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Diffstat (limited to 'Elements_of_Physical_Chemistry_by_Atkins_Peter')
23 files changed, 4098 insertions, 0 deletions
diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter1.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter1.ipynb new file mode 100755 index 00000000..6a51c37c --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter1.ipynb @@ -0,0 +1,216 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ " Chapter 1 - The properties of gases"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the pressure in the glass flask\n",
+ "#Initialzation of variables\n",
+ "m=1.25 #g\n",
+ "MN2=28.02 #g/mol\n",
+ "T=20+273.15 #K\n",
+ "V=0.25#L\n",
+ "#Calculations\n",
+ "P=m*8.31451*T/(MN2*V)\n",
+ "#Results\n",
+ "print '%s %.1f %s' %('Pressure in the gas flask = ',P,'kPa')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pressure in the gas flask = 434.9 kPa\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the partial pressures of Oxygen, Nitrogen and Argon\n",
+ "#Initialzation of variables\n",
+ "xN2=0.780\n",
+ "xO2=0.210\n",
+ "xAr=0.009\n",
+ "P=100 #kPa\n",
+ "#Calculations\n",
+ "PN2=xN2*P\n",
+ "PO2=xO2*P\n",
+ "PAr=xAr*P\n",
+ "#Results\n",
+ "print '%s %.1f' %('Partial pressure of Nitrogen(kPa) = ',PN2)\n",
+ "print '%s %.1f' %('\\n Partial pressure of Oxygen(kPa) = ',PO2)\n",
+ "print '%s %.1f' %('\\n Partial pressure of Argon(kPa) = ',PAr)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Partial pressure of Nitrogen(kPa) = 78.0\n",
+ "\n",
+ " Partial pressure of Oxygen(kPa) = 21.0\n",
+ "\n",
+ " Partial pressure of Argon(kPa) = 0.9\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I3 - Pg 22"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the percentage loss of speed of air molecules\n",
+ "#Initialzation of variables\n",
+ "import math\n",
+ "T1=298. #K\n",
+ "T2=273. #K\n",
+ "#Calculations\n",
+ "factor=math.sqrt(T2/T1)\n",
+ "percentage=(1-factor)*100\n",
+ "#Results\n",
+ "print '%s %.3f' %('Factor by which speed is reduced = ',factor)\n",
+ "print '%s %d' %('Percentage loss of speed of air molecules = ',percentage)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Factor by which speed is reduced = 0.957\n",
+ "Percentage loss of speed of air molecules = 4\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I4 - Pg 24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the ratio of rates of effusion\n",
+ "#Initialzation of variables\n",
+ "import math\n",
+ "MH2=2.016 #g/mol\n",
+ "MCO2=44.01 #g/mol\n",
+ "#calculations\n",
+ "ratio=math.sqrt(MCO2/MH2)\n",
+ "#results\n",
+ "print '%s %.3f' %('ratio of rates of effusion = ',ratio)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ratio of rates of effusion = 4.672\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I5 - Pg 25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the mean free path and collision frequency\n",
+ "#Initialzation of variables\n",
+ "import math\n",
+ "T=25+273. #K\n",
+ "sigma=0.4*math.pow(10,(-18)) #m^2\n",
+ "P=math.pow(10,5) #Pa\n",
+ "c=481.8 #m/sec\n",
+ "#Calculations\n",
+ "Lambda=8.31451*T/(math.pow(2,0.5) *6.022*math.pow(10,23) *sigma*P)\n",
+ "frequency=math.pow(2,0.5) *6.022*math.pow(10,23) *sigma*P*c/(8.31451*T)\n",
+ "#Results\n",
+ "print '%s %.1e %s' %('Mean free path =',Lambda,'m')\n",
+ "print '%s %.1e %s' %('\\n Collision frequency =',frequency,'m')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mean free path = 7.3e-08 m\n",
+ "\n",
+ " Collision frequency = 6.6e+09 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter10.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter10.ipynb new file mode 100755 index 00000000..d07870df --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter10.ipynb @@ -0,0 +1,322 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ " Chapter 10 - The rates of reactions"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 222"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the overall rate constant and hence the overall rate law\n",
+ "#Initialization of variables\n",
+ "%matplotlib inline\n",
+ "import math\n",
+ "import numpy as np\n",
+ "from numpy import linalg\n",
+ "import matplotlib\n",
+ "from matplotlib import pyplot\n",
+ "def fun(x,n):\n",
+ "\tfor i in range(0,len(x)):\n",
+ "\t\tx[i]=x[i]*math.pow(10,-n)\n",
+ "\treturn\n",
+ "\n",
+ "I=([1., 2., 4., 6.])\n",
+ "r1=([1.070, 3.48, 13.9, 31.3])\n",
+ "r2=([4.35, 17.4, 69.6, 157])\n",
+ "r3=([10.69, 34.7, 138, 313])\n",
+ "Ar=([1., 5., 10.])\n",
+ "fun(r1,3)\n",
+ "fun(I,5)\n",
+ "fun(r2,3)\n",
+ "fun(r3,3)\n",
+ "fun(Ar,3)\n",
+ "#calculations\n",
+ "def fun1(x):\n",
+ "\tfor i in range(0,len(x)):\n",
+ "\t\tx[i]=math.log10(x[i])\n",
+ "\treturn x\n",
+ "\n",
+ "logI=fun1(I)\n",
+ "logr1=fun1(r1)\n",
+ "logr2=fun1(r2)\n",
+ "logr3=fun1(r3)\n",
+ "#The calculations are approximate.hence the value differs from textbook a bit.\n",
+ "x=logI\n",
+ "y=logr1\n",
+ "A = np.vstack([x, np.ones(len(x))]).T\n",
+ "m1, b1 = np.linalg.lstsq(A, y)[0]\n",
+ "\n",
+ "y=logr2\n",
+ "A = np.vstack([x, np.ones(len(x))]).T\n",
+ "m2, b2 = np.linalg.lstsq(A, y)[0]\n",
+ "\n",
+ "y=logr3\n",
+ "A = np.vstack([x, np.ones(len(x))]).T\n",
+ "m3, b3 = np.linalg.lstsq(A, y)[0]\n",
+ "\n",
+ "logAr=fun1(Ar)\n",
+ "kdash=([b1, b2, b3])\n",
+ "pyplot.plot(logAr,kdash)\n",
+ "pyplot.xlabel('log(Ar)')\n",
+ "pyplot.ylabel('log(kdash)')\n",
+ "x=logAr\n",
+ "y=kdash\n",
+ "\n",
+ "A = np.vstack([x, np.ones(len(x))]).T\n",
+ "m4, b4 = np.linalg.lstsq(A, y)[0]\n",
+ "\n",
+ "logk=b4\n",
+ "k=math.pow(10,logk)\n",
+ "#results\n",
+ "print '%s %.1e %s' %(\"Overall rate law is r =\",k,\" [I]^2 [Ar]\")\n",
+ "pyplot.show()\n",
+ "print '%s' %('The answers in the textbook are rounded and hence a bit different from the code.')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Populating the interactive namespace from numpy and matplotlib\n",
+ "Overall rate law is r = 8.1e+09 [I]^2 [Ar]"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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hISKeQkHhAAkJZj/i4YdhxQqoXdvVFYmIOI6mnirAMCAyEmbMgI8+ggcecHVF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\nc+SMsRQAAAAASUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x6097fd0>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The answers in the textbook are rounded and hence a bit different from the code.\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 224"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calcualte the rate constant for the given first order reaction\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy as np\n",
+ "import matplotlib\n",
+ "from matplotlib import pyplot\n",
+ "t=([0, 1000., 2000., 3000., 4000.])\n",
+ "p=[10.20, 5.72, 3.99, 2.78, 1.94]\n",
+ "def fun2(x):\n",
+ "\tfor i in range(0,len(x)):\n",
+ "\t\tx[i]=math.log(x[i])\n",
+ "\treturn x\n",
+ "lnp=fun2(p)\n",
+ "x=t\n",
+ "y=lnp\n",
+ "#hence the value differs from textbook a bit.\n",
+ "A = np.vstack([x, np.ones(len(x))]).T\n",
+ "m1, b1 = np.linalg.lstsq(A, y)[0]\n",
+ "k= -m1\n",
+ "pyplot.plot(x,y)\n",
+ "pyplot.xlabel('time t/sec ')\n",
+ "pyplot.ylabel('ln(P/Torr)')\n",
+ "#Since first order reaction\n",
+ "#results\n",
+ "print '%s %.2e %s' %(\"rate constant =\",k,\" s^-1\")\n",
