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-rwxr-xr-xTextbook_of_Engineering_Chemistry/README.txt10
-rwxr-xr-xTextbook_of_Engineering_Chemistry/dChapter_10.ipynb70
-rwxr-xr-xTextbook_of_Engineering_Chemistry/dChapter_12.ipynb58
-rwxr-xr-xTextbook_of_Engineering_Chemistry/dChapter_2.ipynb1402
-rwxr-xr-xTextbook_of_Engineering_Chemistry/dChapter_3.ipynb1100
-rwxr-xr-xTextbook_of_Engineering_Chemistry/dChapter_6.ipynb92
-rwxr-xr-xTextbook_of_Engineering_Chemistry/dChapter_7.ipynb850
-rwxr-xr-xTextbook_of_Engineering_Chemistry/dChapter_8.ipynb979
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diff --git a/Textbook_of_Engineering_Chemistry/README.txt b/Textbook_of_Engineering_Chemistry/README.txt
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--- /dev/null
+++ b/Textbook_of_Engineering_Chemistry/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Deepak Shakya
+Course: btech
+College/Institute/Organization: DCRUST
+Department/Designation: Chemical Engg
+Book Title: Textbook of Engineering Chemistry
+Author: R. N. Goyal And H. Goel
+Publisher: Ane Books Pvt. Ltd., New Delhi
+Year of publication: 2009
+Isbn: 9788180520631
+Edition: 1 \ No newline at end of file
diff --git a/Textbook_of_Engineering_Chemistry/dChapter_10.ipynb b/Textbook_of_Engineering_Chemistry/dChapter_10.ipynb
new file mode 100755
index 00000000..86454a4e
--- /dev/null
+++ b/Textbook_of_Engineering_Chemistry/dChapter_10.ipynb
@@ -0,0 +1,70 @@
+{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10:Polymer Chemistry"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:1,Page no:272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#Variable declaration\n",
+ "N1=5.0 #no of molecules#\n",
+ "N2=10.0 \n",
+ "N3=20.0 \n",
+ "N4=20.0 \n",
+ "N5=10.0 \n",
+ "M1=5000.0 #molecular mass of each molecule#\n",
+ "M2=6000.0 \n",
+ "M3=10000.0 \n",
+ "M4=15000.0 \n",
+ "M5=25000.0 \n",
+ "\n",
+ "#Calculation\n",
+ "M=(M1*N1+M2*N2+M3*N3+M4*N4+M5*N5)/(N1+N2+N3+N4+N5) #formula for number average molecular mass#\n",
+ "Mw=(N1*M1**2+N2*M2**2+N3*M3**2+N4*M4**2+N5*M5**2)/(M1*N1+M2*N2+M3*N3+M4*N4+M5*N5) #formula of weight-average molecular mass#\n",
+ "\n",
+ "#Result\n",
+ "print\"The number average molecular mass is %.2e\"%M\n",
+ "print\"\\nThe weight average molecular mass is %.2e\"%Mw"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number average molecular mass is 1.28e+04\n",
+ "\n",
+ "The weight average molecular mass is 1.59e+04\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Textbook_of_Engineering_Chemistry/dChapter_12.ipynb b/Textbook_of_Engineering_Chemistry/dChapter_12.ipynb
new file mode 100755
index 00000000..7d0296aa
--- /dev/null
+++ b/Textbook_of_Engineering_Chemistry/dChapter_12.ipynb
@@ -0,0 +1,58 @@
+{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12:Coordination Chemistry"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:1,Page no:310"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "atomic_no=27 #Atomic on of Co\n",
+ "oxdn_state=3 #Oxidation state of Co\n",
+ "ele_donated=6*27 #Electron donated by ligand\n",
+ "\n",
+ "#Calculation\n",
+ "EAN=atomic_no-oxdn_state+ele_donated #Effective atomic no\n",
+ "\n",
+ "#Result\n",
+ "print\"Effective Atomic no of [Co(NH3)6]Cl3 is \",EAN,\"which is equal to atomic no of Kr\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Effective Atomic no of [Co(NH3)6]Cl3 is 186 which is equal to atomic no of Kr\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Textbook_of_Engineering_Chemistry/dChapter_2.ipynb b/Textbook_of_Engineering_Chemistry/dChapter_2.ipynb
new file mode 100755
index 00000000..9db18147
--- /dev/null
+++ b/Textbook_of_Engineering_Chemistry/dChapter_2.ipynb
@@ -0,0 +1,1402 @@
+{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:Acids and Bases"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:1,Page no:38"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "OH=0.0025 #OH- concentration#\n",
+ "K=1*10**-14#water ionization constant#\n",
+ "\n",
+ "#Calculation\n",
+ "H=K/OH \n",
+ "H=H/10**-12 \n",
+ "\n",
+ "#Result\n",
+ "print\"The concentration of H+ ions is\",H*10**-12,\"M\" \n",
+ " \n",
+ "print\"\\nAs concentration of H+ is lesser than the concentration of OH-(0.0025) the cleaning solution will be basic in nature\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The concentration of H+ ions is 4e-12 M\n",
+ "\n",
+ "As concentration of H+ is lesser than the concentration of OH-(0.0025) the cleaning solution will be basic in nature\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2,Page no:40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "pH=7.3 #pH value of human blood\n",
+ "H=10**-pH \n",
+ "\n",
+ "#Calculation\n",
+ "H1=H\n",
+ "k=1*10**-14 #water ionization constant\n",
+ "OH=k/H \n",
+ "OH=OH\n",
+ "\n",
+ "#Result\n",
+ "print\"H+ concentration of human blood is %.e\"%H1,\"M\" \n",
+ "print\"\\nOH- concentration of human blood is %.3g\"%OH,\"M(in scientiifc form) or 0.2*10**-6 M\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "H+ concentration of human blood is 5e-08 M\n",
+ "\n",
+ "OH- concentration of human blood is 2e-07 M(in scientiifc form) or 0.2*10**-6 M\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3,Page no:41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Variable declaration\n",
+ "N1=0.2 #normality of HCl#\n",
+ "V1=25 #volume of HCl in ml#\n",
+ "M2=0.25 #molarity of NaOH#\n",
+ "N2=M2*1 #normality of NaOH#\n",
+ "V2=50 #volume of NaOH in ml#\n",
+ "\n",
+ "#Calculation\n",
+ "V=V1+V2 #volume of resulting solution#\n",
+ "N=(N2*V2-N1*V1)/V #normality of resulting solution#\n",
+ "\n",
+ "K=1*10**-14 #ionization constant of water#\n",
+ "H=K/N \n",
+ "H1=H/10**-13 \n",
+ "\n",
+ "pH=-math.log10(H) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print\"\\npH of the mixure will be\",pH"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "pH of the mixure will be 13.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:4,Page no:44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "S=0.2 #salt concentration#\n",
+ "A=0.2 #acid concentration#\n",
+ "k=1.8*10**-5 #dissociation constant of acetic acid#\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "pH=-math.log10(k)+math.log10(S/A) \n",
+ "v=1*10**-3 #amount of HCl added in lit#\n",
+ "M=1 #molarity of HCl added#\n",
+ "n=v*M #no of moles of HCl added per litre#\n",
+ "A1=A+n \n",
+ "S1=S-n \n",
+ "pH2=-math.log10(k)+math.log10(S1/A1) \n",
+ "p=pH-pH2 \n",
+ "\n",
+ "#Result\n",
+ "print\"pH of the buffer solution before adding HCl is\",round(pH ,4)\n",
+ "\n",
+ "print\"\\npH of the buffer solution after adding HCl is\",round(pH2,3)\n",
+ "print\"\\nAns: Change in pH is\",round(p,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "pH of the buffer solution before adding HCl is 4.7447\n",
+ "\n",
+ "pH of the buffer solution after adding HCl is 4.74\n",
+ "\n",
+ "Ans: Change in pH is 0.004\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.1,Page no:46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#(a)#\n",
+ "#Variable declaration\n",
+ "N1=1.0/1000.0 #normality of HCl#\n",
+ "a=100.0 #percentage of ionization#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "C1=N1*a/100.0 \n",
+ "pH1=-math.log10(C1) \n",
+ "N2=1.0/10000.0 #normality of NaOH solution#\n",
+ "C2=N2*a/10.0 \n",
+ "C2a=C2/10.0**-4 \n",
+ "k=10.0**-14 #dissociation constant of water#\n",
+ "H2=k/C2 \n",
+ "H2a=H2/10.0**-10 \n",
+ "pH2=-math.log10(H2) \n",
+ "N3=1.0/1000.0 #normality of NaOH solution#\n",
+ "C3=N3*a/1000.0 \n",
+ "C3a=C3/10.0**-3 \n",
+ "H3=k/C3 \n",
+ "H3a=H3/10.0**-11 \n",
+ "pH3=-math.log10(H3) \n",
+ "\n",
+ "#Result\n",
+ "print\"Ans(a)\\n(i)\\tThe pH of N/1000 HCl solution is\",pH1 \n",
+ "print\"\\n(ii)\\tThe pH of the N/10000 solution is\",pH2 \n",
+ "print\"\\n(iii)\\tThe pH of the N/1000 solution is\",pH3\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "#(b)#\n",
+ "#Variable declaration\n",
+ "N=0.1 #normality of given weak base#\n",
+ "pH=9.0 #pH of the base#\n",
+ "H=10.0**(-pH) \n",
+ "Ha=H/10.0**-9\n",
+ "\n",
+ "#Calculation\n",
+ "OH=k/H \n",
+ "OHa=OH/10.0**-5 \n",
+ "a1=OH/N \n",
+ "a1b=a1\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nAns(b)\\nDegree of ionization of given weak base is\",a1b,\"=\",a1b*100,\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ans(a)\n",
+ "(i)\tThe pH of N/1000 HCl solution is 3.0\n",
+ "\n",
+ "(ii)\tThe pH of the N/10000 solution is 11.0\n",
+ "\n",
+ "(iii)\tThe pH of the N/1000 solution is 10.0\n",
+ "\n",
+ "Ans(b)\n",
+ "Degree of ionization of given weak base is 0.0001 = 0.01 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.2,Page no:47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N=0.002 #normality of acetic acid solution#\n",
+ "a=2.3 #percentage of ionization#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "H=N*a/100.0 #concentration of H+ ion#\n",
+ "pH=-math.log10(H) \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nAns(a)\\n pH value of acid solution is\",round(pH,4)\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Part b(b)#\n",
+ "\n",
+ "#Variable declaration\n",
+ "N1=0.01 #normality of acetic acid solution#\n",
+ "a1=60.0 #percentage of ionization#\n",
+ "#ii#\n",
+ "N2=0.1 #normality of acetic acid solution#\n",
+ "a2=1.8 #percentage of ionization#\n",
+ "#iii#\n",
+ "N3=0.04 #normality of HNO3#\n",
+ "a3=100.0 #percentage of ionization#\n",
+ "#iv#\n",
+ "W=4.0 #weight of NaOH dissolved in water in grams#\n",
+ "EW=40.0 #equivalent weight weight of NaOH#\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "#i#\n",
+ "H1=N1*a1/100.0 #concentration of H+ ion#\n",
+ "pH1=-math.log10(H1) \n",
+ "#ii#\n",
+ "H2=N2*a2/100.0 #concentration of H+ ion#\n",
+ "pH2=-math.log10(H2) \n",
+ "#iii#\n",
+ "H3=N3*a3/100.0 \n",
+ "pH3=-math.log10(H3) \n",
+ "N4=0.0001 #normality of Hcl#\n",
+ "a4=100.0 #percentage of ionization#\n",
+ "H4=N4*a4/100.0 \n",
+ "pH4=-math.log10(H4) \n",
+ "N5=1.0 #normality of Hcl#\n",
+ "a5=100.0 #percentage of ionization#\n",
+ "H5=N5*a5/100.0 \n",
+ "pH5=-math.log10(H5) \n",
+ "N6=0.1 #normality of HNO3#\n",
+ "a6=100.0 #percentage of ionization#\n",
+ "OH6=N6*a6/100.0 \n",
+ "Kw=10.0**-14 \n",
+ "H6=Kw/OH6 \n",
+ "pH6=-math.log10(H6)\n",
+ "N7=0.001 #normality of NaOH#\n",
+ "a7=100.0 #percentage of ionization#\n",
+ "OH7=N7*a7/100.0 \n",
+ "Kw=10.0**-14 \n",
+ "H7=Kw/OH7 \n",
+ "pH7=-math.log10(H7) \n",
+ "#iv#\n",
+ "N8=W/EW \n",
+ "a8=100.0 #percentage of ionization#\n",
+ "OH8=N8*a8/100.0 \n",
+ "Kw=10.0**-14 \n",
+ "H8=Kw/OH8 \n",
+ "pH8=-math.log10(H8) \n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print'\\nAns(b)\\n(i) pH value of 0.01N acid solution is',round(pH1,4)\n",
+ "\n",
+ "print\"\\n(ii) pH value of decinormal acid solution is\",round(pH2 ,4)\n",
+ "\n",
+ "print\"\\n(iii) The pH of 0.04N HNO3 solution is\",round(pH3,3)\n",
+ "print\"\\n The pH of 0.0001N Hcl solution is\",pH4 \n",
+ "print\"\\n The pH of 1N Hcl solution is\",pH5 \n",
+ "print\"\\n The pH of 0.1N NaOH solution is \",pH6\n",
+ "print\"\\n The pH of 0.01N NaOH solution is\",pH7\n",
+ "\n",
+ "print\"\\n(iv) The pH of solution containing 4g NaoH solution is \",pH8 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Ans(a)\n",
+ " pH value of acid solution is 4.3372\n",
+ "\n",
+ "Ans(b)\n",
+ "(i) pH value of 0.01N acid solution is 2.2218\n",
+ "\n",
+ "(ii) pH value of decinormal acid solution is 2.7447\n",
+ "\n",
+ "(iii) The pH of 0.04N HNO3 solution is 1.398\n",
+ "\n",
+ " The pH of 0.0001N Hcl solution is 4.0\n",
+ "\n",
+ " The pH of 1N Hcl solution is -0.0\n",
+ "\n",
+ " The pH of 0.1N NaOH solution is 13.0\n",
+ "\n",
+ " The pH of 0.01N NaOH solution is 11.0\n",
+ "\n",
+ "(iv) The pH of solution containing 4g NaoH solution is 13.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.3,Page no:48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "N1=0.1 #normality of acetic acid#\n",
+ "a1=1.3 #percentage of ionization#\n",
+ "M1=10**-8 #molarity of hcl solution#\n",
+ "a=100 #percentage of ionization#\n",
+ "N2=0.05 #normality of Hcl#\n",
+ "a2=100 #percentage of ionization#\n",
+ "\n",
+ "#Calculation\n",
+ "#(a)#\n",
+ "H1=N1*a1/100 \n",
+ "#(b)#\n",
+ "H=M1*a/100 \n",
+ "pH=-math.log10(H) \n",
+ "#(c)#\n",
+ "pH2=-math.log10(N2*a2/100) \n",
+ "M3=0.05 #molarity os H2SO4#\n",
+ "a3=100 #percentage of ionization#\n",
+ "pH3=-math.log10(M3*a3/100.0) \n",
+ "\n",
+ "#Result\n",
+ "print\"(a).The hydrogen ion concentration of solution is %.2e\"%H1,\"g.ion/lit\"\n",
+ "print'\\n(b).The pH of the Hcl solution is',pH\n",
+ "print\"Theoretically the pH should be 8,however,the value will be close to 7 because H+ ions of water also plays a role\"\n",
