m\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12 - pg 373" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the Solubility of Agcl\n", - "#Initialization of variables\n", - "print(\"From table 14.3\")\n", - "import math\n", - "K1=-13.5089\n", - "K2=-22.9792\n", - "K3=19.2218\n", - "c=0.1 #m\n", - "#calculations\n", - "logK=K1-K2-K3\n", - "K=10**logK\n", - "mu=0.5*(c*1**2 + c*1**2)\n", - "print(\"From tables 14-5 and 14-6,\")\n", - "gAg=0.745\n", - "gCl=0.755\n", - "x2=K/(gAg*gCl)\n", - "x=math.sqrt(x2)\n", - "#results\n", - "print '%s %.2e %s' %(\"Solubility of Agcl =\",x,\"m\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "From table 14.3\n", - "From tables 14-5 and 14-6,\n", - "Solubility of Agcl = 1.78e-05 m\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 13 - pg 376" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the Concentration of Na and Cl in both cases\n", - "#Initialization of variables\n", - "import numpy\n", - "Cna=0.11\n", - "Ccl=0.1\n", - "#calculations\n", - "p=([99, - 2.1, Cna*Ccl])\n", - "z=numpy.roots(p)\n", - "alpha=z[1]\n", - "Na1=Cna-10*alpha\n", - "Cl1=Ccl-10*alpha\n", - "#results\n", - "print '%s %.4f %s' %(\" Concentration of Na in 1 =\",Na1,\"M\")\n", - "print '%s %.4f %s' %(\"\\n Concentration of Cl in 1 =\",Cl1,\" M\")\n", - "print '%s %.4f %s' %(\"\\n Concentration of Na in 2 =\",alpha,\"M\")\n", - "print '%s %.4f %s' %(\"\\n Concentration of Cl in 2 =\",alpha,\"M\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Concentration of Na in 1 = 0.0157 M\n", - "\n", - " Concentration of Cl in 1 = 0.0057 M\n", - "\n", - " Concentration of Na in 2 = 0.0094 M\n", - "\n", - " Concentration of Cl in 2 = 0.0094 M\n" - ] - } - ], - "prompt_number": 14 - } - ], - "metadata": {} - } - ] -} \ No newline at end of file -- cgit