+ "pyplot.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "rate constant = 4.04e-04 s^-1\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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VFoqCggK0adPG9NrHx8f0L4JSNBoN+vTpg/DwcPzzn/8EABQVFUGr1QIAtFot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\nRGaxUBARkVksFEREZBYLBRERmfX/emWTHUtSxxAAAAAASUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5d4f970>"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the Activation energy and Arrhenius factor\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy as np\n",
+ "from numpy import linalg\n",
+ "T=[700, 730, 760, 790, 810, 840, 910, 1000]\n",
+ "k=[0.011, 0.035, 0.105, 0.343, 0.789, 2.17, 20, 145]\n",
+ "#calculations\n",
+ "def fun1(x):\n",
+ "\tfor i in range(0,len(x)):\n",
+ "\t\tx[i]=1000./(x[i])\n",
+ "\treturn x\n",
+ "x= fun1(T)\n",
+ "def fun2(x):\n",
+ "\tfor i in range(0,len(x)):\n",
+ "\t\tx[i]=math.log(x[i])\n",
+ "\treturn x\n",
+ "y=fun2(k)\n",
+ "A = np.vstack([x, np.ones(len(x))]).T\n",
+ "m1, b1 = np.linalg.lstsq(A, y)[0]\n",
+ "print '%s' %('from graph')\n",
+ "Ea=-m1*8.3145/1000. *1000.\n",
+ "A=math.pow(math.e,(b1))\n",
+ "#results\n",
+ "print '%s %d %s' %(\"Activation energy =\",Ea,\" kJ/mol\")\n",
+ "print '%s %.2e %s' %(\"\\n Arrhenius factor =\",A,\"L/ mol s\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "from graph\n",
+ "Activation energy = 188 kJ/mol\n",
+ "\n",
+ " Arrhenius factor = 1.08e+12 L/ mol s\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I1 - Pg 228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the net time required\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "t=28.4 #min\n",
+ "#calculations\n",
+ "n=math.log10(8.) / math.log10(2.)\n",
+ "time=n*t\n",
+ "#results\n",
+ "print '%s %.1f %s' %(\"Time required =\",time,\"min\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time required = 85.2 min\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the value of kdash\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "E=50*1000. #J/mol\n",
+ "T1=25+273. #K\n",
+ "T2=37+273. #K\n",
+ "#calculations\n",
+ "ln=E/8.3145 *(1./T1-1./T2)\n",
+ "factor=math.pow(math.e,(ln))\n",
+ "#results\n",
+ "print '%s %.2f %s' %(\"kdash =\",factor,\"k\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "kdash = 2.18 k\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter11.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter11.ipynb new file mode 100755 index 00000000..5b2332ec --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter11.ipynb @@ -0,0 +1,236 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 11 - Accounting for the rate laws"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Max. velocity and Michaelis constant\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy as np\n",
+ "from numpy import linalg\n",
+ "S=[10, 20, 40, 80, 120, 180, 300]\n",
+ "v=[0.32, 0.58, 0.9, 1.22, 1.42, 1.58, 1.74]\n",
+ "#calculations\n",
+ "n=len(S)\n",
+ "def fun1(x):\n",
+ "\tfor i in range(0,len(x)):\n",
+ "\t\tx[i]=1000./(x[i])\n",
+ "\treturn x\n",
+ "\n",
+ "\n",
+ "def fun2(x):\n",
+ "\tfor i in range(0,len(x)):\n",
+ "\t\tx[i]=1./(x[i])\n",
+ "\treturn x\n",
+ "bys=fun1(S)\n",
+ "byv=fun2(v)\n",
+ "x=bys\n",
+ "y=byv\n",
+ "A = np.vstack([x, np.ones(len(x))]).T\n",
+ "m1, b1 = np.linalg.lstsq(A, y)[0]\n",
+ "print '%s' %(\"From graph,\")\n",
+ "vmax=1/b1\n",
+ "Km=m1*1000./b1\n",
+ "#results\n",
+ "print '%s %.2f %s' %(\"Max. velocity =\",vmax,\" mumol/L s\")\n",
+ "print '%s %.1f %s' %(\"\\n Michaelis constant =\",Km,\" mumol/L\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "From graph,\n",
+ "Max. velocity = 2.10 mumol/L s\n",
+ "\n",
+ " Michaelis constant = 55.0 mumol/L\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Equilibrium constant\n",
+ "#Initialization of variables\n",
+ "c=1.234\n",
+ "m=2.044\n",
+ "#calculations\n",
+ "Ki=c/m\n",
+ "#results\n",
+ "print '%s %.2f' %(\"KI = \",Ki)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "KI = 0.60\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 265"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the no. of diheptane molecules destroyed \n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "P=50. #J/s\n",
+ "l=313.*math.pow(10,-9) #m\n",
+ "h=6.62608*math.pow(10,-34) #Js\n",
+ "N=6.023*math.pow(10,23)\n",
+ "c=2.99792*math.pow(10,8) #m/s\n",
+ "yiel=0.21\n",
+ "#calculations\n",
+ "rate=P*l/(h*c)\n",
+ "Frate=yiel*rate\n",
+ "molrate=Frate/N\n",
+ "#results\n",
+ "print '%s %.1e %s' %(\"No.of diheptane molecules destroyed =\",Frate,\" s^-1\")\n",
+ "print '%s %.2e %s' %(\"\\n Moles of diheptane molecules destroyed =\",molrate,\"mol s^-1\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No.of diheptane molecules destroyed = 1.7e+19 s^-1\n",
+ "\n",
+ " Moles of diheptane molecules destroyed = 2.75e-05 mol s^-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I1 - Pg 243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Equilibrium constant for dimerization\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "kf=8.18*math.pow(10,8) #L/mol s\n",
+ "kb=2*math.pow(10,6) #s^-1\n",
+ "#calculations\n",
+ "K=kf/kb\n",
+ "#results\n",
+ "print '%s %.1e' %(\"Equilibrium constant for dimerization = \",K)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equilibrium constant for dimerization = 4.1e+02\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 248"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate if the reaction step is far from equilibrium and calculate the heat generated\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "F16bP=1.9*math.pow(10,-5) #mmol/L\n",
+ "ADP=1.3/1000. #mmol/L\n",
+ "ATP=11.4/1000. #mmol/L\n",
+ "F6P=8.9*math.pow(10,-5) #mmol/L\n",
+ "k=1.2*1000.\n",
+ "#calculations\n",
+ "Q=F16bP*ADP/(F6P*ATP)\n",
+ "if(Q<k):\n",
+ " print '%s %.3f' %(\"The reaction step is far from equilibrium and Q= \",Q)\n",
+ "else:\n",
+ " print '%s %.3f' %(\"The reaction step is at equilibrium and Q= \",Q)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reaction step is far from equilibrium and Q= 0.024\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter12.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter12.ipynb new file mode 100755 index 00000000..7ec0d96f --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter12.ipynb @@ -0,0 +1,239 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12 - Quantum theory"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 276"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the number of photons required\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "P=100. #W\n",
+ "t=10. #s\n",
+ "l=560. #nm\n",
+ "#calculations\n",
+ "TE=P*t\n",
+ "E1=6.626*math.pow(10,-34) *2.998*math.pow(10,8) /(l*math.pow(10,-9))\n",
+ "N=TE/E1\n",
+ "#results\n",
+ "print '%s %.2e' %(\"No. of photons required = \",N)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No. of photons required = 2.82e+21\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the wavelength of electrons\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "V=1000. #V\n",
+ "#calculations\n",
+ "l=6.626*math.pow(10,-34) /math.sqrt(2*9.11*math.pow(10,-31) *1.602*math.pow(10,-19) *V)\n",
+ "#results\n",
+ "print '%s %.2e %s' %(\"Wavelength of electrons =\",l,\" m\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of electrons = 3.88e-11 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the no. of times the electron would be more propable at r2 than at r1\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "r1=0 #multiply by a0\n",
+ "r2=1 #multiply by a0\n",
+ "#calculations\n",
+ "ratio=math.pow(math.e,r1) /math.pow(math.e,(-2.*r2))\n",
+ "#results\n",
+ "print '%s %.2f %s' %(\"It is more propable that electron would be found\",ratio,\"times more at r1\")\n",