+ "print\"\\n(c).The pH of 0.05 Hcl solution is\",round(pH2,3)\n",
+ "print\"The pH of 0.05M H2SO4 solution is\",round(pH3,3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a).The hydrogen ion concentration of solution is 1.30e-03 g.ion/lit\n",
+ "\n",
+ "(b).The pH of the Hcl solution is 8.0\n",
+ "Theoretically the pH should be 8,however,the value will be close to 7 because H+ ions of water also plays a role\n",
+ "\n",
+ "(c).The pH of 0.05 Hcl solution is 1.301\n",
+ "The pH of 0.05M H2SO4 solution is 1.301\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.4,Page no:49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "H1=0.005 #H+ ion concentration of solution in g.ion/lit#\n",
+ "H2=3*10**-4 #H+ concentration of the solution#\n",
+ "k=10**-14 #dissociation constant of water#\n",
+ "OH3=0.1#hydroxyl concentration of a solution#\n",
+ "k4=1.8*10**-5#dissociation constant of acetic acid at 180C#\n",
+ "N4=0.1 #normality of acetic acid#\n",
+ "N5=0.01 #normality of acetic acid#\n",
+ "N6=0.001 #normality of acetic acid#\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "#Part-a#\n",
+ "pH1=-math.log10(H1) \n",
+ "\n",
+ "#Part-b#\n",
+ "pH2=-math.log10(H2) \n",
+ "pOH2=14-pH2 \n",
+ "OH2=k/H2\n",
+ "\n",
+ "#Part-c#\n",
+ "H3=k/OH3 \n",
+ "pH3=-math.log10(H3) \n",
+ "V4=1/N4 \n",
+ "\n",
+ "#Part-d#\n",
+ "a4=math.sqrt(k4*V4) #formula for degree of dissociation#\n",
+ "H4=N4*a4 #H+ ion concentration#\n",
+ "pH4=-math.log10(H4) \n",
+ "V5=1/N5 \n",
+ "a5=sqrt(k4*V5) #formula for degree of dissociation#\n",
+ "H5=N5*a5 #H+ ion concentration#\n",
+ "pH5=-math.log10(H5) \n",
+ "V6=1/N6 \n",
+ "a6=sqrt(k4*V6) #formula for degree of dissociation#\n",
+ "H6=N6*a6 #H+ ion concentration#\n",
+ "pH6=-math.log10(H6) \n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"\\n\\n(a) The pH value of solution whose H+ ion concentration is 0.005g.ion/lit is\",round(pH1 ,3)\n",
+ "print\"\\n(b) The pH of a solution in which H+ is 3*10**-4 is\",round(pH2 ,2)\n",
+ "print\"\\n pOH of the solution is\",round(pOH2,2)\n",
+ "print\"\\n OH- concentration for a solution is%.1e\"%OH2,\"M\"\n",
+ "print\"\\n(c) pH of the solution whose hydroxyl concentration is N/10g.ion/lit is\",pH3\n",
+ "print\"\\n(d) pH of 0.1N acetic acid solution is\",round(pH4,3)\n",
+ "print\"\\n pH of 0.01N acetic acid solution is\",round(pH5,4)\n",
+ "print\"\\n pH of 0.001N acetic acid solution is\",round(pH6,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ "(a) The pH value of solution whose H+ ion concentration is 0.005g.ion/lit is 2.301\n",
+ "\n",
+ "(b) The pH of a solution in which H+ is 3*10**-4 is 3.52\n",
+ "\n",
+ " pOH of the solution is 10.48\n",
+ "\n",
+ " OH- concentration for a solution is3.3e-11 M\n",
+ "\n",
+ "(c) pH of the solution whose hydroxyl concentration is N/10g.ion/lit is 13.0\n",
+ "\n",
+ "(d) pH of 0.1N acetic acid solution is 2.872\n",
+ "\n",
+ " pH of 0.01N acetic acid solution is 3.3724\n",
+ "\n",
+ " pH of 0.001N acetic acid solution is 3.8724\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.5,Page no:51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "K1=10.0**-8 #dissociation constant of weak mono basic acid#\n",
+ "N1=0.01 #normality of the acid#\n",
+ "a2=4.0/100.0 #percentage of dissociation of acid at 20C#\n",
+ "N2=0.1 #normality of acid#\n",
+ "N3=0.1 #normality of HCl#\n",
+ "N4=1.0/50.0 #normality of HCl#\n",
+ "N5=0.01 #normality of H2SO4#\n",
+ "\n",
+ "#Calculation\n",
+ "#Part-a#\n",
+ "V1=1.0/N1 \n",
+ "a1=math.sqrt(K1*V1) #degree of dissociation for weak acids#\n",
+ "H1=N1*a1 #H+ concentration of the solution#\n",
+ "pH1=-math.log10(H1) \n",
+ "#Part-b#\n",
+ "V2=1.0/N2 \n",
+ "K2=(a2**2)/V2 \n",
+ "#Part-c#\n",
+ "pH3=-math.log10(N3) \n",
+ "pH4=-math.log10(N4) \n",
+ "pH5=-math.log10(N5) \n",
+ "\n",
+ "#Result\n",
+ "print\"(a) pH value of 0.01N solution of a weak mono basic acid is\",pH1 \n",
+ "print\"\\n(b) The dissociation constant of the acid is %.1e\"%K2\n",
+ "print\"\\n(c) The pH of the 0.1N HCl solution is\",pH3\n",
+ "print\"\\n The pH of the 1/50N HCl solution is\",round(pH4,1)\n",
+ "print\"\\n The pH of the 0.01N H2SO4 solution is \",pH5"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) pH value of 0.01N solution of a weak mono basic acid is 5.0\n",
+ "\n",
+ "(b) The dissociation constant of the acid is 1.6e-04\n",
+ "\n",
+ "(c) The pH of the 0.1N HCl solution is 1.0\n",
+ "\n",
+ " The pH of the 1/50N HCl solution is 1.7\n",
+ "\n",
+ " The pH of the 0.01N H2SO4 solution is 2.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.6,Page no:52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V1=50.0 #volume of Hcl in ml#\n",
+ "V2=30.0 #volume of NaOH in ml#\n",
+ "N1=1.0 #normality of Hcl#\n",
+ "N2=1.0 #nomality of NaOH#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "V=V1+V2 #total volume of mixure of solutions#\n",
+ "a=100.0 #percentage of ionization#\n",
+ "N=(N1*V1-N2*V2)/V \n",
+ "H=N*a/100 \n",
+ "pH=-math.log10(H) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print'\\nThe pH of resultant solution is',round(pH,3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The pH of resultant solution is 0.602\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.7,Page no:52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N1=1.0/10.0 #normality of NaOH#\n",
+ "N2=1.0/20.0 #normality of HCl#\n",
+ "V1=1.0 #volume of NaOH in lit#\n",
+ "V2=1.0 #volume of HCl in lit#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "V=V1+V2 #volume of resultant solution#\n",
+ "N=(N1*V1-N2*V2)/V \n",
+ "k=1.0*10.0**-14 #ionization constant of water#\n",
+ "H1=k/N \n",
+ "H=H1/10.0**-13 \n",
+ "pH=-math.log10(H1)\n",
+ "\n",
+ "#Result\n",
+ "print\"\\npH of the solution is\",round(pH,1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "pH of the solution is 12.4\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.8,Page no:53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W=2.0 #weight of NaOH dissolved in water in grams#\n",
+ "M=40.0 #molecular weight of NaOH#\n",
+ "N=W/M #normality#\n",
+ "a=100.0 #percentage of ionization#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "OH=N*a/100.0 #the OH- ion concentration of solution#\n",
+ "Kw=10.0**-14 \n",
+ "H=Kw/OH \n",
+ "pH=-math.log10(H) \n",
+ "\n",
+ "#Result\n",
+ "print'\\n The pH of the NaOH solution is',round(pH,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The pH of the NaOH solution is 12.7\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.9,Page no:53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "M=0.001 #molarity of benzoic acid#\n",
+ "N=M #normality of benzoic acid#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "V=1/N \n",
+ "K=7.3*10**-5 #dissociation constant of benzoic acid#\n",
+ "a=math.sqrt(K*V) #since benzoic acid is very weak#\n",
+ "\n",
+ "#Result\n",
+ "H=N*a \n",
+ "print\"\\n The H+ concentration of the solution is%.3e\"%H,\"g.ion/litre\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The H+ concentration of the solution is2.702e-04 g.ion/litre\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.10,Page no:53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W=0.092 #weight of Formic acid per litre in grams#\n",
+ "M=46 #molecular weight of Formic acid#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "N=W/M \n",
+ "V=1/N \n",
+ "K=2.4*10**-4 #Dissociation constant of Formic acid at 25C#\n",
+ "a=math.sqrt(K*V) #For weak acids#\n",
+ "\n",
+ "#Result\n",
+ "H=a*N \n",
+ "print'\\n The H+ concentration of the solution is %.3e'%H"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The H+ concentration of the solution is 6.928e-04\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.11,Page no:54"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "k=2.5*10**-5 #dissociation constant of NH4OH#\n",
+ "N=1.0/100.0 #normality of NH4OH#\n",
+ "V=100\n",
+ "#Calculation\n",
+ "C=N #since volume of solution is one litre#\n",
+ "NH=C \n",
+ "NHOH=C \n",
+ "OH1=k*NHOH/NH \n",
+ "a=math.sqrt(k*V)\n",
+ "\n",
+ "#Result\n",
+ "print\"Cong of OH- ions in solution is\",a*N,\"g ion per litre\"\n",
+ "print\"\\nWhen 1/100 part of a g mol NH4Cl are dissolverd in a litre ,then ,\\nHydroxyl ion concentration in the solution is\",OH1\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Cong of OH- ions in solution is 0.0005 g ion per litre\n",
+ "\n",
+ "When 1/100 part of a g mol NH4Cl are dissolverd in a litre ,then ,\n",
+ "Hydroxyl ion concentration in the solution is 2.5e-05\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.12,Page no:56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "K=1.7*10**-5 #Dissociation constant of NH4OH#\n",
+ "N=0.01 #Normality of NH4OH solution#\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "V=1.0/N \n",
+ "a=math.sqrt(K*V) #since a is very small#\n",
+ "OH=a*N \n",
+ "\n",
+ "NH4=0.05 #concentration of NH4+ in g.ion/lit#\n",
+ "NH4OH=0.01 #concentration of NH4OH in g.mol/lit#\n",
+ "OH2=K*NH4OH/NH4 \n",
+ "#Result\n",
+ "print\"\\nConcentration of OH- ions before addition of NH4Cl is %.2e\"%OH,\"g.ion/litre\"\n",
+ "print\"\\nThe concentration of hydroxyl ions after adding NH4Cl is\",OH2,\"g.ion/litre\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Concentration of OH- ions before addition of NH4Cl is 4.12e-04 g.ion/litre\n",
+ "\n",
+ "The concentration of hydroxyl ions after adding NH4Cl is 3.4e-06 g.ion/litre\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.13,Page no:56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "k=1.8*10**-5 #dissociation constant of acetic acid at 18C#\n",
+ "N=0.25 #normality of acetic acid solution#\n",
+ "N2=0.25#normality os sodium acetate added#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "#Part-a#\n",
+ "V=1/N \n",
+ "a=math.sqrt(k*V) #formula of degree of dissociation for weak acids#\n",
+ "H=N*a \n",
+ "#Part-b#\n",
+ "CH3COO=N2 \n",
+ "CH3COOH=N2 \n",
+ "H2=k*CH3COOH/CH3COO \n",
+ "H3=H2/10**-5 \n",
+ "a2=H2/N2 \n",
+ "a3=a2/10**-5 \n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) 0.25 N acetic acid solution---\"\n",
+ "print\"\\tDegree of dissociation of acetic acid is %.3e\"%a\n",
+ "print\"\\tH+ concentration of the solution is %.3e\"%H,\"g.ion/litre\"\n",
+ "\n",
+ "print\"\\n(b) 0.25 N acetic acid solution containing 0.25N sodium acetate----\"\n",
+ "print\"\\tH+ ion concentration after adding sodium acetate is\",H3*10**-5\n",
+ "print\"\\tDegree of dissociation after adding sodium acetate is\",a3*10**-5"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) 0.25 N acetic acid solution---\n",
+ "\tDegree of dissociation of acetic acid is 8.485e-03\n",
+ "\tH+ concentration of the solution is 2.121e-03 g.ion/litre\n",
+ "\n",
+ "(b) 0.25 N acetic acid solution containing 0.25N sodium acetate----\n",
+ "\tH+ ion concentration after adding sodium acetate is 1.8e-05\n",
+ "\tDegree of dissociation after adding sodium acetate is 7.2e-05\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.14,Page no:57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C1=0.06 #concentration od acetic acid in g.mol/lit#\n",
+ "C2=0.04 #concentration of sodium acetate in g.mol/li#\n",
+ "K=1.8*10**-5 #dissociation constant of acetic acid#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "H=K*C1/C2 \n",
+ "pH=-math.log10(H) \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nThe pH of solution is\",round(pH,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The pH of solution is 4.5686\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.15,Page no:58"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "M1=0.2 #molarity of acetic acid#\n",
+ "M2=0.2 #molarity of sodium acetate#\n",
+ "K=1.8*10**-5 \n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "pH=-math.log10(K)+math.log10(M2/M1) #by using Henderson's equation#\n",
+ "\n",
+ "#Result\n",
+ "print\"The pH value of buffer solution is\",round(pH,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The pH value of buffer solution is 4.7447\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.16,Page no:58"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N=1.0/100.0 #normality of acetic acid#\n",
+ "V=1.0/N \n",
+ "k=1.8*10**-5 #dissociation constant of acetic acid#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "#Part-a#\n",
+ "a=math.sqrt(k*V) #formula of degree of dissociation for weak acids#\n",
+ "H=a*N \n",
+ "\n",
+ "#Part-b#\n",
+ "n=0.01 #sodium acetate added in moles to one litre of acetic acid solution#\n",
+ "CH3COO=n \n",
+ "CH3COOH=n \n",
+ "H1=k*CH3COOH/CH3COO \n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) H+ concentration of N/100 acetic acid solution is %.2e\"%H,\"g ion/litre\"\n",
+ "print\"\\n(b) H+ ion concentration in the solution after adding the sodium acetate is\",H1,\"g.ions/litre\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) H+ concentration of N/100 acetic acid solution is 4.24e-04 g ion/litre\n",
+ "\n",
+ "(b) H+ ion concentration in the solution after adding the sodium acetate is 1.8e-05 g.ions/litre\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.17,Page no:59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V=10 #volume of water in litres#\n",
+ "N1=0.10 #moles of HCN added in solution#\n",
+ "N2=0.10 #moles of NaCN added in solution#\n",
+ "K=7.2*10**-10 #dissociation constant of HCN#\n",
+ "CN=0.1 #CN- concentration#\n",
+ "HCN=0.1 #HCN concentration#\n",
+ "\n",
+ "#Calculation\n",
+ "H1=K*HCN/CN \n",
+ "H=H1/10**-10 \n",
+ "k=1*10**-14 #ionization constant of water#\n",
+ "\n",
+ "#Result\n",
+ "print\"H+ concentration in the solution is\",H*10**-10\n",
+ "OH=k/H1 \n",
+ "print\"\\nOH- concentration in the solution is %.1e\"%OH"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "H+ concentration in the solution is 7.2e-10\n",
+ "\n",
+ "OH- concentration in the solution is 1.4e-05\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.18,Page no:59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "K=1.7*10**-5 #dissociation constant of acid#\n",
+ "pH=3.77#pH value of buffer solution#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "M=pH+math.log10(K) \n",
+ "N=10**M #ratio of salt to acid#\n",
+ "L=1/N\n",
+ "\n",
+ "#Result\n",
+ "print\"The ratio of salt to acid in buffer is\",round(L)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of salt to acid in buffer is 10.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.19,Page no:59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "k=1.8*10**-5 #dissociation constant of acetic acid#\n",
+ "M=0.01 #molarity of acetic acid#\n",
+ "N=M*1 #normality of acetic acid#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "V=1/N \n",
+ "a=math.sqrt(k*V)#degree of dissociation for weak acids#\n",
+ "H1=a/V \n",
+ "H=H1/10**-4 \n",
+ "pH=-math.log10(H1) \n",
+ "\n",
+ "#Result\n",
+ "print\"Degree of dissociation of solution is %.2e\"%a \n",
+ "print\"pH of the solution is\",round(pH ,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Degree of dissociation of solution is 4.24e-02\n",
+ "pH of the solution is 3.3724\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.20,Page no:60"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N1=0.2#concentration of acetic acid in g.molecule/lit#\n",
+ "N2=0.25#concentration of sodium acetate in g.molecule/lit#\n",
+ "K=1.8*10**-5#ionization constant of acetic acid at room temparature#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "pH1=-math.log10(K)+math.log10(N2/N1) \n",
+ "N=1.0#normality of HCl added#\n",
+ "V=0.5*10**-3#amount of HCl added in lit#\n",
+ "M=N*V\n",
+ "C1=N1+M#concentration of CH3COOH in moles/lit#\n",
+ "C2=N2-M#concentration of CH3COONa in moles/lit#\n",
+ "pH2=-math.log10(K)+math.log10(C2/C1)\n",
+ "pH=pH1-pH2 \n",
+ "\n",
+ "#Result\n",
+ "print\"pH value of the solution before adding HCl is\",round(pH1,4)\n",
+ "print\"\\nThe pH of the solution after adding HCl is\",round(pH2,4)\n",
+ "print\"\\nThe change of pH is\",round(pH,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "pH value of the solution before adding HCl is 4.8416\n",
+ "\n",
+ "The pH of the solution after adding HCl is 4.8397\n",
+ "\n",
+ "The change of pH is 0.002\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.21,Page no:61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "K=18*10**-6 #dissociation constant of NH4OH#\n",
+ "N1=0.1 #normality of NH4OH solution#\n",
+ "V=1.0/N1 \n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "a=math.sqrt(K*V)#since a is very small#\n",
+ "OH=a/V \n",
+ "W=2.0#weight of added NH4Cl in grams#\n",
+ "M=53.0#molecular weight of NH4Cl#\n",
+ "C=W/M \n",
+ "C1=0.1 #concentration of NH4OH in g.mol/lit#\n",
+ "OH2=K*C1/C\n",
+ "CH3COO=0.02 #g ion per litre\n",
+ "CH3COOH=0.2 #g mol per litre\n",
+ "H_plus=K*CH3COOH/CH3COO\n",
+ "pH=math.log10(H_plus)\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nThe concentration of hydroxyl ion before adding of NH4Cl is %.3e\"%OH,\"g ion per litre\"\n",
+ "print\"\\nThe concentration of hydroxyl ion after adding 2g of NH4Cl is %.1e\"%OH2,\"g ion per litre\"\n",
+ "print\"NOTE:Calculation mistake in book.(wrongly written as 48.6*10**-5,it should be 10**-6\"\n",
+ "print \"\\npH is\",round(-pH,4)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The concentration of hydroxyl ion before adding of NH4Cl is 1.342e-03 g ion per litre\n",
+ "\n",
+ "The concentration of hydroxyl ion after adding 2g of NH4Cl is 4.8e-05 g ion per litre\n",
+ "NOTE:Calculation mistake in book.(wrongly written as 48.6*10**-5,it should be 10**-6\n",
+ "\n",
+ "pH is 3.7447\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.22,Page no:62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "ly=11.92 #equivalent conductvity of 0.02acetic acid solution in mho at 20C#\n",
+ "lih=360 #the equivalent ionic conductance of an infinite dillution of hydrogen ion in mho#\n",
+ "lic=40 #of acetate ion#\n",
+ "li=lih+lic #of acetic acid#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "a=ly/li #degree of dissociation#\n",
+ "N=0.02 #normality of acetic acid#\n",
+ "V=1/N \n",
+ "K=(a**2)/V \n",
+ "W=82 #mol.wt of CH3COONa#\n",
+ "M=8.2#amount of sodium acetate added in g per litre solution#\n",
+ "CH3COO=M/W \n",
+ "H=K*N/CH3COO \n",
+ "pH=-math.log10(H) \n",
+ "\n",
+ "#Result\n",
+ "print\"Dissociation constant of acetic acid is %.2e\"%K\n",
+ "print\"\\npH of the solution is\",round(pH,2)\n",
+ "print\"\\nNOTE:\\n(i)Calculation istake in calculation of K in book,exponent wrongly written as -6\"\n",
+ "print\"(ii)pH is wrongly calculated in book as 3.45\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dissociation constant of acetic acid is 1.78e-05\n",
+ "\n",
+ "pH of the solution is 5.45\n",
+ "\n",
+ "NOTE:\n",
+ "(i)Calculation istake in calculation of K in book,exponent wrongly written as -6\n",
+ "(ii)pH is wrongly calculated in book as 3.45\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Textbook_of_Engineering_Chemistry/dChapter_3.ipynb b/Textbook_of_Engineering_Chemistry/dChapter_3.ipynb
new file mode 100755
index 00000000..5406df23
--- /dev/null
+++ b/Textbook_of_Engineering_Chemistry/dChapter_3.ipynb
@@ -0,0 +1,1100 @@
+{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3:Chemical Kinetics & Catalysis"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:1,Page no:69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from scipy.optimize import fsolve\n",
+ "#Variable declaration\n",
+ "A0=0.25 #[M]\n",
+ "At=0.15 #[M]\n",
+ "k=6.7*10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "def f(t):\n",
+ " x=math.log10(A0/At)-(k*t)/2.303\n",
+ " return(x)\n",
+ "t=fsolve(f,1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Time taken to decrease concentration is%.3e\"%t[0],\"s (approx)\"\n",
+ "print\"NOTE: Approximate value taken in book\"\n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time taken to decrease concentration is7.626e+02 s (approx)\n",
+ "NOTE: Approximate value taken in book\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2,Page no:71"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "#Initially:\n",
+ "A=0.1 #Concentration of A\n",
+ "B=0.1 #Concentration of B\n",
+ "rate=5.5*10**-6 #Rate of reaction initial in M/s\n",
+ "x=2 #exponent\n",
+ "y=1 #exponent\n",
+ "\n",
+ "#Calculation\n",
+ "order=x+y #Reaction order\n",
+ "k=rate/((A**x)*(B**y)) #Rate law constant in [M**2/s]\n",
+ "\n",
+ "#Result\n",
+ "print\"The rate law is k=rate/(A**2)*(B)\"\n",
+ "print\"\\nRate constant is %.2e\"%k,\"M**2/s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rate law is k=rate/(A**2)*(B)\n",
+ "\n",
+ "Rate constant is 5.50e-03 M**2/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3,Page no:76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "\n",
+ "T=[700.0,730.0,760.0,790.0] #Temperature in [K]\n",
+ "k=[0.011,0.035,0.195,0.343] #rate constants in [L/mol s]\n",
+ "import numpy \n",
+ "onebyT=numpy.reciprocal(T) #Reciprocal of temperature\n",
+ "log_k=numpy.log10(k) #log of k\n",
+ "R=8.30*10**-3 #[kJ/kmol]\n",
+ "\n",
+ "#Calculation\n",
+ "%pylab inline\n",
+ "import matplotlib.pyplot as plt\n",
+ "plt.plot(onebyT,log_k)\n",
+ "plt.ylabel('$log k$')\n",
+ "plt.xlabel('$1/T$')\n",
+ "plt.title('1/T vs log k\\n')\n",
+ "slope,intercept=np.polyfit(onebyT,log_k,1)\n",
+ "print\"Slope is\",slope\n",
+ "plt.show()\n",
+ "slope=-9.9*10**3 #Slope given in book [K]\n",
+ "E_star=slope*(-2.303*R) #Activation energy in [kJ/mol]\n",
+ "\n",
+ "#Result\n",
+ "print\"Activation energy of reaction is\",round(E_star,1),\"kJ/mol\"\n",
+ "print\"NOTE that in book slope is approximated as 9.9*10**3 K\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Populating the interactive namespace from numpy and matplotlib\n",
+ "Slope is"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " -9663.82327366\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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o2iwsgJdfFntFnTgBeHkBP/wgd1VU3zj0RES1IknAtm1ikd6oUUBUFNCihdxV\nUV1w6ImI9EqlAsaMEVNp798XU2kTEuSuiuoDexREpBcHDoiFen37ioV6HTrIXRHVFHsURFSvAgKA\nU6cAOzvAwwPYsIEL9UwFexREpHfHjwNTp4rV3atXA126yF0RVQd7FERkMD4+wNGj4ryL3r3FUBQX\n6hkvRQRFbGws3N3dYWlpibS0tErb5ebmYvTo0ejevTvc3NyQwg30iRTLygpYuFBMn42PBwYMEENT\nZHwUERQeHh6Ii4vDwIEDdbabM2cOhg4dirNnz+LkyZPo3r27gSokotpydgaSksSeUYMHA4sXAw8e\nyF0V1YQigsLV1RXOzs4629y+fRuHDx/GlClTAAANGjRAq1atDFEeEdWRhQUwfTqQni6m03p5AUeO\nyF0VVZcigqI6MjMz0b59e0yePBne3t6YPn068vLy5C6LiGrA1haIixNndI8bB8ycCdy5I3dVVJUG\nhnoijUaDnJyccvdHRUVh2LBhVf5+UVER0tLSsGrVKvTp0wdz585FTEwMlixZUmH7yMjI0u/VajXU\nanVtSyciPRs5UlzoXrBAbC748cdANT4GSM+0Wi20Wm2V7RQ1PdbPzw8rVqyAt7d3uZ/l5OSgf//+\nyMzMBAAcOXIEMTEx2L17d7m2nB5LZDySksSutI+2Mbexkbsi82U002Mr+4Dv2LEjHBwckJGRAQA4\ncOAA3LnPMZHRGzxYzIZ68kmgZ0/gP//hQj2lUUSPIi4uDrNnz8aNGzfQqlUreHl5Yc+ePcjKysL0\n6dOR8McGMunp6Zg2bRoePnwIR0dHrF27tsIL2uxREBmntDQxO8raWizU69pV7orMS2WfnYoICn1j\nUBAZr6Ii4P33gXffBd58U+xO28BgV1PNG4OCiIzKL7+Iaxd374oT9Tw95a7I9BnNNQoiIgDo1g04\neFAclKTRAH//OxfqyYVBQUSKpVKJzQXT04Hz50Wv4r//lbsq88OhJyIyGnFxwKxZQGgosGwZwM0Z\n9ItDT0Rk9J57TmwBIklioV58vNwVmQf2KIjIKB06JPaP8vQE/v1voGNHuSsyfuxREJFJGTRIXLtw\nchIL9b78kgv16gt7FERk9NLTxUXvVq2Azz4DHB3lrsg4sUdBRCbL0xNISQGGDgX69gWWLxcL90g/\n2KMgIpPy22/ASy8BN2+KhXpeXnJXZDzYoyAis9C1K7Bvn5hGO2SIOI41P1/uqowbg4KITI5KBbzw\ngtiVNjNEWT86AAALn0lEQVRTXOyuxrELVAkOPRGRydu1S5ymFxQkrl+0bi13RcrEoSciMlvPPisW\n6llZiYV6O3bIXZFxYY+CiMzKkSPizAt3d7FQz9ZW7oqUgz0KIiIATz8N/N//AW5uYlrtmjVcqFcV\n9iiIyGydOiV6F02bioV6Tk5yVyQv9iiIiP7CwwP44Qdg+HCgf3+xI21hodxVKY8igiI2Nhbu7u6w\ntLREWlpape2io6Ph7u4ODw8PhIWFoaCgwIBVEpEpsrQE5s4Fjh4FkpIAX1/g+HG5q1IWRQSFh4cH\n4uLiMHDgwErbXLhwAWvWrEFaWhpOnTqF4uJibN682YBVEpEpe/JJYO9e4LXXxFYgCxYAeXlyV6UM\niggKV1dXODs762zTsmVLWFlZIS8vD0VFRcjLy4OdnZ2BKiQic6BSAZMmiWsXV66IhXpJSXJXJT9F\nBEV1tG3bFq+//jo6deoEW1tbtG7dGgEBAXKXRUQmqEMH4OuvgZUrxQrvqVOBW7fkrko+BgsKjUYD\nDw+Pcl/ffPNNtX7/119/xcqVK3HhwgVkZWXh3r172LhxYz1XTUTmLDQUOH1azIpydwe2bTPPqbQN\nDPVE+/fvr9PvHzt2DAMGDIC1tTUAYOTIkfjhhx8wceLECttHRkaWfq9Wq6FWq+v0/ERknlq0EAvz\nJkwQU2m/+gr46CPAFEa+tVottNXYBEtR6yj8/Pzw3nvvwcfHp9zP0tPTMXHiRBw9ehSNGzfGCy+8\nAF9fX8ycObNcW66jIKL6UFAAREUBH38MvPOOOIrVwmgG8Kum6HUUcXFxcHBwQEpKCkJCQhAcHAwA\nyMrKQkhICADA09MT4eHh6N27N3r27AkAePHFF2WrmYjMT6NGwD//CXz3HbB2LeDnB5w/L3dV9U9R\nPQp9YY+CiOpbcbEYglqyREypnT9fbDpozCr77GRQEBHVwcWLwMsvA1lZwBdfAL17y11R7Sl66ImI\nyFh17gwkJooFeqGhwN/+Bty/L3dV+sWgICKqI5UKmDhRLNTLyRF7SNVxoqeicOiJiEjP9uwBZswA\n1Grg/feBtm3lrqh6OPRERGQgwcHiRL1WrcRCvS1bjHuhHnsURET1KCVFLNTr2lWsv7C3l7uiyrFH\nQUQkg379gLQ0MRvKy0uERUmJ3FXVDHsUREQGcuaMWM2tUgGffw64uspdUVnsURARyczNDTh8WOwb\n9cwzYhuQhw/lrqpqDAoiIgOysABmzhSn6CUnAz4+wI8/yl2Vbhx6IiKSiSSJGVHz5gHjxwNvvw00\nby5fPRx6IiJSGJVKBMRPPwE3b4qFet9+K3dV5bFHQUSkEN9+K/aNeuYZ4IMPgD+O3zEY9iiIiBRu\nyBCxDUi7dkCPHsCmTcpYqMceBRGRAqWmirO6O3UCPvlE/Le+sUdBRGREfH3FzKgBA8TMqFWr5Fuo\nxx4FEZHCnTsnFuoVF4uFem5u9fM87FEQERkpV1fg0CEgPBwYNEgcx2rI3gWDgojICFhYiBlRJ04A\nTZuK2wZ7bsM9VeXmz5+P7t27w9PTEyNHjsTt27crbLd37164urrCyckJy5YtM3CVRETys7cX53Mb\nkiKCIjAwEKdPn0Z6ejqcnZ0RHR1drk1xcTFeffVV7N27F2fOnMGmTZtw9uxZGao1HVqtVu4SzA7f\nc8Pje153iggKjUYDiz/6UX379sWVK1fKtUlNTUW3bt3QpUsXWFlZYfz48di5c6ehSzUp/B/I8Pie\nGx7f87pTRFA87ssvv8TQoUPL3X/16lU4ODiU3ra3t8fVq1cNWRoRkVlqYKgn0mg0yMnJKXd/VFQU\nhg0bBgBYunQpGjZsiLCwsHLtVCpVvddIREQVkBRi7dq10oABA6T8/PwKf56cnCwNGTKk9HZUVJQU\nExNTYVtPT08JAL/4xS9+8asGX56enhV+pipiwd3evXvx+uuv49ChQ2jXrl2FbYqKiuDi4oKDBw/C\n1tYWvr6+2LRpE7p3727gaomIzIsirlHMmjUL9+7dg0ajgZeXF1555RUAQFZWFkJCQgAADRo0wKpV\nqzBkyBC4ublh3LhxDAkiIgNQRI+CiIiUSxE9Cqq+6iw6nD17NpycnODp6YkTJ05U+buxsbFwd3eH\npaUl0tLSSu9PTU2Fl5cXvLy80LNnT2zZsqX0Z8ePH4eHhwecnJwwZ86cenilyqGU91ytVsPV1bX0\n5zdu3KiHV6sMhnzPH7l06RKaN2+OFStWlN5nTn/nOtXtEjQZUlFRkeTo6ChlZmZKDx8+lDw9PaUz\nZ86UaZOQkCAFBwdLkiRJKSkpUt++fav83bNnz0rnz5+X1Gq1dPz48dLHysvLk4qLiyVJkqTs7GzJ\n2tpaKioqkiRJkvr06SP9+OOPkiRJUnBwsLRnz576ffEyUdJ7/te2psrQ7/kjo0aNksaOHSu99957\npfeZy995VdijMCLVWXS4a9cuREREABCLF3Nzc5GTk6Pzd11dXeHs7Fzu+Zo0aVK6EDI/Px+tWrWC\npaUlsrOzcffuXfj6+gIAwsPDER8fX58vXTZKec8fkcxgpNjQ7zkAxMfHo2vXrnB7bFtWc/o7rwqD\nwohUZ9FhZW2ysrJqtWAxNTUV7u7ucHd3x/vvv1/6HPb29qVt7OzsTHbxo1Le80ciIiLg5eWFd955\np7YvSfEM/Z7fu3cP7777LiIjI8s9h7n8nVeFQWFEqrvoUJ//6vT19cXp06eRlpaGOXPmVLpho6lS\n0nu+ceNG/PTTTzh8+DAOHz6MDRs26O05lcTQ73lkZCTmzZuHpk2bmkWPrTYMtjKb6s7Ozg6XL18u\nvX358uUy/+KpqM2VK1dgb2+PwsLCKn9XF1dXVzg6OuKXX36Bvb19mf24rly5Ajs7u9q8JMVTynvu\n4+MDW1tbAEDz5s0RFhaG1NRUTJo0qbYvTbEM/Z6npqZi+/btWLBgAXJzc2FhYYEmTZpg5MiRZvN3\nXiVZr5BQjRQWFkpdu3aVMjMzpYKCgiov8iUnJ5de5KvO76rVaunYsWOltzMzM6XCwkJJkiTpwoUL\nkoODg3T79m1JkiTJ19dXSklJkUpKSkz6Ip9S3vOioiLpf//7nyRJkvTw4UNp1KhR0urVq+vtdcvJ\n0O/54yIjI6UVK1aU3jaXv/OqMCiMTGJiouTs7Cw5OjpKUVFRkiRJ0qeffip9+umnpW1mzpwpOTo6\nSj179iwzu6Oi35UkSdqxY4dkb28vNW7cWLKxsZGCgoIkSZKk9evXS+7u7lKvXr2kPn36lPmf5Nix\nY1KPHj0kR0dHadasWfX9smWlhPf83r17ko+Pj9SzZ0/J3d1dmjt3rlRSUmKIly8LQ77nj/trUJjT\n37kuXHBHREQ68WI2ERHpxKAgIiKdGBRERKQTg4KIiHRiUBARkU4MCiIi0olBQUREOjEoiIhIJwYF\nkQEVFRXh/PnzcpdBVCMMCiI9KykpwWuvvVbhz7RaLSwsLJCRkYHg4GCsXr0aAQEBmDp1KlavXg0f\nHx+UlJQYuGIi3bh7LJEe3bp1C2vXrsWhQ4cq/Pn58+cREBCArVu3YteuXbCyskJcXBwWLFgAFxcX\ntGrVqvTgIiKl4F8kkR61adM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+ "text": [