+ "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "It is more propable that electron would be found 7.39 times more at r1\n",
+ "The answer is a bit different due to rounding off error in textbook\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 - Pg 284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the uncertainity in position\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "m=1 #g\n",
+ "v=math.pow(10,-6) #m/s\n",
+ "#calculations\n",
+ "dx=1.054*math.pow(10,-34) /(2*m*math.pow(10,-3) *v)\n",
+ "#results\n",
+ "print '%s %.1e %s' %(\"Uncertainity in position =\",dx,\" m\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uncertainity in position = 5.3e-26 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I1 - Pg 271"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the surface temperature\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "lmax=4.9*math.pow(10,-7) #m\n",
+ "#calculations\n",
+ "T=2.9*math.pow(10,-3) /lmax\n",
+ "#results\n",
+ "print '%s %d %s' %(\"Surface temperature must be close to\",T,\"K\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Surface temperature must be close to 5918 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 290"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the separation between adjacent levels frequency and energy\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "k=516. #N/m\n",
+ "m=1.67*math.pow(10,-27) #kg\n",
+ "#calculations\n",
+ "v=math.sqrt(k/m) /(2*math.pi)\n",
+ "E=6.624*math.pow(10,-34) *v\n",
+ "#results\n",
+ "print '%s %.2e %s' %(\"Separation between adjacent levels frequency,\",v,\"Hz\")\n",
+ "print '%s %.2e %s' %(\"\\n Energy =\",E,\"J\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Separation between adjacent levels frequency, 8.85e+13 Hz\n",
+ "\n",
+ " Energy = 5.86e-20 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter13.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter13.ipynb new file mode 100755 index 00000000..4a426e50 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter13.ipynb @@ -0,0 +1,97 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 13 - Atomic structure"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the probability of finding the electron and the chance that the electron would be found\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "dv=1 #pm^3\n",
+ "a0=52.9 #pm\n",
+ "#calculations\n",
+ "Probability=dv/(math.pi*a0*a0*a0)\n",
+ "#results\n",
+ "print '%s %.1e' %(\"probability of finding electron = \",Probability)\n",
+ "print '%s %d %s' %(\"\\n Chance that electron would be found is one in\",1./Probability,\"times\")\n",
+ "#The answer is a bit different due to rounding off error in textbook"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "probability of finding electron = 2.2e-06\n",
+ "\n",
+ " Chance that electron would be found is one in 465068 times\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I3 - Pg 304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the probability of finding the electron\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "dr=1 #pm\n",
+ "r=52.9 #pm\n",
+ "#calculations\n",
+ "Probability=4*math.pow(math.e,(-2)) *dr/r\n",
+ "#results\n",
+ "print '%s %.1f' %(\"About 1 inspection in \",1./Probability)\n",
+ "#The answer is a bit different due to rounding off error in textbook"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "About 1 inspection in 97.7\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter15.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter15.ipynb new file mode 100755 index 00000000..ada2a951 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter15.ipynb @@ -0,0 +1,140 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 15 - Metallic and Ionic Solids"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the lattice energy\n",
+ "#Initialization of variables\n",
+ "Hs=89 #kJ/mol\n",
+ "HI=418 #kJ/mol\n",
+ "HD=244 #kJ/mol\n",
+ "HE=-349 #kJ/mol\n",
+ "Hf=-437 #kJ/mol\n",
+ "#calculations\n",
+ "HL=Hs+HD/2. +HI+HE-Hf\n",
+ "#results\n",
+ "print '%s %d %s' %(\"Lattice energy =\",HL,\"kJ/mol\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Lattice energy = 717 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 369"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the separation between the molecules in both the cases\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "a=0.82 #nm\n",
+ "b=0.94 #nm\n",
+ "c=0.75 #nm\n",
+ "h=1.\n",
+ "k=2.\n",
+ "l=3.\n",
+ "#calculations\n",
+ "invd=math.sqrt(h*h/(a*a) + k*k/(b*b) + l*l/(c*c))\n",
+ "d=1./invd\n",
+ "invd2=math.sqrt(h*h*4/(a*a) + k*k*4/(b*b) + l*l*4/(c*c))\n",
+ "d2=1./invd2\n",
+ "#results\n",
+ "print '%s %.2f %s' %(\"In case 1, separation =\",d,\" nm\")\n",
+ "print '%s %.2f %s' %(\"\\n In case 2, separation =\",d2,\" nm\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "In case 1, separation = 0.21 nm\n",
+ "\n",
+ " In case 2, separation = 0.11 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 371"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the length of the side of the unit cell\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "l=154. #pm\n",
+ "theta=11.2 #degrees\n",
+ "#calculations\n",
+ "d=l/(2*math.sin(theta*math.pi/180.))\n",
+ "a=d*math.sqrt(3)\n",
+ "#results\n",
+ "print '%s %.1f %s' %(\"Length of the side of the unit cell =\",a,\"pm\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Length of the side of the unit cell = 686.6 pm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter16.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter16.ipynb new file mode 100755 index 00000000..c958cdd7 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter16.ipynb @@ -0,0 +1,170 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 16 - Molecular substances"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the net dipole moment\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "e=1.609*math.pow(10,-19) #C\n",
+ "#calculations\n",
+ "mux=(-0.36*e*(-0.8) + 0.45*e*(2.1) )*math.pow(10,-12) /(3.33564*math.pow(10,-30))\n",
+ "muy=-0.96\n",
+ "muz=0\n",
+ "mux=-1.1\n",
+ "mu=math.sqrt(mux*mux+muy*muy+muz*muz)\n",
+ "#results\n",
+ "print '%s %.1f %s' %(\"Net dipole moment =\",mu,\"D\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Net dipole moment = 1.5 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 390"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the potential energy of the system\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "Na=6.023*math.pow(10,23) # /mol\n",
+ "e=1.60228*math.pow(10,-19) #C\n",
+ "e0=8.85419*math.pow(10,-12) #C^2/J m\n",
+ "#calculations\n",
+ "factor=Na*e*e /(4*math.pi*e0)\n",
+ "#Multiply by Z^2/R to get the value of potential energy. Plot the graph\n",
+ "#results\n",
+ "print '%s %.3e %s' %(\"Potential energy =\",factor,\" Z*Z/R kJ/mol\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potential energy = 1.390e-04 Z*Z/R kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I1 - Pg 383"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calcualte the predicted dipole moment\n",
+ "#Initialization of variables\n",
+ "EH=2.1\n",
+ "EBr=2.8\n",
+ "#calculations\n",
+ "diff=-EH+EBr\n",
+ "#results\n",
+ "print '%s %.1f %s' %(\"Prediced dipole moment =\",diff,\"D\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Prediced dipole moment = 0.7 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the potential energy\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "mu1=1.4 #D\n",
+ "mu2=1.4 #D\n",
+ "angle=180. #degrees\n",
+ "d=3 #nm\n",