+ "<matplotlib.figure.Figure at 0x7a19438>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Activation energy of reaction is 189.2 kJ/mol\n",
+ "NOTE that in book slope is approximated as 9.9*10**3 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.2,Page no:86"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "T1=10.0 #in min\n",
+ "T2=20.0 #in min\n",
+ "a=25.0 #amount of KMnO4 in ml at t=0min#\n",
+ "a1=20.0 #amount of KMnO4 in ml at t=10min or a-x value at t=10#\n",
+ "a2=15.7 #a-x value at t=20min#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "k1=(2.303/T1)*math.log10(a/a1) #formula of rate constant for first order reaction#\n",
+ "print\"At t=10min rate constant k=\",round(k1,5),\"/min\"\n",
+ "k2=(2.303/T2)*math.log10(a/a2) #rate constant formula#\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nAt t=20min rate constant k=\",round(k2,5),\"/min\" \n",
+ "print\"\\nNOTE:Calculation mistake in book\"\n",
+ "print\"\\nIf we calculate the rate constant at other t values we will see that k values are almost constnat\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "At t=10min rate constant k= 0.02232 /min\n",
+ "\n",
+ "At t=20min rate constant k= 0.02326 /min\n",
+ "\n",
+ "NOTE:Calculation mistake in book\n",
+ "\n",
+ "If we calculate the rate constant at other t values we will see that k values are almost constnat\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.3,Page no:87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "T=40.5 #in min#\n",
+ "R1=25.0 #percentage of decomposed reactant#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "R2=100.0-R1 #percentage of left out reactant which is a-x value#\n",
+ "R3=100.0/R2 #value of a/(a-x)#\n",
+ "K=(2.303/T)*math.log10(R3) #formula of rate constant for first order reaction#\n",
+ "\n",
+ "#Result\n",
+ "print\"The rate constant of the reaction is %.2e\"%K,\"/min\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rate constant of the reaction is 7.10e-03 /min\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.4,Page no:87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "pi=0.0 #pressure of N2 at t=0#\n",
+ "t1=2.0 \n",
+ "t2=8.0 \n",
+ "t3=16.0 \n",
+ "t4=24.0 \n",
+ "t5=50.0 \n",
+ "pf=34.0 #pressure of N2 at infinity#\n",
+ "p1=1.6 #pressure of N2 at t=2min#\n",
+ "p2=6.2 #pressure of N2 at t=8min#\n",
+ "p3=11.2 #pressure Of N2 at t=16min#\n",
+ "p4=15.5 #pressure of N2 at t=24min#\n",
+ "p5=24.4 #pressure of N2 at t=50min#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "a=pf-pi #value of a#\n",
+ "a1=pf-p1 #a-x value at t=2min#\n",
+ "a2=pf-p2 #a-x value at t=8min#\n",
+ "a3=pf-p3 #a-x value at t=16min#\n",
+ "a4=pf-p4 #a-x value at t=24min#\n",
+ "a5=pf-p5 #a-x value at t=50min#\n",
+ "k1=(1/t1)*math.log(a/a1) #rate constant at t=2min#\n",
+ "k2=(1/t2)*math.log(a/a2) #rate constant at t=8min#\n",
+ "k3=(1/t3)*math.log(a/a3) #rate constant at t=16min#\n",
+ "k4=(1/t4)*math.log(a/a4) #rate constant at t=24min#\n",
+ "k5=(1/t5)*math.log(a/a5) #rate constant at t=50min#\n",
+ "k=(k1+k2+k3+k4+k5)/5 \n",
+ "\n",
+ "#Result\n",
+ "print\"Time(min): 2\\t\\t8\\t\\t16\\t\\t24\\t\\t50\"\n",
+ "print\"k1 per min %.2e\\t\"%k1,\"%.2e\\t\"%k2,\"%.2e\\t\"%k3,\"%.3e\\t\"%k4,\"%.2e\"%k5\n",
+ "print\"\\nAverage rate constant is %.3e\"%k,\"min^-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time(min): 2\t\t8\t\t16\t\t24\t\t50\n",
+ "k1 per min 2.41e-02\t2.52e-02\t2.50e-02\t2.536e-02\t2.53e-02\n",
+ "\n",
+ "Average rate constant is 2.498e-02 min^-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example no:3.5.,Page no:88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t1=0 \n",
+ "t2=4.89 \n",
+ "t3=10.07 \n",
+ "t4=23.66 \n",
+ "v1=47.65 #ml of alkali used at t=0min or a value#\n",
+ "v2=38.92 #ml of alkali used or a-x value at t=4.89min#\n",
+ "v3=32.62 #ml of alkali used or a-x value at t=10.07min#\n",
+ "v4=22.58 #ml of alkali used or a-x value at t=23.66min#\n",
+ "\n",
+ "#Calculation\n",
+ "x2=v1-v2 #x value at t=4.89min#\n",
+ "x3=v1-v3 #x value at t=10.07min#\n",
+ "x4=v1-v4 #x value at t=23.66min#\n",
+ "k22=(1/t2)*(x2/(v1*v2)) #rate constant for second order equation#\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate constant k2 value at t=\",t2,\"min is \",round(k22,6),\"/min\"\n",
+ "k23=(1/t3)*(x3/(v1*v3)) #rate constant for second order equation#\n",
+ "print\"\\nRate constant k2 value at t=\",t3,\"min is \",round(k23,6),\"/min\"\n",
+ "k24=(1/t4)*(x4/(v1*v4)) #rate constant for second order equation#\n",
+ "print\"\\nRate constant k2 value at t=\",t4,\"min is\",round(k24,5),\"/min\" \n",
+ "print\"\\nAlmost constant values of k2 indicate that reaction is second order\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate constant k2 value at t= 4.89 min is 0.000963 /min\n",
+ "\n",
+ "Rate constant k2 value at t= 10.07 min is 0.00096 /min\n",
+ "\n",
+ "Rate constant k2 value at t= 23.66 min is 0.00098 /min\n",
+ "\n",
+ "Almost constant values of k2 indicate that reaction is second order\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.6,Page no:88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t=1590 #half life of given radio active element in years#\n",
+ "\n",
+ "#Calculation\n",
+ "k=0.693/t #formula of decay constant for first order reactions#\n",
+ "\n",
+ "#Result\n",
+ "print\"the value of decay constant is \",round(k,6),\"/year\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of decay constant is 0.000436 /year\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.8,Page no:89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t1=5.0 \n",
+ "t2=15.0 \n",
+ "t3=25.0 \n",
+ "t4=45.0 \n",
+ "a=37.0 #volume of KMnO4 in cm**3 at t=0 or value of a#\n",
+ "a1=29.8 #volume of KMnO4 in cm**3 or a-x value at t=5min#\n",
+ "a2=19.6 #volume of KMnO4 in cm**3 or a-x value at t=15min#\n",
+ "a3=12.3 #volume of KMnO4 in cm**3 or a-x value at t=25min#\n",
+ "a4=5.0 #volume of KMnO4 in cm**3 or a-x value at t=45min#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "k1=(2.303/t1)*math.log10(a/a1) \n",
+ "print\"\\nRate constant value at t=5min is %.3e\"%k1,\"min**-1\"\n",
+ "k2=(2.303/t2)*math.log10(a/a2) \n",
+ "print\"\\nRate constant value at t=15min is %.3e\"%k2,\"min**-1\"\n",
+ "k3=(2.303/t3)*math.log10(a/a3) \n",
+ "print\"\\nRate constant value at t=25min is %.3e\"%k3,\"min**-1\"\n",
+ "k4=(2.303/t4)*math.log10(a/a4) \n",
+ "print\"\\nRate constant value at t=45min is %.3e\"%k4,\"min**-1\"\n",
+ "print\"\\nAs the different values of k are nearly same,the reaction is of first oredr.\"\n",
+ "k=(k1+k2+k3+k4)/4.0 \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nThe average value of k is %.3e\"%k,\"min**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Rate constant value at t=5min is 4.329e-02 min**-1\n",
+ "\n",
+ "Rate constant value at t=15min is 4.237e-02 min**-1\n",
+ "\n",
+ "Rate constant value at t=25min is 4.406e-02 min**-1\n",
+ "\n",
+ "Rate constant value at t=45min is 4.449e-02 min**-1\n",
+ "\n",
+ "As the different values of k are nearly same,the reaction is of first oredr.\n",
+ "\n",
+ "The average value of k is 4.355e-02 min**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.9,Page no:89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "k=6.0*10**-4 #rate constant of first order decomposition of N2O5 in CCl4 in /min#\n",
+ "\n",
+ "#Calculation\n",
+ "#Part-a#\n",
+ "k1=k/60.0 \n",
+ "#Part-b#\n",
+ "t=0.693/k \n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Rate constant in terms of seconds is \",k1,\"/s\"\n",
+ "print\"\\n(b) Half life of the reaction is %.2e\"%t,\"min\"\n",
+ "print\"NOTE:Slight rounding off in book in final answer\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Rate constant in terms of seconds is 1e-05 /s\n",
+ "\n",
+ "(b) Half life of the reaction is 1.15e+03 min\n",
+ "NOTE:Slight rounding off in book in final answer\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.10,Page no:90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t1=40.0 \n",
+ "t2=80.0 \n",
+ "t3=120.0 \n",
+ "t4=160.0 \n",
+ "t5=240.0 \n",
+ "vi=0.0 #volume of oxygen collected at constant pressure in ml at t=0#\n",
+ "v1=15.6 #volume of oxygen collected at constant pressure in ml at t=40#\n",
+ "v2=27.6 #volume of oxygen collected at constant pressure in ml at t=80#\n",
+ "v3=37.7 #volume of oxygen collected at constant pressure in ml at t=120#\n",
+ "v4=45.8 #volume of oxygen collected at constant pressure in ml at t=160#\n",
+ "v5=58.3 #volume of oxygen collected at constant pressure in ml at t=200#\n",
+ "vf=84.6 #volume of oxygen collected at constant pressure in ml at t=infinity#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "a=vf-vi #the initial concentration of N2O5 in solution i.e a#\n",
+ "a1=vf-v1 #a-x value at t=40min#\n",
+ "a2=vf-v2 #a-x value at t=80min#\n",
+ "a3=vf-v3 #a-x value at t=120min#\n",
+ "a4=vf-v4 #a-x value at t=160min#\n",
+ "a5=vf-v5 #a-x value at t=200min#\n",
+ "k1=(1.0/t1)*math.log(a/a1) \n",
+ "k2=(1.0/t2)*math.log(a/a2) \n",
+ "k3=(1.0/t3)*math.log(a/a3) \n",
+ "k4=(1.0/t4)*math.log(a/a4) \n",
+ "k5=(1.0/t5)*math.log(a/a5) \n",
+ "\n",
+ "#Result\n",
+ "print\"Time(min): 40\\t\\t80\\t\\t120\\t\\t160\\t\\t240\"\n",
+ "print\"k1 per min %.2e\\t\"%k1,\"%.2e\\t\"%k2,\"%.2e\\t\"%k3,\"%.3e\\t\"%k4,\"%.2e\"%k5\n",
+ "print\"\\nNOTE:Calculation mistake in book in calculating a4,it should be 38.8\"\n",
+ "print\"\\nAs k value is fairly constant the reaction is first order\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time(min): 40\t\t80\t\t120\t\t160\t\t240\n",
+ "k1 per min 5.10e-03\t4.94e-03\t4.92e-03\t4.872e-03\t4.87e-03\n",
+ "\n",
+ "NOTE:Calculation mistake in book in calculating a4,it should be 38.8\n",
+ "\n",
+ "As k value is fairly constant the reaction is first order\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.11,Page no:90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t1=120.0 #time in sec#\n",
+ "t2=240.0 \n",
+ "t3=530.0 \n",
+ "t4=600.0 \n",
+ "a=0.05 #initial concentration#\n",
+ "x1=32.95 #extent of reaction or x value at t=120sce#\n",
+ "x2=48.8 #extent of reaction or x value at t=240sce#\n",
+ "x3=69.0 #extent of reaction or x value at t=530sce#\n",
+ "x4=70.35 #extent of reaction or x value at t=600sce#\n",
+ "a1=100.0-x1 #extent of left out or a-x value at t=120sec#\n",
+ "a2=100.0-x2 #extent of left out or a-x value at t=240sec#\n",
+ "a3=100.0-x3 #extent of left out or a-x value at t=530sec#\n",
+ "a4=100.0-x4 #extent of left out or a-x value at t=600sec#\n",
+ "\n",
+ "#Calculation\n",
+ "k1=(1.0/(a*t1))*(x1/a1) \n",
+ "print\"Rate constant value at t=120sec is %.2e\"%k1,\"dm**3 mol**-1.s**-1\"\n",
+ "k2=(1.0/(a*t2))*(x2/a2) \n",
+ "print\"\\nRate constant value at t=240sec is %.2e\"%k2,\"dm**3 mol**-1.s**-1\"\n",
+ "k3=(1.0/(a*t3))*(x3/a3) \n",
+ "print\"\\nRate constant value at t=530sec is %.2e\"%k3,\"dm**3 mol**-1.s**-1\"\n",
+ "k4=(1.0/(a*t4))*(x4/a4) \n",
+ "print\"\\nRate constant value at t=600sec is %.2e\"%k4,\"dm**3 mol**-1.s**-1\"\n",
+ "k=(k1+k2+k3+k4)/4.0 \n",
+ "\n",
+ "#Result\n",
+ "print\"\\n\\nAverage value of rate constant is %.1e\"%k,\"dm**3 mol**-1.s**-1\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate constant value at t=120sec is 8.19e-02 dm**3 mol**-1.s**-1\n",
+ "\n",
+ "Rate constant value at t=240sec is 7.94e-02 dm**3 mol**-1.s**-1\n",
+ "\n",
+ "Rate constant value at t=530sec is 8.40e-02 dm**3 mol**-1.s**-1\n",
+ "\n",
+ "Rate constant value at t=600sec is 7.91e-02 dm**3 mol**-1.s**-1\n",
+ "\n",
+ "\n",
+ "Average value of rate constant is 8.1e-02 dm**3 mol**-1.s**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.13,Page no:91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t1=75.0 #time in min#\n",
+ "t2=119.0 \n",
+ "t3=183.0 \n",
+ "vi=9.62 #volume of alkali used in ml at t=0min#\n",
+ "v1=12.10 #volume of alkali used in ml at t=75min#\n",
+ "v2=13.10 #volume of alkali used in ml at t=119min#\n",
+ "v3=14.75 #volume of alkali used in ml at t=183min#\n",
+ "vf=21.05 #volume of alkali used in ml at t=infinity#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "k1=(1.0/t1)*math.log((vf-vi)/(vf-v1)) #formula of rate constant for first order reactions#\n",
+ "k2=(1.0/t2)*math.log((vf-vi)/(vf-v2)) \n",
+ "k3=(1.0/t3)*math.log((vf-vi)/(vf-v3)) \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nRate constant value at t=75min is \",round(k1,6),\"min**-1\"\n",
+ "print\"\\nRate constant value at t=119min is \",round(k2,6),\"min**-1\"\n",
+ "print\"\\nRate constant value at t=183min is \",round(k3,6),\"min**-1\"\n",
+ "\n",
+ "print\"\\nNOTE:Slight Calculation mistake in book in k calculation above\" \n",
+ "print\"\\nAn almost constant value of k shows that the hydrolysis of ethyl acetateis a first order reaction\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Rate constant value at t=75min is 0.003261 min**-1\n",
+ "\n",
+ "Rate constant value at t=119min is 0.003051 min**-1\n",
+ "\n",
+ "Rate constant value at t=183min is 0.003255 min**-1\n",
+ "\n",
+ "NOTE:Slight Calculation mistake in book in k calculation above\n",
+ "\n",
+ "An almost constant value of k shows that the hydrolysis of ethyl acetateis a first order reaction\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.14,Page no:92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t=15 #the half time of given first order reaction in min#\n",
+ "k=0.693/t #formula of rate constant#\n",
+ "print\"The rate constant value of the given first order reaction is is\",k,\"min**-1\"\n",
+ "a=100 #percentage of initial concentration#\n",
+ "x=80 #percentage of completed reaction#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "a1=a-x #percentage of left out concentration#\n",
+ "t1=(2.303/k)*(math.log10(a/a1)) #formula to find time taken#\n",
+ "t2=t1*60 \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nThe time taken to complete 80 percentage of the reaction is \",round(t1,2),\"min or\",round(t2),\"sec\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rate constant value of the given first order reaction is is 0.0462 min**-1\n",
+ "\n",
+ "The time taken to complete 80 percentage of the reaction is 34.84 min or 2091.0 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.15,Page no:92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t1=6.18 #time in min#\n",
+ "t2=18.0 \n",
+ "t3=27.05 \n",
+ "ri=24.09 #rotation in degrees when t=0min#\n",
+ "r1=21.4 #rotation in degrees when t=6.18min#\n",
+ "r2=17.7 #rotation in degrees when t=18min#\n",
+ "r3=15.0 #rotation in degrees when t=27.05min#\n",
+ "rf=-10.74 #rotation in degrees when t=infinity#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "a=ri-rf #a value#\n",
+ "a1=r1-rf #a-x value at t=6.18min#\n",
+ "a2=r2-rf #a-x value at t=18min#\n",
+ "a3=r3-rf #a-x value at t=27.05min#\n",
+ "k1=(2.303/t1)*math.log10(a/a1) \n",
+ "k2=(2.303/t2)*math.log10(a/a2) \n",
+ "k3=(2.303/t3)*math.log10(a/a3) \n",
+ "\n",
+ "#Result\n",
+ "print\"Rate constant value at t=\",t1,\"min %.3e\"%k1,\"min**-1\"\n",
+ "print\"\\nRate constant value at t=\",t2,\"min %.3e\"%k2,\"min**-1\"\n",
+ "print\"\\nRate constant value at t=\",t3,\"min %.3e\"%k3,\"min**-1\"\n",
+ "\n",
+ "print\"\\nNOTE:Again,Calculation mistake in book\"\n",
+ "print\"\\nSince rate constant values are nearly same,hence reaction is of first order\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate constant value at t= 6.18 min 1.301e-02 min**-1\n",
+ "\n",
+ "Rate constant value at t= 18.0 min 1.126e-02 min**-1\n",
+ "\n",
+ "Rate constant value at t= 27.05 min 1.118e-02 min**-1\n",
+ "\n",
+ "NOTE:Again,Calculation mistake in book\n",
+ "\n",
+ "Since rate constant values are nearly same,hence reaction is of first order\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.16,Page no:93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t1=10.0#time in min#\n",
+ "t2=20.0 \n",
+ "t3=30.0 \n",
+ "t4=40.0 \n",
+ "ri=32.4 #rotation in degrees when t=0min#\n",
+ "r1=28.8 #rotation in degrees when t=10min#\n",
+ "r2=25.5 #rotation in degrees when t=20min#\n",
+ "r3=22.4 #rotation in degrees when t=30min#\n",
+ "r4=19.6 #rotation in degrees when t=40min#\n",
+ "rf=-11.1 #rotation in degrees when t=0min#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "a=ri-rf #a value#\n",
+ "a1=r1-rf #a-x value at t=10min#\n",
+ "a2=r2-rf #a-x value at t=20min#\n",
+ "a3=r3-rf #a-x value at t=30min#\n",
+ "a4=r4-rf #a-x value at t=40min#\n",
+ "k1=(1.0/t1)*math.log(a/a1) \n",
+ "k2=(1.0/t2)*math.log(a/a2) \n",
+ "k3=(1.0/t3)*math.log(a/a3) \n",
+ "k4=(1.0/t4)*math.log(a/a4) \n",
+ "\n",
+ "#Result\n",
+ "print\"Rate constant value at t=10min \",round(k1,6),\"min**-1\"\n",
+ "print\"\\nRate constant value at t=20min \",round(k2,6),\"min**-1\"\n",
+ "print\"\\nRate constant value at t=30min \",round(k3,6),\"min**-1\"\n",
+ "print\"\\nRate constant value at t=40min \",round(k4,6),\"min**-1\"\n",
+ "print\"\\nSince rate constant values are nearly same,hence inversion of sucrose is of first order\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate constant value at t=10min 0.008638 min**-1\n",
+ "\n",
+ "Rate constant value at t=20min 0.008636 min**-1\n",
+ "\n",
+ "Rate constant value at t=30min 0.008707 min**-1\n",
+ "\n",
+ "Rate constant value at t=40min 0.008712 min**-1\n",
+ "\n",
+ "Since rate constant values are nearly same,hence inversion of sucrose is of first order\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.17,Page no:93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "T1=27.0 #initial temparature in C#\n",
+ "T1=T1+273 #in kelvin#\n",
+ "Tr=10.0 #rise in temparature#\n",
+ "T2=T1+Tr #final temparature in kelvin#\n",
+ "r=2.0 #ratio of final to initial rates of chemical reactions(k1/k2)#\n",
+ "R=8.314 #value of constant R in J/K.mol#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "E=math.log(r)*R*305*295/Tr #from equation k=A*e**(-E/R*T)#\n",
+ "\n",
+ "#Result\n",
+ "print\"Activation energy of the reaction is \",round(E/1000,2),\"kJ/mol\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Activation energy of the reaction is 51.85 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.18,Page no:94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "k=4.5*10**3 #value of k in /sec of a first order reaction at 1C#\n",
+ "E=58*10**3 #activation energy in J/mol#\n",
+ "T=1 #temperature in C#\n",
+ "T1=T+273 #in kelvin#\n",
+ "R=8.314 #value of constant R in J/K.mol#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "lA=math.log10(k)+(E/(2.303*R*T1)) \n",
+ "k1=10**4 #value of k in /sec at some temperature#\n",
+ "a=math.log10(k1) \n",
+ "b=lA-a \n",
+ "T2=E/(2.303*R*b) \n",
+ "\n",
+ "#Result\n",
+ "print\"The temperature at which k=1*10**4/sec is\",round(T2),\"K\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature at which k=1*10**4/sec is 283.0 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.19,Page no:94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "T1=300.0 #temperature in kelvin#\n",
+ "t1=20.0 #half time of chemical reaction in min at T=300K#\n",
+ "T2=350.0 #temperature in kelvin#\n",
+ "t2=5.0 #half time of chemical reaction in min at T=350K#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "k1=0.6932/t1 \n",
+ "k2=0.6932/t2 \n",
+ "l=math.log10(k2/k1) \n",
+ "R=8.314 #value of constant R in J/K.mol#\n",
+ "E=l*2.303*R*T1*T2/(T2-T1) \n",
+ "\n",
+ "#Result\n",
+ "print\"Rate constant of the reaction at T=300k is \",k1,\"/min\" \n",
+ "print\"\\nRate constant of the reaction at T=350k is \",k2,\"/min\"\n",
+ "print\"\\nActivation energy of the reaction is\",E,\"J/mol OR\",round(E/1000,1),\"kJ/mol\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate constant of the reaction at T=300k is 0.03466 /min\n",
+ "\n",
+ "Rate constant of the reaction at T=350k is 0.13864 /min\n",
+ "\n",
+ "Activation energy of the reaction is 24208.2291076 J/mol OR 24.2 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.20,Page no:94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "R=8.314 #value of constant R in J/K.mol#\n",
+ "H=1.25*10**4 #value of E/(2.303*R).It is given in the question#\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "#Part-i#\n",
+ "E=H*2.303*R \n",
+ "la=14.34 #value of math.log(a)#\n",
+ "T=670 #temperature in kelvin#\n",
+ "#Part-ii#\n",
+ "lk=la-(H/T) \n",
+ "k=10**lk \n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Activation energy is %.2e\"%E,\"J mol**-1 or\",round(E/1000),\"kJ mol**-1\"\n",
+ "print\"\\n(ii) Rate constant at 670K is %.1e\"%k,\"s**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Activation energy is 2.39e+05 J mol**-1 or 239.0 kJ mol**-1\n",
+ "\n",
+ "(ii) Rate constant at 670K is 4.8e-05 s**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.21,Page no:95"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Ti=27.0 #given temperature in C#\n",
+ "T1=Ti+273.0 #in kelvin#\n",
+ "t1=T1-5\n",
+ "Tr=10.0 #rise in temperature#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "T2=T1+Tr \n",
+ "t2=T2-5\n",
+ "k=3.0 #value of k1/k2#\n",
+ "R=8.314 #value of constant R in J/K.mol#\n",
+ "E=math.log(k)*R*t1*t2/(T2-T1) \n",
+ "\n",
+ "#Result\n",
+ "print\"Activation energy of the reaction is\",round(E),\"J mol**-1 or\",round(E/1000,2),\"kJ mol**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Activation energy of the reaction is 82182.0 J mol**-1 or 82.18 kJ mol**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Textbook_of_Engineering_Chemistry/dChapter_6.ipynb b/Textbook_of_Engineering_Chemistry/dChapter_6.ipynb
new file mode 100755
index 00000000..7428ce43
--- /dev/null
+++ b/Textbook_of_Engineering_Chemistry/dChapter_6.ipynb
@@ -0,0 +1,92 @@
+{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6:Lubricants"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:1,Page no:141"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "d=760.0 #viscocity of Pennysylvanian oil in s at 37C#\n",
+ "a=528.0 #viscocity of lubricating oil in s at 37C#\n",
+ "c=480.0 #viscocity of Gulf oil in s at 37C#\n",
+ "\n",
+ "#Calculation\n",
+ "V=((d-a)/(d-c))*(100) #formula of viscocity index#\n",
+ "\n",
+ "#Result\n",
+ "print\"Viscocity index of the lubricating oil is \",round(V ,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Viscocity index of the lubricating oil is 82.9\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2,Page no:144"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "s=0.86 #specific gravity of lubricating oil#\n",
+ "\n",
+ "#Calculation\n",
+ "A=(141.5/s)-131.5 #formula of API gravity#\n",
+ "\n",
+ "#Result\n",
+ "print\"The gravity of lubricating oil is \",round(A)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The gravity of lubricating oil is 33.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Textbook_of_Engineering_Chemistry/dChapter_7.ipynb b/Textbook_of_Engineering_Chemistry/dChapter_7.ipynb
new file mode 100755
index 00000000..3854b5db
--- /dev/null
+++ b/Textbook_of_Engineering_Chemistry/dChapter_7.ipynb
@@ -0,0 +1,850 @@
+{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7:Water Chemistry"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:1,Page no:152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "MgSO4=60.0 #[mg]\n",
+ "M_MgSO4=64.0 #Molecular weight of MgSO4\n",
+ "M_CaCO3=48.0 #Molecular wt of CaCO3\n",
+ "m_mgso4=120.0 #Weight of MgSO4 eq to CaCO3\n",
+ "m_caco3=100.0 #Weight of CaCO3 eq to MgSO4\n",
+ "\n",
+ "#Calculation\n",
+ "hard=(m_caco3/m_mgso4)*MgSO4 #Hardness of water in mg\n",
+ "\n",
+ "#Result\n",