+ "D=4.7*math.pow(10,-30) #C m\n",
+ "#calculations\n",
+ "Vmol=D*D*(1-3*math.cos(angle*math.pi/180.)*math.cos(angle*math.pi/180.))/(4*math.pi*8.854*math.pow(10,-12) *math.pow((d*math.pow(10,-9)),3))\n",
+ "V=Vmol*(6.023*math.pow(10,23))\n",
+ "#results\n",
+ "print '%s %.1f %s' %(\"Potential energy =\",V,\" J/mol\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potential energy = -8.9 J/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter17.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter17.ipynb new file mode 100755 index 00000000..ab0a46f5 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter17.ipynb @@ -0,0 +1,99 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 17 - Molecular rotations and vibrations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 423"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the frequency of transistion\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "mH=1.673*math.pow(10,-27) #kg\n",
+ "mCl=5.807*math.pow(10,-26) #kg\n",
+ "R=127.4 *math.pow(10,-12) #m\n",
+ "#calculations\n",
+ "mu=mH*mCl/(mH+mCl)\n",
+ "I=mu*R*R\n",
+ "B=1.05457*math.pow(10,-34) /(4*math.pi*I)\n",
+ "f=2*B/math.pow(10,9)\n",
+ "#results\n",
+ "print '%s %.1f %s' %(\"Frequency of transistion =\",f,\"GHz\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency of transistion = 635.9 GHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I1 - Pg 428"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the wavenumber and wavelength\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "v=89.6*math.pow(10,12) #Hz\n",
+ "#calculations\n",
+ "l=3*math.pow(10,8) /v\n",
+ "wn=math.pow(10,-2) /l\n",
+ "#results\n",
+ "print '%s %d %s' %(\"Wavenumber =\",wn,\" cm^-1\")\n",
+ "print '%s %.2f %s' % (\"\\n Wavelength =\",l*math.pow(10,6),\"mu m\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavenumber = 2986 cm^-1\n",
+ "\n",
+ " Wavelength = 3.35 mu m\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter18.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter18.ipynb new file mode 100755 index 00000000..36e81ed5 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter18.ipynb @@ -0,0 +1,111 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 18 - Electronic Transitions"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 441"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the transmittance\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "wl=256*math.pow(10,-9) #m\n",
+ "t=1 #mm\n",
+ "C=0.050 #mol/L\n",
+ "T=0.16\n",
+ "t2=2 #mm\n",
+ "#calculations\n",
+ "E=-math.log10(T) /(C*t)\n",
+ "A1=-math.log10(T)\n",
+ "A2=E*C*t2\n",
+ "Tr=math.pow(10,(-A2))\n",
+ "#results\n",
+ "print '%s %.3f' %(\"Transmittance = \",Tr)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Transmittance = 0.026\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 450"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Quenching rate constant and half life\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy as np\n",
+ "from numpy import linalg\n",
+ "Q=[1., 2., 3., 4., 5]\n",
+ "t1=[5.2, 9.4, 13.7, 18., 22.2]\n",
+ "t2=[1.1, 2., 2.9, 4., 4.5]\n",
+ "#calculations\n",
+ "A = np.vstack([Q, np.ones(len(Q))]).T\n",
+ "kqbykf=np.linalg.lstsq(A,t1)[0]\n",
+ "slope1=kqbykf[0] *1000.\n",
+ "kq=np.linalg.lstsq(A,t2)[0]\n",
+ "slope2=kq[0] *math.pow(10,10)\n",
+ "kq=slope2\n",
+ "kf=kq/slope1\n",
+ "thalf=math.log (2) /kf\n",
+ "#results\n",
+ "print '%s %.1e %s' %(\"Quenching rate constant =\",kq,\"L ml^-1 s^-1\")\n",
+ "print '%s %.1e %s' %(\"\\n Half life=\",thalf,\"s\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Quenching rate constant = 8.8e+09 L ml^-1 s^-1\n",
+ "\n",
+ " Half life= 3.4e-07 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter19.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter19.ipynb new file mode 100755 index 00000000..89f694e9 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter19.ipynb @@ -0,0 +1,64 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 19 - Magnetic resonance"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 467"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the spin-spin coupling constants\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "A=5.1 #Hz\n",
+ "B=-1.4 #Hz\n",
+ "C=3.2 #Hz\n",
+ "an1=120*math.pi/180. #radians\n",
+ "an2=180*math.pi/180. #radians\n",
+ "#calculations\n",
+ "j1=A+B*math.cos(an1) + C*math.cos(2*an1)\n",
+ "j2=A+B*math.cos(an2) + C*math.cos(2*an2)\n",
+ "#results\n",
+ "print '%s %d %s' %(\"Spin-spin coupling constant =\",j1,\"Hz\")\n",
+ "print '%s %.1f %s' %(\"\\n Spin-spin coupling constant =\",j2,\"Hz\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Spin-spin coupling constant = 4 Hz\n",
+ "\n",
+ " Spin-spin coupling constant = 9.7 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter2.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter2.ipynb new file mode 100755 index 00000000..1730ffc1 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter2.ipynb @@ -0,0 +1,172 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2 - Thermodynamics : The first law"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I1 - Pg 46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the change in temperature\n",
+ "#Initialization of variables\n",
+ "Cpm=75 #J/k mol\n",
+ "n=5.55 #mol\n",
+ "q=1 #kJ\n",
+ "#Calculations\n",
+ "deltaT=q*1000/(n*Cpm)\n",
+ "#results\n",
+ "print '%s %.1f %s' %('Change in temperature =',deltaT,'K')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in temperature = 2.4 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I4 - Pg 52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the change in enthalpy of the sample\n",
+ "#Initialization of variables\n",
+ "n=5.55 #mol\n",
+ "T1=20 #C\n",
+ "T2=80 #K\n",
+ "Cpm=75.29 #J/K mol\n",
+ "#Calculations\n",
+ "H=n*Cpm*(T2-T1)/1000.\n",
+ "#results\n",
+ "print '%s %d %s' %('Enthalpy of the sample changes by',H,'kJ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Enthalpy of the sample changes by 25 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the heat supplied, heat capacity and heat output of the calorimeter\n",
+ "#Initialization of variables\n",
+ "A=1.23 #A\n",
+ "V=12 #V\n",
+ "t=123 #s\n",
+ "Temp=4.47 #C\n",
+ "rise=3.22 #C\n",
+ "#Calculations\n",
+ "q=A*V*t\n",
+ "C=q/Temp\n",
+ "Output= C*rise/1000.\n",
+ "#Results\n",
+ "print '%s %.1f %s' %('heat supplied during calibration =',q,'J')\n",
+ "print '%s %.1f %s' %('\\n Heat capacity of the calorimeter =',C,'J/C')\n",
+ "print '%s %.2f %s' %('\\n Heat output =',Output,' kJ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "heat supplied during calibration = 1815.5 J\n",
+ "\n",
+ " Heat capacity of the calorimeter = 406.1 J/C\n",
+ "\n",
+ " Heat output = 1.31 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the amount by which the person's internal energy falls\n",
+ "#Initialization of variables\n",
+ "work=-622 #kJ\n",
+ "heat=-82 #kJ\n",
+ "#Calculations\n",
+ "U=work+heat\n",
+ "#results\n",
+ "print '%s %d %s' %('The persons internal energy falls by',-U,'kJ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The persons internal energy falls by 704 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter20.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter20.ipynb new file mode 100755 index 00000000..cb95fd4e --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter20.ipynb @@ -0,0 +1,250 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 20 - Statistical thermodynamics"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I1 - Pg 477"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the relative populations of boat and chair conformations\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "E=22*1000. #kJ/mol\n",