+ "print\"Hardness of water is \",hard,\"ppm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Hardness of water is 50.0 ppm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.1,Page no:172"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=16.2 #Ca(HCO3)2 in water in mg/lit#\n",
+ "W2=7.3 #MgHCO3 in water in mg/lit#\n",
+ "W3=13.6 #CaSO4 in water in mg/lit#\n",
+ "W4=9.5 #MgCl2 in water in mg/lit#\n",
+ "M1=100/162.0 #multiplication factor of Ca(HCO3)2#\n",
+ "M2=100/146.0 #multiplication factor of MgHCO3#\n",
+ "M3=100/136.0 #multiplication factor of CaSO4#\n",
+ "M4=100/95.0 #multiplication factor of MgCl2#\n",
+ "\n",
+ "#Calculation\n",
+ "P1=W1*M1 #Ca(HCO3)2 in terms of CaCO3 or #\n",
+ "P2=W2*M2 #MgHCO3 in terms of CaCO3 or #\n",
+ "P3=W3*M3 #CaSO4 in terms of CaCO3 or #\n",
+ "P4=W4*M4 #MgCl2 in terms of CaCO3 or #\n",
+ "T=P1+P2 #Temporary hardness\n",
+ "P=P3+P4 #Permanent hardness\n",
+ "To=T+P #Total hardness\n",
+ "\n",
+ "#Result\n",
+ "print\"Temporary hardness is\",T,\"mg/l or ppm\"\n",
+ "print\"\\nPermanant hardness is \",P,\"mg/l or ppm\"\n",
+ "print\"\\nTotal hardness is \",To,\"mg/l or ppm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Temporary hardness is 15.0 mg/l or ppm\n",
+ "\n",
+ "Permanant hardness is 20.0 mg/l or ppm\n",
+ "\n",
+ "Total hardness is 35.0 mg/l or ppm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.2,Page no:172"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "F=56.0 #atomic weight of ferrus#\n",
+ "S=32.0 #atomic weight of sulphur#\n",
+ "O=16.0 #atomic weight of oxygen#\n",
+ "Ca=40.0 #atomic weight of calsium#\n",
+ "C=12.0 #atomic weight of carbon#\n",
+ "\n",
+ "#Calculation\n",
+ "W1=136\n",
+ "P=210.5 #required ppm of hardness#\n",
+ "B=(W1/100.0)*P \n",
+ "\n",
+ "#Result\n",
+ "print\"Required FeSO4 for 100ppm of hardness is\",W1,\"ppm pf FeSO4\"\n",
+ "print\"\\nRequired FeSO4 for 210.5ppm of hardness is \",round(B,1),\"ppm of FeSO4\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Required FeSO4 for 100ppm of hardness is 136 ppm pf FeSO4\n",
+ "\n",
+ "Required FeSO4 for 210.5ppm of hardness is 286.3 ppm of FeSO4\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.3,Page no:173"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=162.0 #Ca(HCO3)2 in water in mg/lit#\n",
+ "W2=73.0 #MgHCO3 in water in mg/lit#\n",
+ "W3=136.0 #CaSO4 in water in mg/lit#\n",
+ "W4=95.0 #MgCl2 in water in mg/lit#\n",
+ "W5=111.0 #CaCl2 in water in mg/lit#\n",
+ "W6=100.0 #NaCl in water in mg/lit#\n",
+ "M1=100/162.0 #multiplication factor of Ca(HCO3)2#\n",
+ "M2=100/146.0 #multiplication factor of MgHCO3#\n",
+ "M3=100/136.0 #multiplication factor of CaSO4#\n",
+ "M4=100/95.0 #multiplication factor of MgCl2#\n",
+ "M5=100/111.0 #multiplication factor of CaCl2#\n",
+ "M6=100/100.0 #multiplication factor of NaCl#\n",
+ "\n",
+ "#Calculation\n",
+ "P1=W1*M1 #Ca(HCO3)2 in terms of CaCO3 or #\n",
+ "P2=W2*M2 #MgHCO3 in terms of CaCO3 or #\n",
+ "P3=W3*M3 #CaSO4 in terms of CaCO3 or #\n",
+ "P4=W4*M4 #MgCl2 in terms of CaCO3 or #\n",
+ "P5=W5*M5 #CaCl2 in terms of CaCO3 or #\n",
+ "T=P1+P2 \n",
+ "P=P3+P4+P5 \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nTemporary hardness is\",T,\"mg/l or ppm\" \n",
+ "print\"\\nPermanant hardness is\",P,\"mg/l or ppm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Temporary hardness is 150.0 mg/l or ppm\n",
+ "\n",
+ "Permanant hardness is 300.0 mg/l or ppm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.4,Page no:173"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N=0.08 #normality of MgSO4#\n",
+ "V1=12.5 #volume of MgSO4 in ml#\n",
+ "V2=100 #volume of water sample#\n",
+ "\n",
+ "#Calculation\n",
+ "M=N/2 #molarity of MgSO4#\n",
+ "N1=(M*12.5)/1000 #no of moles of MgSO4 in 100 ml water#\n",
+ "N2=(N1*1000)/100 #no of moles of MgSO4 in one litre water#\n",
+ "W=100 #molecular weight of CaCO3\n",
+ "W1=N2*W*1000 #MgSO4 in terms of CaCO3 in mg/lit#\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nThe hardness due to MgSO4 is \",W1,\"mg/l CaCO3 or ppm of CaCO3\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The hardness due to MgSO4 is 500.0 mg/l CaCO3 or ppm of CaCO3\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.5,Page no:173"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=144.0 #MgCO3 in water in mg/lit#\n",
+ "W2=25.0 #CaCO3 in water in mg/lit#\n",
+ "W3=111.0 #CaCl2 in water in mg/lit#\n",
+ "W4=95.0 #MgCl2 in water in mg/lit#\n",
+ "M1=100/84.0 #multiplication factor of MgCO3#\n",
+ "M2=100/100.0 #multiplication factor of CaCO3#\n",
+ "M3=100/111.0 #multiplication factor of CaCl2#\n",
+ "M4=100/95.0 #multiplication factor of MgCl2#\n",
+ "\n",
+ "#Calculation\n",
+ "P1=W1*M1 #MgCO3 in terms of CaCO3 or ppm#\n",
+ "P2=W2*M2 #CaCO3 in terms of CaCO3 or ppm#\n",
+ "P3=W3*M3 #CaCl2 in terms of CaCO3 or ppm#\n",
+ "P4=W4*M4 #MgCl2 in terms of CaCO3 or ppm#\n",
+ "V=50000 #volume of water in lit#\n",
+ "L=0.74*(2*P1+P2+P4)*V \n",
+ "S=1.06*(P1+P3+P4)*V \n",
+ "\n",
+ "#Result\n",
+ "print\"Requirement of lime is \",L,\"mg=\",round(L/1000000,1),\"kg\" \n",
+ "print\"\\nRequirement of soda is \",S,\"mg=\",round(S/1000000,1),\"kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Requirement of lime is 17310714.2857 mg= 17.3 kg\n",
+ "\n",
+ "Requirement of soda is 19685714.2857 mg= 19.7 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.6,Page no:174"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=12.0 #Mg2+ in water in ppm or mg/l#\n",
+ "W2=40.0 #Ca2+ in water in ppm or mg/l#\n",
+ "W3=164.7 #HCO3- in water in ppm or mg/l#\n",
+ "W4=30.8 #CO2 in water in ppm or mg/l#\n",
+ "M1=100.0/24.0 #multiplication factor of Mg2+#\n",
+ "M2=100.0/40.0 #multiplication factor of Mg2+#\n",
+ "M3=100.0/61.0 #multiplication factor of Mg2+#\n",
+ "M4=100.0/44.0 #multiplication factor of Mg2+#\n",
+ "\n",
+ "#Calculation\n",
+ "P1=W1*M1 # in terms of CaCO3#\n",
+ "P2=W2*M2 # in terms of CaCO3#\n",
+ "P3=W3*M3/2 # in terms of CaCO3#\n",
+ "P4=W4*M4 # in terms of CaCO3#\n",
+ "V=50000.0#volume of water in lit#\n",
+ "L=0.74*(P1+P3+P4)*V \n",
+ "\n",
+ "#Result\n",
+ "print\"Lime required is %fmg\",round(L/10**6,1),\"kg\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Lime required is %fmg 9.4 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.7,Page no:174"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=160.0 #Ca2+ in water in mg/l or ppm#\n",
+ "W2=72.0 #Mg2+ in water in mg/l or ppm#\n",
+ "W3=732.0 #HCO3- in water in mg/l or ppm#\n",
+ "W4=44.0 #CO2 in water in mg/l or ppm#\n",
+ "W5=16.4 #NaAlO2 in water in mg/l or ppm#\n",
+ "W6=30.0 #(CO3)2- in water in mg/l or ppm#\n",
+ "W7=17.0 #OH- in water in mg/l or ppm#\n",
+ "\n",
+ "#Calculation\n",
+ "M1=100/40.0 #multiplication factor of Ca2+#\n",
+ "M2=100/24.0 #multiplication factor of Ca2+#\n",
+ "M3=100/(61.0*2.0) #multiplication factor of Ca2+#\n",
+ "M4=100/44.0 #multiplication factor of Ca2+#\n",
+ "M5=100/(82.0*2.0) #multiplication factor of Ca2+#\n",
+ "M6=100/60.0 #multiplication factor of Ca2+#\n",
+ "M7=100/(17.0*2.0) #multiplication factor of Ca2+#\n",
+ "P1=W1*M1 #in terms of CaCO3#\n",
+ "P2=W2*M2 #in terms of CaCO3#\n",
+ "P3=W3*M3 #in terms of CaCO3#\n",
+ "P4=W4*M4 #in terms of CaCO3#\n",
+ "P5=W5*M5 #in terms of CaCO3#\n",
+ "P6=W6*M6 #in terms of CaCO3#\n",
+ "P7=W7*M7 #in terms of CaCO3#\n",
+ "V=200000.0 #volume of water in lit#\n",
+ "L=0.74*(P2+P3+P4-P5+P7)*V \n",
+ "L=L/10.0**6.0 #in kgs#\n",
+ "S=1.06*(P1+P2-P3-P5-P6+P7)*V \n",
+ "S=S/10.0**6 #in kgs#\n",
+ "\n",
+ "#Result\n",
+ "print\"Lime required is \",L,\"kg\"\n",
+ "print\"\\nSoda required is \",S,\"kg\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Lime required is 153.92 kg\n",
+ "\n",
+ "Soda required is 19.08 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.8,Page no:175"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N=150.0 #amount of NaCl in solution in g/l#\n",
+ "V=8.0 #volume of NaCl solution#\n",
+ "\n",
+ "#Calculation\n",
+ "M=N*V \n",
+ "V=10000.0 #volume of hard water#\n",
+ "W=58.5 #molecular weight of NaCl#\n",
+ "K=(M*100.0/(W*2))/V \n",
+ "J=K*1000.0 \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print\"\\nHardness of water is \",round(J,1),\"mg/l or ppm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Hardness of water is 102.6 mg/l or ppm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.9,Page no:176"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=219.0 #amount of Mg(HCO3)2 in water in ppm#\n",
+ "W2=36.0 #amount of Mg2+ in water in ppm#\n",
+ "W3=18.3 #amount of (HCO3)- in water in ppm#\n",
+ "W4=1.5 #amount of H+_in water in ppm#\n",
+ "M1=100/146.0 #multiplication factor of Mg(HCO3)2#\n",
+ "M2=100/24.0 #multiplication factor of Mg(HCO3)2#\n",
+ "M3=100/122.0 #multiplication factor of Mg(HCO3)2#\n",
+ "M4=100/2.0 #multiplication factor of Mg(HCO3)2#\n",
+ "\n",
+ "#Calculation\n",
+ "P1=W1*M1 #in terms of CaCO3#\n",
+ "P2=W2*M2 #in terms of CaCO3#\n",
+ "P3=W3*M3 #in terms of CaCO3#\n",
+ "P4=W4*M4 #in terms of CaCO3#\n",
+ "L=0.74*((2*P1)+P2+P3+P4) \n",
+ "\n",
+ "R=1.0 #water supply rate in m**3/s#\n",
+ "D=R*60.0*60.0*24.0*L \n",
+ "K=D*1000.0 #in lit/day#\n",
+ "T=K/10.0**9 #in tonnes#\n",
+ "S=1.06*(P2+P4-P3) \n",
+ "D2=R*60*60*24*S \n",
+ "A=D2*1000 #in lit/day#\n",
+ "B=A/10.0**9 #in tonnes#\n",
+ "J1=90/100.0 #purity of lime#\n",
+ "J2=95/100.0 #purity of soda#\n",
+ "C1=500.0 #cost of one tonne lime#\n",
+ "C2=7000.0 #cost of one tonne soda#\n",
+ "CL=round(T,1)*C1/J1 \n",
+ "print\"\\ncost of lime is\",CL,\"Rs\"\n",
+ "CS=round(B,1)*C2/J2 \n",
+ "print\"\\ncost of soda is \",CS,\"Rs\"\n",
+ "C=CL+CS \n",
+ "\n",
+ "#Result\n",
+ "print\"\\ntotal cost is \",round(C) ,\"Rs\"\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "cost of lime is 19166.6666667 Rs\n",
+ "\n",
+ "cost of soda is 141473.684211 Rs\n",
+ "\n",
+ "total cost is 160640.0 Rs\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.10,Page no:176"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=40.0 #amount of Ca2+ in water in mg/l#\n",
+ "W2=24.0 #amount of Mg2+ in water in mg/l#\n",
+ "W3=8.05 #amount of Na+ in water in mg/l#\n",
+ "W4=183.0 #amount of (HCO3)- in water in mg/l#\n",
+ "W5=55.68 #amount of (SO4)2- in water in mg/l#\n",
+ "W6=6.74 #amount of Cl- in water in mg/l#\n",
+ "M1=100/40.0 #multiplication factor of Ca2+#\n",
+ "M2=100/24.0 #multiplication factor of Mg2+#\n",
+ "M3=100/(23.0*2) #multiplication factor of Na+#\n",
+ "M4=100/(61.0*2) #multiplication factor of (HCO3)-#\n",
+ "M5=100/96.0 #multiplication factor of (SO4)2-#\n",
+ "M6=100/(35.5*2) #multiplication factor of Cl-#\n",
+ "\n",
+ "#Calculation\n",
+ "P1=W1*M1 #in terms of CaCO3#\n",
+ "P2=W2*M2 #in terms of CaCO3#\n",
+ "P3=W3*M3 #in terms of CaCO3#\n",
+ "P4=W4*M4 #in terms of CaCO3#\n",
+ "P5=W5*M5 #in terms of CaCO3#\n",
+ "P6=W6*M6 #in terms of CaCO3#\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nCalcium alkalinity =\",P1,\"ppm\" \n",
+ "print\"\\nMagnesium alkalinity =\",P4-P1,\"ppm\"\n",
+ "print\"\\n total alkalinity = \",P1+P4-P1,\"ppm\"\n",
+ "print\"\\n total hardness = \",P1+P2,\"ppm\"\n",
+ "print\"\\nCa temporary hardness = \",P1,\"ppm\"\n",
+ "print\"\\nMg temporary hardness = \",P4-P1,\"ppm\"\n",
+ "print\"\\nMg permanant hardness = \",P2-(P4-P1),\"ppm\"\n",
+ "print\"\\nSalts are:\"\n",
+ "print\"\\nCa(HCO3)2 salt = \",P1,\"ppm\"\n",
+ "print\"\\nMg(HCO3)2 salt = \",P4-P1,\"ppm\"\n",
+ "print\"\\nMgSO4 salt = \",P2-(P4-P1),\"ppm\"\n",
+ "print\"\\nNaCl salt = \",P6,\"ppm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Calcium alkalinity = 100.0 ppm\n",
+ "\n",
+ "Magnesium alkalinity = 50.0 ppm\n",
+ "\n",
+ " total alkalinity = 150.0 ppm\n",
+ "\n",
+ " total hardness = 200.0 ppm\n",
+ "\n",
+ "Ca temporary hardness = 100.0 ppm\n",
+ "\n",
+ "Mg temporary hardness = 50.0 ppm\n",
+ "\n",
+ "Mg permanant hardness = 50.0 ppm\n",
+ "\n",
+ "Salts are:\n",
+ "\n",
+ "Ca(HCO3)2 salt = 100.0 ppm\n",
+ "\n",
+ "Mg(HCO3)2 salt = 50.0 ppm\n",
+ "\n",
+ "MgSO4 salt = 50.0 ppm\n",
+ "\n",
+ "NaCl salt = 9.49295774648 ppm\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.11,Page no:177"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P=0.0 #phenolplthalein alkalinity in water sample#\n",
+ "V=16.9 #required HCl in ml for 100 ml water sample#\n",
+ "N=0.02 #normality of HCl#\n",
+ "print\"Since P=0 the alkalinity is due to HCO3- ions\" \n",
+ "C=50.0 #equivalent of CaCO3 in mg for 1 ml 1N of HCl#\n",
+ "\n",
+ "#Calculation\n",
+ "A=C*V*N \n",
+ "print\"\\nIn 100ml water sample the alkalinity is\",A,\"mg/s\"\n",
+ "B=A*1000.0/100.0\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nFor 1 litre of water the alkalinity is \",B,\"mg/l\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Since P=0 the alkalinity is due to HCO3- ions\n",