+ "T=293. #K \n",
+ "#calculations\n",
+ "ratio=math.pow(math.e,(-E/(8.31451*T)))\n",
+ "#results\n",
+ "print '%s %.1e' %(\"Relative populations of boat and chair conformations is \",ratio)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Relative populations of boat and chair conformations is 1.2e-04\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 478"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the required ratio\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "g2=5.\n",
+ "g1=3.\n",
+ "E2=6.\n",
+ "E1=2.\n",
+ "k=1.38*math.pow(10,-23) #J/K\n",
+ "h=6.626*math.pow(10,-34) #J s\n",
+ "B=3.18*math.pow(10,11) #Hz\n",
+ "T =298 #K\n",
+ "#calculations\n",
+ "ratio=g2/g1 *(math.pow(math.e,((E1-E2)*h*B/(k*T))))\n",
+ "#results\n",
+ "print '%s %.2f' %(\"Ratio= \",ratio)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ratio= 1.36\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I3 - Pg 481"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the translational partition function\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "T=298 #K\n",
+ "m=32*1.66054*math.pow(10,-27) #kg\n",
+ "k=1.38066*math.pow(10,-23) #j/k\n",
+ "V=math.pow(10,-4) #m^3\n",
+ "h=6.62608*math.pow(10,-34) #J/s\n",
+ "#calculations\n",
+ "q=math.pow((2*math.pi*m*k*T),1.5) *V/h/h/h \n",
+ "#results\n",
+ "print '%s %.2e' %(\"Translational partition function = \",q)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Translational partition function = 1.75e+28\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the partition function at 20 C\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "E=22 #kJ/mol\n",
+ "R=8.214 #J/K mol\n",
+ "T=293 #K\n",
+ "#Calculations\n",
+ "q=1+math.pow(math.e,(-E*1000. /(R*T)))\n",
+ "#results\n",
+ "print '%s %.4f' %(\"At 20 C, partition function = \",q)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "At 20 C, partition function = 1.0001\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 485"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the contribution to rotational motion\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "k=1.38*math.pow(10,-23) #J/K\n",
+ "h=6.626*math.pow(10,-34) #J s\n",
+ "B=3.18*math.pow(10,11) #Hz\n",
+ "T=298 #K\n",
+ "R=8.314 #J/K mol\n",
+ "#calculations\n",
+ "Sm=R*(1+math.log(k*T/(h*B)))\n",
+ "#results\n",
+ "print '%s %.1f %s' %(\"Contribution to rotational motion=\",Sm,\"J/ K mol\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Contribution to rotational motion= 33.0 J/ K mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 - Pg 488"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Equilibrium constant\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "me=9.10939*math.pow(10,-31) #kg\n",
+ "k=1.38*math.pow(10,-23) #J/K\n",
+ "h=6.626*math.pow(10,-34) #J s\n",
+ "p=math.pow(10,5) #Pa\n",
+ "T=1000 #K\n",
+ "R=8.314 #J/K mol\n",
+ "I=376*1000. #J/mol\n",
+ "#calculations\n",
+ "K=math.pow((2*math.pi*me),1.5) *math.pow((k*T),2.5) /(p*h*h*h) *math.pow(math.e,(-I/(R*T)))\n",
+ "#results\n",
+ "print '%s %.2e' %(\"Equilibrium constant = \",K)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equilibrium constant = 2.41e-19\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter3.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter3.ipynb new file mode 100755 index 00000000..87d165d7 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter3.ipynb @@ -0,0 +1,291 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3 - Thermochemistry"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I1 - Pg 67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the change in enthalpy\n",
+ "#Initialization of variables\n",
+ "dU=-969.6 #kJ/mol\n",
+ "nN2=1/2.\n",
+ "nCO2=2.\n",
+ "nO2=9./4.\n",
+ "T=298.15 #K\n",
+ "#Calculations\n",
+ "n=nCO2+nN2-nO2\n",
+ "H=dU+n*8.3145*T/1000.\n",
+ "#results\n",
+ "print '%s %.1f %s' %('Enthalpy change = ',H,'kJ/mol')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Enthalpy change = -969.0 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 58"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the Molar Enthalpy Change\n",
+ "#Initialization of variables\n",
+ "I=0.682 #A\n",
+ "V=12 #V\n",
+ "t=500. #s\n",
+ "m=4.33 #g\n",
+ "MW=46.07 #g/mol\n",
+ "#Calculations\n",
+ "q=I*V*t\n",
+ "n=m/MW\n",
+ "H=q/n/1000.\n",
+ "#Results\n",
+ "print '%s %.1f %s' %('Molar enthalpy change =',H,'kJ/mol')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Molar enthalpy change = 43.5 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the Heat supplied\n",
+ "#Initialization of variables\n",
+ "m=1 #g\n",
+ "MW=24.31 #g/mol\n",
+ "H=2337 #kJ/mol\n",
+ "#Calculations\n",
+ "n=m/MW\n",
+ "q=n*H\n",
+ "#results\n",
+ "print '%s %.1f %s' %('Heat supplied =',q,'kJ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat supplied = 96.1 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the sum of enthalpy changes\n",
+ "#Initialization of variables\n",
+ "HC=716.68 #kJ\n",
+ "HH=871.88 #kJ\n",
+ "HO=249.17 #kJ\n",
+ "Hcond=-38 #kJ\n",
+ "HCH=-412\n",
+ "HCO=-360\n",
+ "HOH=-463\n",
+ "#Calculations\n",
+ "H1=HC+HH+HO\n",
+ "H2=3*HCH+HCO+HOH\n",
+ "H3=Hcond\n",
+ "H=H1+H2+H3\n",
+ "#results\n",
+ "print '%s %d %s' %('Sum of enthalpy changes =',H,' kJ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Sum of enthalpy changes = -259 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 - Pg 68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the standard enthalpy of combustion of propene\n",
+ "#Initialization of variables\n",
+ "Hf=-124 #kJ\n",
+ "Hoxi=-2220 #kJ\n",
+ "Hwater=286 #kJ\n",
+ "#Calculations\n",
+ "H=Hf+Hoxi+Hwater\n",
+ "#results\n",
+ "print '%s %d %s' %('Standard enthalpy of combustion of propene =',H,'kJ/mol')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Standard enthalpy of combustion of propene = -2058 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 - Pg 70"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the standard enthalpy of combustion of benzene\n",
+ "#Initialization of variables\n",
+ "nCO2=6 #mol\n",
+ "nH2O=3 #mol\n",
+ "nO2=15/2. #mol\n",
+ "nC6H6=1 #mol\n",
+ "HC6H6=49 #kJ/mol\n",
+ "HH2O=-285.83\n",
+ "HO2=0\n",
+ "HCO2=-393.51 \n",
+ "#Calculations\n",
+ "H=nCO2*HCO2+nH2O*HH2O-nC6H6*HC6H6-nO2*HO2\n",
+ "#results\n",
+ "print '%s %.1f %s' %('Standard enthalpy of combustion of benzene is',H,'kJ/mol')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Standard enthalpy of combustion of benzene is -3267.6 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 - Pg 74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the enthalpy of formation of water\n",
+ "#Initialization of variables\n",
+ "HH2O=-241.82 #kJ/mol\n",
+ "T1=25 #C\n",
+ "T2=100 #C\n",
+ "CpH2O=33.58 #J/K mol\n",
+ "CpH2=28.84 #J/K mol\n",
+ "CpO2=29.37 #J/K mol\n",
+ "#calculations\n",
+ "dCp=CpH2O-CpH2-0.5*CpO2\n",
+ "dH=HH2O+dCp*(T2-T1)/1000.\n",
+ "#results\n",
+ "print '%s %.2f %s' %('Enthalpy of formation of water at 100 C is',dH,'kJ/mol')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Enthalpy of formation of water at 100 C is -242.57 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter4.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter4.ipynb new file mode 100755 index 00000000..b5b8ff90 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter4.ipynb @@ -0,0 +1,208 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4 - Thermodynamics: the Second Law"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I1 - Pg 80"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the entropy change at 273 and 373 K\n",