+ "\n",
+ "In 100ml water sample the alkalinity is 16.9 mg/s\n",
+ "\n",
+ "For 1 litre of water the alkalinity is 169.0 mg/l\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.12,Page no:178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P=4.7 #required HCl in ml using HpH indicator #\n",
+ "H=10.5 #required HCl im ml using MeOH indicator#\n",
+ "M=P+H \n",
+ "N=0.02 #normality of HCl#\n",
+ "\n",
+ "print\"\\nSince P<0.5*M sample contain (CO3)2- and (HCO3)- alkalinity\"\n",
+ "C=50 #equivalent of CaCO3 in mg for 1ml 1N HCl#\n",
+ "\n",
+ "#Calculation\n",
+ "A=C*(2*P)*N #amount of (CO3)2- alkalinity in mg in 100 ml of water#\n",
+ "B=A*1000/100 \n",
+ "D=C*(M-2*P)*N #the amount of (HCO3)- alkalinity in mg in 100 ml of water#\n",
+ "E=D*1000/100 \n",
+ "T=B+E \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nTotal alkalinity is \",T,\"mg/l or ppm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Since P<0.5*M sample contain (CO3)2- and (HCO3)- alkalinity\n",
+ "\n",
+ "Total alkalinity is 152.0 mg/l or ppm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.13,Page no:178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=160.0 #amount of Ca2+ in ppm#\n",
+ "W2=88.0 #amount of Mg2+ in ppm#\n",
+ "W3=72.0 #amount of CO2 in ppm#\n",
+ "W4=488.0 #amount of (HCO3)- in ppm#\n",
+ "W5=139.0 #amount of (FeSO4).7H2O in ppm#\n",
+ "M1=100/40.0 #multiplication factor of Ca2+#\n",
+ "M2=100/24.0 #multiplication factor of Mg2+#\n",
+ "M3=100/44.0 #multiplication factor of CO2#\n",
+ "M4=100/(61.0*2.0) #multiplication factor of (HCO3)-#\n",
+ "M5=100/278.0 #multiplication factor of (FeSO4).7H2O#\n",
+ "\n",
+ "P1=400 #in terms of CaCO3#\n",
+ "P2=300 #in terms of CaCO3#\n",
+ "P3=200 #in terms of CaCO3#\n",
+ "P4=400 #in terms of CaCO3#\n",
+ "P5=50 #in terms of CaCO3#\n",
+ "V=100000.0 #volume of water in litres#\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "L=0.74*(P2+P3+P4+P5)*V #lime required in mg#\n",
+ "L=L/10.0**6 \n",
+ "S=1.06*(P1+P2+P5-P4)*V #soda required in mg#\n",
+ "S=S/10.0**6 \n",
+ "\n",
+ "#Result\n",
+ "print\"Lime required is \",L,\"kg\"\n",
+ "print\"\\nSoda required is \",S,\"kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Lime required is 70.3 kg\n",
+ "\n",
+ "Soda required is 37.1 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.14,Page no:179"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W=50 #amount of NaCl in g/l in NaCl solution#\n",
+ "V=200 #volume of NaCl solution in litres#\n",
+ "\n",
+ "#Calculation\n",
+ "A=W*V \n",
+ "V=10000 #volume of hard water passed through Zeolite softener#\n",
+ "M=100/(58.5*2) #multiplication factor of NaCl#\n",
+ "P=M*A \n",
+ "B=P*1000/V \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nIn terms of CaCO3=\",round(P),\"g CaCO3\"\n",
+ "print\"\\nFor 1 litre of hard water=\",round(B,1),\"mg/l or ppm\"\n",
+ "print\"NOTE:In book answer wrongly written as 845.7\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "In terms of CaCO3= 8547.0 g CaCO3\n",
+ "\n",
+ "For 1 litre of hard water= 854.7 mg/l or ppm\n",
+ "NOTE:In book answer wrongly written as 845.7\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.15,Page no:179"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=0.28 #amount of CaCO3 in grams dissolved in 1 litre of water#\n",
+ "V1=28 #required EDTA in ml on titration of 100ml of CaCO3 solution#\n",
+ "V2=33 #required EDTA in ml for 100ml of unknown hard water sample#\n",
+ "V3=10 #required EDTA in ml for 100 ml of unknown sample after boiling and cooling#\n",
+ "M1=100/100 #multiplication factor of CaCO3#\n",
+ "\n",
+ "#Calculation\n",
+ "C=W1*M1 \n",
+ "A=C*100#for 100 ml of sample equivalent to 28 ml of EDTA#\n",
+ "B=A/V1 \n",
+ "D=V2*B #for 100 ml#\n",
+ "D=D*1000/100 \n",
+ "E=V3*B #for 100 ml#\n",
+ "E=E*1000/100 \n",
+ "T=D-E \n",
+ "\n",
+ "#Result\n",
+ "print\"Total hardness is \",D,\"mg CaCO3 eq\"\n",
+ "print\"\\nPermanant hardness is \",E,\"mg CaCO3 eq\"\n",
+ "print\"\\nTemporary hardness is \",T,\"mg CaCO3\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total hardness is 330.0 mg CaCO3 eq\n",
+ "\n",
+ "Permanant hardness is 100.0 mg CaCO3 eq\n",
+ "\n",
+ "Temporary hardness is 230.0 mg CaCO3\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Textbook_of_Engineering_Chemistry/dChapter_8.ipynb b/Textbook_of_Engineering_Chemistry/dChapter_8.ipynb
new file mode 100755
index 00000000..06e61c83
--- /dev/null
+++ b/Textbook_of_Engineering_Chemistry/dChapter_8.ipynb
@@ -0,0 +1,979 @@
+{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8: Fuels & Combustion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:1,Page no:188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "WC=1.508 #weight of coal sample in grams#\n",
+ "WH110=1.478 #weight of sample after heating at 110 degrees in grams#\n",
+ "m=WC-WH110 #weight of moisture in the sample#\n",
+ "\n",
+ "#Calculation\n",
+ "pm=m*100/WC #percentage of moisture in the sample#\n",
+ "WH950=1.068 #weight of sample after heating at 950 degrees in grams#\n",
+ "vm=WH110-WH950 #volatile matter in grams#\n",
+ "pvm=vm*100/WC #percentage of voltaile matter#\n",
+ "\n",
+ "#Result\n",
+ "print'Weight of moisture in the sample=',m,\"g\"\n",
+ "print'Percentage of moisture in the sample=',round(pm,2),\"%\"\n",
+ "print'\\nWeight of volatile matter in the sample=',vm,\"g\"\n",
+ "print'Percentage of volatile matter in the sample=',round(pvm,2),\"%\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Weight of moisture in the sample= 0.03 g\n",
+ "Percentage of moisture in the sample= 1.99 %\n",
+ "\n",
+ "Weight of volatile matter in the sample= 0.41 g\n",
+ "Percentage of volatile matter in the sample= 27.19 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2,Page no:192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "CR=7.8 #compression ratio for first case#\n",
+ "E1=1.0-(1.0/CR)**0.258 #Energy efficiency corresponding to CR value 7.8#\n",
+ "CR=9.5 #compreesion ratio for second case#\n",
+ "\n",
+ "#Calculation\n",
+ "E2=1.0-(1.0/CR)**0.258 #Energy efficiency corresponding to CR value 9.5#\n",
+ "IE=E2-E1 #Increase in efficiency#\n",
+ "PIE=round(IE,2)*100.0/round(E2,3) #percentage of increase in efficiency#\n",
+ "\n",
+ "#Result\n",
+ "print'\\nIncrease in efficiency=IE=%f',round(IE,2)\n",
+ "print'\\nPercentage of increase in efficiency=PIE=%f',PIE,\"%\"\n",
+ "print \"NOTE:Calculation mistake in book for % increase\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Increase in efficiency=IE=%f 0.03\n",
+ "\n",
+ "Percentage of increase in efficiency=PIE=%f 6.80272108844 %\n",
+ "NOTE:Calculation mistake in book for % increase\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3,Page no:196"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=3.0 #weight of carbon in 1kg of coal sample in Kimath.lograms#\n",
+ "WO2=C*32/12.0 #weight of oxygen in carbon sample in Kimath.lograms#\n",
+ "\n",
+ "#Calculation\n",
+ "WA=WO2*100/23.0 #weight of air in the carbon sample in Kimath.lograms#\n",
+ "MA=WA/28.92 #mol of air in kimath.lograms#\n",
+ "VA=MA*22.4 #Volume of air required in m3 air#\n",
+ "\n",
+ "#Result\n",
+ "print'weight of air required for combustion of carbon=',round(WA,1),\"kg\"\n",
+ "print'\\nVolume of air required=',round(VA,1),\"m**3 air\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "weight of air required for combustion of carbon= 34.8 kg\n",
+ "\n",
+ "Volume of air required= 26.9 m**3 air\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:4,Page no:196"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "CO=0.46 #volume of carbon monoxide in 1kg of gas sample in m3#\n",
+ "C2H2=0.020 #volume of C2H2 in 1kg of gas sample in m3#\n",
+ "CH4=0.1 #volume of CH4 in 1kg of gas sample in m3#\n",
+ "N2=0.01 #volume of nytrogen in 1kg of gas sample in m3#\n",
+ "H2=0.40 #volume of hydrogen in 1kg of gas sample in m3#\n",
+ "\n",
+ "#Calculation\n",
+ "VA=0.68*(100/21.0) #volume of air needed in m3#\n",
+ "\n",
+ "#Result\n",
+ "print'\\nVolume of air needed=',round(VA,3),\"m**3\"\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Volume of air needed= 3.238 m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5,Page no:197"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=624.0 #weight of carbon in 1kg of coal sample in grams#\n",
+ "O=69.0 #weight of oxygen in 1kg of coal sample in grams#\n",
+ "S=8.0 #weight of Sulphur in 1kg of coal sample in grams#\n",
+ "N=12.0 #weight of nytrogen in 1kg of coal sample in grams#\n",
+ "H=41.0 #weight of hydrogen in 1kg of coal sample in grams#\n",
+ "CO2=129 #weight of CO2 in 1kg of coal sample in grams#\n",
+ "CO=2.0 #weight of CO in 1kg of coal sample in grams#\n",
+ "\n",
+ "#Calculation\n",
+ "MO=C*32/12.0+H*16/2.0+S*32/32.0-O #minimum weight of oxygen needed in grams#\n",
+ "MA=MO*0.1/23 #minimum weight of air needed in kimath.lograms#\n",
+ "\n",
+ "WC=CO2*(12/44.0)+CO*(12/28.0) #weight of C in fuel gas/kg#\n",
+ "WF=C/WC #Weight of fuel gas/kg of coal in g#\n",
+ "O2=2*16/28.0 #O2 needed to convert CO to CO2 in Kg#\n",
+ "RWO2=(61.0-O2)/1000.0 #remaining weight of O2/kg of fuel gas in Kg#\n",
+ "WO2=WF*RWO2 #weight of O2 obtained by burning 1kg coal in kg#\n",
+ "AR=WO2*100/23.0 #air required in kimath.lograms#\n",
+ "WAS=MA+AR #weight of air actually supplied/kg coal burnt in kg#\n",
+ "\n",
+ "#Result\n",
+ "print'(i) Weight of air theoretically needed=',round(MA,3),\"kg\"\n",
+ "print'\\n(ii) Weight of C in fuel gas/kg=',round(WC ,2),\"g\"\n",
+ "print'\\n Weight of fuel gas/kg of coal=',round(WF,3),\"kg\"\n",
+ "print'\\n(iii) Weight of air actually supplied/kg coal burnt',round(WAS,1),\"kg air\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Weight of air theoretically needed= 8.396 kg\n",
+ "\n",
+ "(ii) Weight of C in fuel gas/kg= 36.04 g\n",
+ "\n",
+ " Weight of fuel gas/kg of coal= 17.315 kg\n",
+ "\n",
+ "(iii) Weight of air actually supplied/kg coal burnt 12.9 kg air\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:6,Page no:202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "w=1080.0 #quantity of water in grams#\n",
+ "W=150.0 #Water equivalent of calorimeter in grams#\n",
+ "x=0.681 #quantity of fuel carried out in combustion in grams#\n",
+ "dt=3.61 #rise in temperature of water in degree C#\n",
+ "\n",
+ "#Calculation\n",
+ "Q=(w+W)*(dt)/x #calorific value of the fuel in cal per grams#\n",
+ "\n",
+ "#Result\n",
+ "print'Calorific value of the fuel=',round(Q,1),\"cal/g =\",round(Q/1000,3),\"kcal/g\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Calorific value of the fuel= 6520.3 cal/g = 6.52 kcal/g\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7,Page no:202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "w=1080.0 #quantity of water taken in grams#\n",
+ "W=150.0 #Water equivalent of calorimeter in grams#\n",
+ "m=0.681 #weight of coal taken or mass of fuel in grams#\n",
+ "dt=3.61 #rise in temperature of water in degree C#\n",
+ "AC=50.0 #Acid correction in calories#\n",
+ "FC=5.0 #Fuse wire correction in calories#\n",
+ "CC=0.05 #cooling correction in calories#\n",
+ "\n",
+ "#Calculation\n",
+ "GCV=((w+W)*(dt+CC)-(AC+FC))/m #Gross calorific value of the sample in cal per grams#\n",
+ "\n",
+ "#Result\n",
+ "print'Gross Calorific value of the fuel=',round(GCV,1),\"cal/g =\",round(GCV/1000,3),\"kcal/g\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Gross Calorific value of the fuel= 6529.8 cal/g = 6.53 kcal/g\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8,Page no:203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=90.2 #percentage of carbon#\n",
+ "O=2.9 #percentage of oxygen#\n",
+ "H=2.40 #percentage of hydrogen#\n",
+ "\n",
+ "#Calculation\n",
+ "GCV=(8080.0*C+34400.0*(H-O/8.0))/100.0 #Gross calorific value of the sample in cal per grams#\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nGross Calorific value of the fuel=%.2e\"%GCV,\"cal/g\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Gross Calorific value of the fuel=7.99e+03 cal/g\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:9,Page no:210"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "a=0.9 #absorptivity#\n",
+ "e=0.04 #emissivity#\n",
+ "P=750 #Sun light energy available in W/m2#\n",
+ "Q=5.67*10**-8 #conductivity \n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "T4=a*P/(Q*e) \n",
+ "T=math.pow(T4,1.0/4.0)\n",
+ "\n",
+ "#Result\n",
+ "print'Maximum temeperature that can be achieved=',round(T,1),\"K\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum temeperature that can be achieved= 738.6 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.1,Page no:212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "w=1500 #quantity of water in grams#\n",