+ "#Initialization of variables\n",
+ "H=100. #kJ\n",
+ "T1=273. #K\n",
+ "T2=373.#K\n",
+ "#calculations\n",
+ "S1=H*1000/T1\n",
+ "S2=H*1000/T2\n",
+ "#results\n",
+ "print '%s %d %s' %('Entropy change at 273 K is',S1,'J/K ')\n",
+ "print '%s %d %s' %('\\n Entropy change at 373 K is',S2,'J/K ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Entropy change at 273 K is 366 J/K \n",
+ "\n",
+ " Entropy change at 373 K is 268 J/K \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 84"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the enthalpy of vaporization\n",
+ "#Initialization of variables\n",
+ "T=59.2 #K\n",
+ "#calculations\n",
+ "Hvap=85*(273.2+T)/1000.\n",
+ "#results\n",
+ "print '%s %d %s' %('Enthalpy of vaportization =',Hvap,'kJ/mol')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Enthalpy of vaportization = 28 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I3 - Pg 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the change in entropy\n",
+ "#Initialization of variables\n",
+ "SH2O=70 #J/K mol\n",
+ "SH2=131 #J/K mol\n",
+ "SO2=205 #J/K mol\n",
+ "#calculations\n",
+ "deltaS=2*SH2O-2*SH2-SO2\n",
+ "print '%s %d %s' %('Change in entropy =',deltaS,'J/K mol')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in entropy = -327 J/K mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 84"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the net heat transferred and entropy production per day\n",
+ "#Initialization of variables\n",
+ "Power=100. #W\n",
+ "time=1 #day\n",
+ "T=20 #C\n",
+ "#calculations\n",
+ "timeins=1*24*3600\n",
+ "qsurr=timeins*Power\n",
+ "Ssurr=qsurr/(T+273.)\n",
+ "#results\n",
+ "print '%s %d %s' %('Heat transferred to surroundings =',qsurr,'J')\n",
+ "print '%s %.2e %s' %('\\n Entropy production per day =',Ssurr,'J/k')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat transferred to surroundings = 8640000 J\n",
+ "\n",
+ " Entropy production per day = 2.95e+04 J/k\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the amount of food bird must consume\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "g=9.81 #m/s^2\n",
+ "m=30*math.pow(10,-3) #kg\n",
+ "d=10 #m\n",
+ "H=2.828*math.pow(10,6) #j/mol\n",
+ "M=180. #g/mol\n",
+ "#calculations\n",
+ "w=g*m*d\n",
+ "n=w/H\n",
+ "m=n*M\n",
+ "#results\n",
+ "print '%s %.1e %s' %('Amount bird must consume =',m,'g')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Amount bird must consume = 1.9e-04 g\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter6.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter6.ipynb new file mode 100755 index 00000000..7705b071 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter6.ipynb @@ -0,0 +1,304 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6 - The properties of mixtures"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the mole fraction of glycerine\n",
+ "#Initialization of variables\n",
+ "m=0.14 #mol/kg\n",
+ "w=1. #kg Assume\n",
+ "#Calculations\n",
+ "ngly=m*w\n",
+ "nwater=w*1000 /18.02\n",
+ "ntotal=ngly+nwater\n",
+ "xgly=ngly/ntotal\n",
+ "#results\n",
+ "print '%s %.2e' %('Mole fraction of glycerine is xgly =',xgly)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mole fraction of glycerine is xgly = 2.52e-03\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the volume of the mixture\n",
+ "#Initialization of variables\n",
+ "mE=50. #g\n",
+ "mW=50. #g\n",
+ "#calculations\n",
+ "nE=mE/46.\n",
+ "nW=mW/18.\n",
+ "ntotal=nE+nW\n",
+ "xE=nE/ntotal\n",
+ "xW=1-xE\n",
+ "print '%s' %('for the observed xE and xW')\n",
+ "vE=55 #cc/mol\n",
+ "vW=18 #cc/mol\n",
+ "V=nE*vE+nW*vW\n",
+ "#results\n",
+ "print '%s %.1f %s' %('\\n Volume of the mixture =',V,'cm^3 ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "for the observed xE and xW\n",
+ "\n",
+ " Volume of the mixture = 109.8 cm^3 \n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 121"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Calculate Ka and Kc from the raoults law line \n",
+ "#Initialization of variables\n",
+ "import matplotlib\n",
+ "from matplotlib import pyplot\n",
+ "xc=([0, 0.20, 0.40, 0.60, 0.80, 1])\n",
+ "pc=([0, 35, 82, 142, 219, 293])\n",
+ "pa=([347, 270, 185, 102, 37, 0])\n",
+ "#calculations\n",
+ "pyplot.plot(xc,pc)\n",
+ "pyplot.plot(xc,pa)\n",
+ "pyplot.xlabel('Mole fraction xc')\n",
+ "pyplot.ylabel('Pressure /Torr')\n",
+ "print '%s' %('From the graph it is clear that KA=175 torr and KC=165 torr. They are plotted with Raoults law lines')\n",
+ "pyplot.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "From the graph it is clear that KA=175 torr and KC=165 torr. They are plotted with Raoults law lines\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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tbY2FhQWbNm3i9ddfZ/To0c+8oedF9erVdY9Hjx5Nr169ALCzsyMxMVH32qVL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\nhBAiR/8PUIILe33jvscAAAAASUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x6121270>"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 - Pg 123"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate if required concentration can be maintained under normal conditions.\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "C=4/1000. #g/L\n",
+ "MO2=32. #g/mol\n",
+ "Mw=18.\n",
+ "w=1 #L\n",
+ "K=3.3*math.pow(10,7) #torr\n",
+ "patm=0.21*760 #torr\n",
+ "#calculations\n",
+ "nO2=C/MO2\n",
+ "nH2O=w*1000. /Mw\n",
+ "xO2=nO2/(nO2+nH2O)\n",
+ "pO2=xO2*K\n",
+ "if(pO2<patm):\n",
+ " print '%s %.1e' %('The required concentration can be maintained under normal conditions and pressure is (Torr)',pO2)\n",
+ "else:\n",
+ " print '%s %.e' %('The required concentration cannot be maintained under normal conditions and pressure is (Torr)',pO2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The required concentration can be maintained under normal conditions and pressure is (Torr) 7.4e+01\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 - Pg 131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the molar mass of the enzyme\n",
+ "#Initialization of variables\n",
+ "%matplotlib inline\n",
+ "import numpy as np\n",
+ "from numpy import linalg\n",
+ "import matplotlib\n",
+ "from matplotlib import pyplot\n",
+ "c=([1, 2, 4, 7, 9])\n",
+ "hbyc=([0.28, 0.36, 0.503, 0.739, 0.889])\n",
+ "R=8.3145 #J/K mol\n",
+ "T=298 #K\n",
+ "g=9.81 #m/s^2\n",
+ "d=0.9998 #g/cm^3\n",
+ "#calculations\n",
+ "pyplot.plot(c,hbyc)\n",
+ "pyplot.xlabel('c')\n",
+ "pyplot.ylabel('hbyc')\n",
+ "A = np.vstack([c, np.ones(len(c))]).T\n",
+ "vector=numpy.linalg.lstsq(A,hbyc)\n",
+ "intercept=vector[0]\n",
+ "intercept1=intercept[1]/100.\n",
+ "M=R*T/(d*g*intercept1)/1000.\n",
+ "#results\n",
+ "print '%s %d %s' %('Molar mass of the enzyme is close to',M,'kDa')\n",
+ "print '%s' %('The answer is a bit different in the textbook due to rounding off error in the textbook')\n",
+ "pyplot.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Populating the interactive namespace from numpy and matplotlib\n",
+ "Molar mass of the enzyme is close to 123 kDa"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The answer is a bit different in the textbook due to rounding off error in the textbook\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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aXZKIiPzM0hBwu93MnDmT7OxsCgoKyMjIYMeOHXXG9OrVi1WrVrF161b++te/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\nH8HDd6tbA3Y3AAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x9caefb0>"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 - Pg 136"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate how many times the rich phase is more abundant in nitrobenzene\n",
+ "#Initialization of variables\n",
+ "nB=0.59 #mol\n",
+ "nNB=0.41 #mol\n",
+ "xN1=0.38\n",
+ "xN2=0.74\n",
+ "xNm=0.41\n",
+ "#calculations\n",
+ "print '%s' %('By lever rule')\n",
+ "ratio=(xNm-xN1)/(xN2-xNm)\n",
+ "percent=ratio*100. +1\n",
+ "#results\n",
+ "print '%s %d %s' %(\"The rich phase is\",percent,\"times more abundant in nitrobenzene\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "By lever rule\n",
+ "The rich phase is 10 times more abundant in nitrobenzene\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter7.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter7.ipynb new file mode 100755 index 00000000..db391867 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter7.ipynb @@ -0,0 +1,346 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7 - Principles of chemical equilibrium"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the reaction gibbs energy\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "G=-31. #kJ/mol\n",
+ "T=37+273. #K\n",
+ "Cadp=1/1000. #mmol/L\n",
+ "Cp=8/1000. #mmol/L\n",
+ "Catp=8/1000. #mmol/L\n",
+ "R=8.314 #J/K mol\n",
+ "#calculations\n",
+ "Q=Cadp*Cp/Catp\n",
+ "deltaG=G+R*T*math.log(Q) /1000.\n",
+ "#results\n",
+ "print '%s %.1f %s' %(\"Reaction Gibbs energy =\",deltaG,\" kJ/mol\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reaction Gibbs energy = -48.8 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the gibbs energy\n",
+ "#Initialization of variables\n",
+ "Hr=-285.83 #kJ/mol\n",
+ "Sr=-163.34 #J/ K mol\n",
+ "T=298.15 #K\n",
+ "#calculations\n",
+ "Gr=Hr-T*Sr/1000.\n",
+ "#results\n",
+ "print '%s %.2f %s' %('Gibbs energy =',Gr,'kJ/mol')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Gibbs energy = -237.13 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 153"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the change in gibbs energy\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "aADP=1 #mol/L\n",
+ "aP=1 #mol/L\n",
+ "aATP=1 #mol/L\n",
+ "aH2O=1 #mol/L\n",
+ "aH=math.pow(10,-7) #mol/L\n",
+ "G=10 #kJ/mol\n",
+ "T=298. #K\n",
+ "R=8.314 #J/K mol\n",
+ "#calculations\n",
+ "Q=aADP*aP*aH/(aATP*aH2O)\n",
+ "Gr=G+R*T*math.log(Q)/1000.\n",
+ "#results\n",
+ "print '%s %.1f %s' %('Change in nGibbs energy =',Gr,'kJ/mol')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in nGibbs energy = -29.9 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 - Pg 155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the equivalent fration\n",
+ "#Initialization of variables\n",
+ "Gr=1.7*1000 #J/mol\n",
+ "T=298. #K\n",
+ "R=8.314 #J/K mol\n",
+ "K=0.5\n",
+ "#calculations\n",
+ "GbyRT=Gr/(R*T)\n",
+ "feq=K/(K+1)\n",
+ "#results\n",
+ "print '%s %.2f' %(\"Equivalent fraction = \",feq)\n",
+ "print '%s' %(\"For the second part, Gr=1.7 + 2.48 ln(f/1-f)\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent fraction = 0.33\n",
+ "For the second part, Gr=1.7 + 2.48 ln(f/1-f)\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 - Pg 157"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the pressure of N2,H2,NH3 gases\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy\n",
+ "species=(['N2', 'H2', 'NH3'])\n",
+ "change=(['-x', '-3x', '2x'])\n",
+ "E=(['1-x', '3-3x', '2x'])\n",
+ "print '%s' %(\"Concentration table\")\n",
+ "print '%s' %(\"species\")\n",
+ "print '%s' %(\"change\")\n",
+ "print(E)\n",
+ "K=977.\n",
+ "#Calculations\n",
+ "g=math.sqrt(27*K/4.)\n",
+ "vector=([g, -(2*g +1), g])\n",
+ "sol=numpy.roots(vector)[1]\n",
+ "\n",
+ "PN2=1-sol\n",
+ "PH2=3-3*sol\n",
+ "PNH3=2*sol\n",
+ "K=math.pow(PNH3,2)/(math.pow(PH2,3) *PN2)\n",
+ "#results\n",
+ "print '%s %.2f %s' %(\"Pressure of N2 gas = \",PN2,\"bar\")\n",
+ "print '%s %.2f %s' %(\"\\n Pressure of H2 gas =\",PH2,\"bar\")\n",
+ "print '%s %.2f %s' %(\"\\n Pressure of NH3 gas =\",PNH3,\"bar\")\n",
+ "print '%s %.2e %s' %(\"\\n K final =\",K,\"> it is close to original value.\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Concentration table\n",
+ "species\n",
+ "change\n",
+ "['1-x', '3-3x', '2x']\n",
+ "Pressure of N2 gas = 0.10 bar\n",
+ "\n",
+ " Pressure of H2 gas = 0.31 bar\n",
+ "\n",
+ " Pressure of NH3 gas = 1.79 bar\n",
+ "\n",
+ " K final = 9.77e+02 > it is close to original value.\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the equilibrium constant\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "Gr=-3.40 #kJ/mol\n",
+ "R=8.314 #J/k mol\n",
+ "T=298. #K\n",
+ "#calculations\n",
+ "lnK=Gr*1000./(R*T)\n",
+ "K=math.exp(lnK)\n",
+ "#results\n",
+ "print '%s %.2f' %('Equilibrium constant K= ',K)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equilibrium constant K= 0.25\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I3 - Pg 149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the decomposition temperature\n",
+ "#Initialization of variables\n",
+ "Hr=178. #kJ/mol\n",
+ "Sr=161. #J/K mol\n",
+ "#calculations\n",
+ "T=Hr*1000 /Sr\n",
+ "#results\n",
+ "print '%s %.2e %s' %(\"Decompostion temperature =\",T,\"K\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Decompostion temperature = 1.11e+03 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I4 - Pg 151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the standard reaction gibbs energy\n",
+ "#Initialization of variables\n",
+ "GCO2=-394. #kJ/mol\n",
+ "GCO=-137. #kJ/mol\n",
+ "GO2=0\n",
+ "#calculations\n",
+ "deltaG=2*GCO2-2*GCO+GO2\n",
+ "#results\n",
+ "print '%s %d %s' %('Standard reaction gibbs energy =',deltaG,' kJ/mol')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Standard reaction gibbs energy = -514 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter8.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter8.ipynb new file mode 100755 index 00000000..a3e364e2 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter8.ipynb @@ -0,0 +1,258 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8 - Consequences of equilibrium"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I1 - Pg 176"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the equlibrium pH\n",
+ "#Initialization of variables\n",
+ "ph1=6.37\n",
+ "ph2=10.25\n",
+ "ph3=7.21\n",
+ "ph4=12.67\n",
+ "#calculations\n",
+ "pH1=0.5*(ph1+ph2)\n",
+ "pH2=0.5*(ph3+ph4)\n",
+ "#results\n",
+ "print '%s %.2f' %(\"Equilibrium pH in case 1 = \",pH1)\n",
+ "print '%s %.2f' %(\"\\n Equilibrium pH in case 2 = \",pH2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equilibrium pH in case 1 = 8.31\n",
+ "\n",
+ " Equilibrium pH in case 2 = 9.94\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the pH of the solution\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "n=2.5/1000. #mol\n",
+ "C=0.2 #mol/L\n",
+ "vbase=37.5/1000. #L\n",
+ "#calculations\n",
+ "V=n/C\n",
+ "base=n/vbase\n",
+ "H=math.pow(10,-14) /base\n",
+ "print '%s' %(\"It follows from example 8.2 that\")\n",
+ "pH=10.2\n",
+ "#results\n",
+ "print '%s %.1f' %(\"\\n pH of the solution = \",pH)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "It follows from example 8.2 that\n",
+ "\n",
+ " pH of the solution = 10.2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the percent of acetic acid molecules that have donated a proton\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "C=0.15 #M\n",