+ "W=125 #Water equivalent of calorimeter in grams#\n",
+ "x=1.050 #quantity of fuel carried out in combustion in grams#\n",
+ "t1=25 #initial temperature of water in degree C#\n",
+ "t2=27.8 #final temperature of water in degree C#\n",
+ "\n",
+ "#Calculation\n",
+ "Q=(w+W)*(t2-t1)/x #calorific value of the fuel in cal per grams#\n",
+ "\n",
+ "#Result\n",
+ "print'Calorific value of the fuel=',round(Q/1000,1),\"kcal/g\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Calorific value of the fuel= 4.3 kcal/g\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.2,Page no:212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from scipy.optimize import fsolve\n",
+ "\n",
+ "#Variable declaration\n",
+ "C=90.0 #percentage of carbon#\n",
+ "O=3.0 #percentage of oxygen#\n",
+ "S=0.5 #percentage of sulphur#\n",
+ "N=0.5 #percentage of nytrogen#\n",
+ "LCV=8500 #Law calorific value#\n",
+ "#Calculation\n",
+ "def f(H):\n",
+ " GCV1=LCV+(9*H/100.0)*587\n",
+ " GCV2=(8080.0*C+34500*(H-O/8.0)+2240*S)/100.0 #Gross calorific value of the sample in cal per grams#\n",
+ " return(GCV1-GCV2)\n",
+ "h=fsolve(f,1)\n",
+ "GCV=LCV+(9*h/100.0)*587\n",
+ "\n",
+ "#Result\n",
+ "print'percentage of hydrogen=H=',round(h[0],1),\"%\"\n",
+ "print'\\nGross Calorific value of the fuel=',round(GCV[0]),\"kcal/g\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "percentage of hydrogen=H= 4.6 %\n",
+ "\n",
+ "Gross Calorific value of the fuel= 8743.0 kcal/g\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.3,Page no:213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "w=500.0 #quantity of water taken in grams#\n",
+ "W=2000.0 #Water equivalent of calorimeter in grams#\n",
+ "m=1.000 #weight of coal taken or mass of fuel in grams#\n",
+ "t1=24.0 #initial temperature of water in degree C#\n",
+ "t2=26.2 #final temperature of water in degree C#\n",
+ "AC=50.0 #Acid correction in calories#\n",
+ "FC=10.0 #Fuse wire correction in calories#\n",
+ "CC=0.0 #cooling correction in calories#\n",
+ "\n",
+ "#Calculation\n",
+ "GCV=((w+W)*(t2-t1+CC)-(AC+FC))/m #Gross calorific value of the sample in cal per grams#\n",
+ "H=6.0 #percentage of hydrogen#\n",
+ "C=93.0 #percentage of carbon#\n",
+ "LCV=GCV-(9*H*580/100.0) #Net calorific value of the sample in cal per gram#\n",
+ "\n",
+ "#Result\n",
+ "print'Gross Calorific value of the fuel=',GCV,\"cal/g\"\n",
+ "print'Net calorific value of the sample=LCV=',LCV,\"cal/g\"\n",
+ "print\"\\nNOTE:Calculation mistake in book,wrongly written as 5540 and 5226.8\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Gross Calorific value of the fuel= 5440.0 cal/g\n",
+ "Net calorific value of the sample=LCV= 5126.8 cal/g\n",
+ "\n",
+ "NOTE:Calculation mistake in book,wrongly written as 5540 and 5226.8\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.4,Page no:213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "WC=1.5642 #weight of coal sample in grams#\n",
+ "WH110=1.5022 #weight of sample after heating at 110 degrees in grams#\n",
+ "\n",
+ "#Calculation\n",
+ "m=WC-WH110 #weight of moisture in the sample#\n",
+ "pm=m*100/WC #percentage of moisture in the sample#\n",
+ "WH950=0.7628 #weight of sample after heating at 950 degrees in grams#\n",
+ "vm=WH110-WH950 #volatile matter in grams#\n",
+ "pvm=vm*100/WC #percentage of voltaile matter#\n",
+ "ac=0.2140 #Ash content left in the last in grams#\n",
+ "pac=ac*100/WC #percentage of Ash content laft#\n",
+ "pfc=100-(pm+pvm-pac) #percentage of fixaed carbon#\n",
+ "\n",
+ "#Result\n",
+ "print'\\npercentage of fixed carbon in the sample=',round(pfc,2),\"%\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "percentage of fixed carbon in the sample= 62.45 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.5,Page no:214"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "WBaSO4=0.0482 #weight of BaSO4 in grams#\n",
+ "W=0.5248 #weight of sample in grams#\n",
+ "\n",
+ "#Calculation\n",
+ "PS=32*WBaSO4*100/(233*W) #percentage of sulphur in the sample#\n",
+ "\n",
+ "#Result\n",
+ "print'Percentage of sulphur in the sample=',round(PS,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage of sulphur in the sample= 1.26 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.6,Page no:215"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W=10 #weight of Water heated of calorimeter in Kimath.lograms#\n",
+ "V=0.1 #volume of gas used in metrecube#\n",
+ "t1=22 #inlet temperature of water in degree C#\n",
+ "t2=30 #outlet temperature of water in degree C#\n",
+ "#Calculation\n",
+ "GCV=W*(t2-t1)/V #Gross calorific value of the sample in Kilocal per metre3#\n",
+ "L=580 #latent heat of water in cal/g#\n",
+ "Ws=0.025 #weight of steam condensed in grams#\n",
+ "LCV=GCV-(Ws*L/V) #Net calorific value of the sample in Kcal per meter3#\n",
+ "\n",
+ "#Result\n",
+ "print'Gross Calorific value of the fuel=',GCV ,\"Kcal/m3\"\n",
+ "print'\\nNet calorific value of the sample=',LCV,\"Kcal/m3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Gross Calorific value of the fuel= 800.0 Kcal/m3\n",
+ "\n",
+ "Net calorific value of the sample= 655.0 Kcal/m3\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.7,Page no:215"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=90.0 #percentage of carbon#\n",
+ "O=3.0 #percentage of oxygen#\n",
+ "S=0.5 #percentage of sulphur#\n",
+ "N=0.5 #percentage of nytrogen#\n",
+ "H=3.5 #percentage of hydrogen#\n",
+ "H2O=0.1 #percentage of H2O#\n",
+ "#Calculation\n",
+ "AO=900*32.0/12.0+35*16.0/2.0+5*32.0/32.0 #amount of oxygen required in grams#\n",
+ "AN=2655*100/23.0 #amount of air needed in grams#\n",
+ "\n",
+ "#Result\n",
+ "print'\\nAmount of air needed=',round(AN/1000,3),\"kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Amount of air needed= 11.543 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.8,Page no:216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "CH4=0.14 #volume of CH4 in 1m3 volume of gaseous fuel in m3#\n",
+ "H2=0.32 #volume of H2 in 1m3 volume of gaseous fuel in m3#\n",
+ "N2=0.40 #volume of N2 in 1m3 volume of gaseous fuel in m3#\n",
+ "O2=0.14 #volume of O2 in 1m3 volume of gaseous fuel in m3#\n",
+ "\n",
+ "#Calculation\n",
+ "V_ch4=O2*2\n",
+ "v_H2=H2*0.5\n",
+ "Total_O2=V_ch4+v_H2\n",
+ "Net_O2=(Total_O2-O2)*1000 # Net O2 needed in L\n",
+ "V_req=Net_O2*(100/21.0)*(125/100.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Volume of air required assuming 21% =\",round(V_req,1),\"L\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Volume of air required assuming 21% = 1785.7 L\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.9,Page no:216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=750.0 #weight of carbon in 1kg of coal sample in grams#\n",
+ "O=121.0 #weight of oxygen in 1kg of coal sample in grams#\n",
+ "A=45.0 #weight of Ash in 1kg of coal sample in grams#\n",
+ "N=32.0 #weight of nytrogen in 1kg of coal sample in grams#\n",
+ "H=52.0 #weight of hydrogen in 1kg of coal sample in grams#\n",
+ "\n",
+ "#Calculation\n",
+ "MO=C*32/12+H*16/2-O #minimum weight of oxygen needed in grams#\n",
+ "MA=MO*100/23 #minimum weight of air needed in grams#\n",
+ "GCV=(808*C+3450*(H-O/8))/100 #Gross calorific value of the sample in cal per grams#\n",
+ "LCV=GCV-0.09*H*0.1*587 #law calorific value of the sample in cal/gram#\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print'\\nGross Calorific value of the fuel=',round(GCV,2),\"kcal/g\"\n",
+ "print'\\nNet calorific value of the sample',round(LCV),\"kcal/g\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Gross Calorific value of the fuel= 7332.19 kcal/g\n",
+ "\n",
+ "Net calorific value of the sample 7057.0 kcal/g\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.10,Page no:217"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=810.0 #weight of carbon in 1kg of coal sample in grams#\n",
+ "O=80.0 #weight of oxygen in 1kg of coal sample in grams#\n",
+ "S=10.0 #weight of Sulphur in 1kg of coal sample in grams#\n",
+ "N=10.0 #weight of nytrogen in 1kg of coal sample in grams#\n",
+ "H=50.0 #weight of hydrogen in 1kg of coal sample in grams#\n",
+ "\n",
+ "#Calculation\n",
+ "MO=C*32/12.0+H*16/2.0+S*32/32.0 #minimum weight of oxygen needed in grams#\n",
+ "MA=2490*100/23.0 #minimum weight of air needed in grams#\n",
+ "print'\\nminimum amount of air needed=',round(MA/1000,3) ,\"kg\"\n",
+ "NF=10+MA*0.77 #weight of nitrogen present in the products in grams#\n",
+ "WD=2970.0+20.0+8346.0 #total weight of dry products in grams#\n",
+ "PCO2=2970*100/WD #percentage composition of CO2#\n",
+ "print'\\nPercentage composition of CO2=',round(PCO2 ,2),\"%\"\n",
+ "PSO2=20*100/WD #percentage composition of SO2#\n",
+ "print'\\nPercentage composition of SO2=',round(PSO2 ,3),\"%\"\n",
+ "PN2=8346*100/WD #percentage composition of N2#\n",
+ "\n",
+ "#Result\n",
+ "print'\\nPercentage composition of N2=',round(PN2,2),\"%\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "minimum amount of air needed= 10.826 kg\n",
+ "\n",
+ "Percentage composition of CO2= 26.2 %\n",
+ "\n",
+ "Percentage composition of SO2= 0.176 %\n",
+ "\n",
+ "Percentage composition of N2= 73.62 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.11,Page no:219"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "CO=0.205 #volume of carbon monoxide in 1kg of gas sample in m3#\n",
+ "CO2=0.060 #volume of CO2 in 1kg of gas sample in m3#\n",
+ "CH4=0.042 #volume of CH4 in 1kg of gas sample in m3#\n",
+ "N=0.501 #volume of nytrogen in 1kg of gas sample in m3#\n",
+ "H2=0.194 #volume of hydrogen in 1kg of gas sample in m3#\n",
+ "\n",
+ "#Calculation\n",
+ "VA=0.283*(100/21)*(130/100) #volume of air needed in m3#\n",
+ "VDCO2=0.06+0.205*1+0.042*1 #volume of dry products containig CO2 formed in m3#\n",
+ "VDN2=0.501+1.752*79/100 #volume of dry products containig N2 formed in m3#\n",
+ "VDO2=1.755*21/100 #volume of dry products containig O2 formed in m3#\n",
+ "TVD=VDCO2+VDN2+VDO2 #total volume of dry products formed in m3#\n",
+ "PDCO2=VDCO2*100/TVD #percentage of dry products containig CO2 formed#\n",
+ "PDN2=VDN2*100/TVD #percentage of dry products containig N2 formed#\n",
+ "PDO2=VDO2*100/TVD #percentage of dry products containig O2 formed#\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print'\\nPercentage of dry products containing CO2 formed=',round(PDCO2,2),\"%\"\n",
+ "print'\\nPercentage of dry products containing N2 formed=',round(PDN2 ,2),\"%\"\n",
+ "print'\\nPercentage of dry products containing O2 formed=',round(PDO2 ,2),\"%\"\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Percentage of dry products containing CO2 formed= 11.99 %\n",
+ "\n",
+ "Percentage of dry products containing N2 formed= 73.62 %\n",
+ "\n",
+ "Percentage of dry products containing O2 formed= 14.39 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.12,Page no:220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=780.0 #weight of carbon in 1kg of coal sample in grams#\n",
+ "O=120.0 #weight of oxygen in 1kg of coal sample in grams#\n",
+ "S=12.0 #weight of Sulphur in 1kg of coal sample in grams#\n",
+ "N=21.0 #weight of nytrogen in 1kg of coal sample in grams#\n",
+ "H=41.0 #weight of hydrogen in 1kg of coal sample in grams#\n",
+ "\n",
+ "#Calculation\n",
+ "MO=C*32/12+H*16/2+S*32/32-O #minimum weight of oxygen needed in grams#\n",
+ "MA=MO*100/23.0 #minimum weight of air needed in grams#\n",
+ "\n",
+ "#Result\n",
+ "print'minimum weight of oxygen needed=',MO/1000,\"kg\"\n",
+ "print'\\nminimum amount of air needed=',MA/1000,\"kg\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "minimum weight of oxygen needed= 2.3 kg\n",
+ "\n",
+ "minimum amount of air needed= 10.0 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.13,Page no:220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=1.5 #weight of carbon in 1kg of coal sample in Kimath.lograms#\n",
+ "\n",
+ "#Calculation\n",
+ "WO2=C*32/12 #weight of oxygen in carbon sample in Kimath.lograms#\n",
+ "WA=WO2*100/23 #weight of air in the carbon sample in Kimath.lograms#\n",
+ "O2_4000=22.4/32.0*4000 #Volume in 4000 g oxygen\n",
+ "V=100/21.0*O2_4000 #Volume of air with 21 % O2\n",
+ "\n",
+ "#Result\n",
+ "print'\\nweight of air in the carbon sample=',round(WA,2),\"kg\" \n",
+ "print \"Volume of air is\",round(V/1000,2),\"m^3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "weight of air in the carbon sample= 17.39 kg\n",
+ "Volume of air is 13.33 m^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
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