+ "Ka=1.8*math.pow(10,-5)\n",
+ "#calculations\n",
+ "x=math.sqrt(C*Ka)\n",
+ "f=x/C\n",
+ "percent=f*100.\n",
+ "#results\n",
+ "print '%s %.1f %s' %(\"percent of acetic acid molecules that have donated a proton =\",percent,\"percent\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "percent of acetic acid molecules that have donated a proton = 1.1 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the fraction proportionated\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "pKa=4.88\n",
+ "C=0.01 #M\n",
+ "pKw=14\n",
+ "#calculations\n",
+ "pKb=pKw-pKa\n",
+ "Kb=math.pow(10,(-pKb))\n",
+ "x=(math.sqrt(C*Kb))\n",
+ "pOH=-math.log(x)\n",
+ "pH=14-pOH\n",
+ "f=x/C\n",
+ "#results\n",
+ "print '%s %.1e' %(\"fraction protonated = \",f)\n",
+ "print '%s %d' %(\"\\n 1 molecule in about \",1./f)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "fraction protonated = 2.8e-04\n",
+ "\n",
+ " 1 molecule in about 3630\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 173"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate te concentration of carbonate ions\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "pKa2=10.25\n",
+ "#calculations\n",
+ "C=math.pow(10,(-pKa2))\n",
+ "#results\n",
+ "print '%s %.1e %s' %(\"Concentration of Carbonate ions =\",C,\" mol/l\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Concentration of Carbonate ions = 5.6e-11 mol/l\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 - Pg 173"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the final pH of the solution\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "vOH=5*math.pow(10,-3) #L\n",
+ "vHClO=25*math.pow(10,-3) #L\n",
+ "C=0.2 #mol/L\n",
+ "#calculations\n",
+ "nOH=vOH*C\n",
+ "nHClO=vHClO*C/2.\n",
+ "nrem=nHClO-nOH\n",
+ "pH=7.53-math.log10(nrem/nOH)\n",
+ "#results\n",
+ "print '%s %.1f' %(\"Final pH= \",pH)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Final pH= 7.4\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter9.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter9.ipynb new file mode 100755 index 00000000..be702d1e --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter9.ipynb @@ -0,0 +1,367 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9 - Electrochemistry"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I1 - Pg 200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the potential of the cell\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "Gr=-math.pow(10,5) #kJ/mol\n",
+ "v=1\n",
+ "F=9.6485*10000. #C/mol\n",
+ "#calculations\n",
+ "E=-Gr/(v*F)\n",
+ "#results\n",
+ "print '%s %d %s' %(\"potential of the cell =\",E,\"V\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "potential of the cell = 1 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the equilibrium constant of the reaction\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "V=1.1 #V\n",
+ "F=9.6485*10000. #C/mol\n",
+ "R=8.314 #J/K mol\n",
+ "T=298.15 #K\n",
+ "#calculations\n",
+ "lnK=2*F*V/(R*T)\n",
+ "k=math.pow(math.e,(lnK))\n",
+ "#results\n",
+ "print '%s %.1e' %(\"Equilibrium constant =\",k)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equilibrium constant = 1.5e+37\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the acidity constant of the acid\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "lw=34.96 #mS m^2 /mol\n",
+ "la=4.09 #mS m^2 /mol\n",
+ "C=0.010 #M\n",
+ "K=1.65 #mS m^2 /mol\n",
+ "#calculations\n",
+ "lmd=lw+la\n",
+ "alpha=K/lmd\n",
+ "Ka=C*alpha*alpha\n",
+ "pKa=-math.log10(Ka)\n",
+ "#results\n",
+ "print '%s %.2f' %(\"Acidity constant of the acid = \",pKa)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Acidity constant of the acid = 4.75\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 - Pg 203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate if the reaction is favoring the products or not. Also calculate the E value\n",
+ "#Initialization of variables\n",
+ "ER=1.23 #V\n",
+ "EL=-0.44 #V\n",
+ "#calculations\n",
+ "E=ER-EL\n",
+ "#results\n",
+ "if(E>0):\n",
+ " print '%s %.2f %s' %(\"The reaction is favouring products and E is\",E,\"V\")\n",
+ "else:\n",
+ " print '%s %.2f %s' %(\"The reaction is not favouring products and E is\",E,\" V\")\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reaction is favouring products and E is 1.67 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7 - Pg 203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the equilibrium constant \n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "ER=0.52 #V\n",
+ "EL=0.15 #V\n",
+ "#calculations\n",
+ "E=ER-EL\n",
+ "lnK=E*1000./(25.69)\n",
+ "K=math.exp(lnK)\n",
+ "#results\n",
+ "print '%s %.1e' %(\"Equilbrum constant K= \",K)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equilbrum constant K= 1.8e+06\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8 - Pg 205"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the biological standard potential\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "E0=-0.11 #V\n",
+ "H=math.pow(10,-7)\n",
+ "#calculations\n",
+ "pH=-math.log10(H)\n",
+ "E=E0-29.59*pH*math.pow(10,-3)\n",
+ "#results\n",
+ "print '%s %.2f %s' %(\"Biological standard potential =\",E,\"V\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Biological standard potential = -0.32 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9 - Pg 206"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the equilibrium constant for the reaction\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "ER=-0.21 #V\n",
+ "EL=-0.6 #V\n",
+ "#calculations\n",
+ "E=ER-EL\n",
+ "lnK=2*E*1000./(25.69)\n",
+ "K=math.exp(lnK)\n",
+ "#results\n",
+ "print '%s %.1e' %(\"Equilibrium constant for the reaction = \",K)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equilibrium constant for the reaction = 1.5e+13\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 - Pg 209"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the electric potential\n",
+ "#Initialization of variables\n",
+ "E1=2*(-0.340)\n",
+ "E2=-0.522 \n",
+ "#calculations\n",
+ "FE=-E1+E2\n",
+ "#results\n",
+ "print '%s %.3f %s' %(\"Electric potential =\",FE,\"V\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric potential = 0.158 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 - Pg 210"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Gibbs enthalpy, Standard entropy and enthalpy of the process\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "v=2\n",
+ "F=9.6485*10000. #C/mol\n",
+ "E=0.2684 #V\n",
+ "V1=0.2699 #V\n",
+ "V2=0.2669 #V\n",
+ "T1=293. #K\n",
+ "T=298. #K\n",
+ "T2=303. #K\n",
+ "#calculations\n",
+ "Gr= -v*F*E/1000.\n",
+ "Sr=v*F*(V2-V1)/(T2-T1)\n",
+ "Hr=Gr+T*Sr/1000.\n",
+ "#results\n",
+ "print '%s %.2f %s' %(\"Gibbs enthalpy =\",Gr,\"kJ/mol\")\n",
+ "print '%s %.1f %s' %(\"\\n Standard Entropy =\",Sr,\"J /K mol\")\n",
+ "print '%s %.1f %s' %(\"\\n Enthalpy =\",Hr,\"kJ/mol\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Gibbs enthalpy = -51.79 kJ/mol\n",
+ "\n",
+ " Standard Entropy = -57.9 J /K mol\n",
+ "\n",
+ " Enthalpy = -69.0 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/README.txt b/Elements_of_Physical_Chemistry_by_Atkins_Peter/README.txt new file mode 100755 index 00000000..f93c44a3 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/README.txt @@ -0,0 +1,10 @@ +Contributed By: Devika Raj +Course: be +College/Institute/Organization: RVR college of Engineering +Department/Designation: Electronics and Communication En +Book Title: Elements of Physical Chemistry +Author: Atkins Peter +Publisher: Oxford University Press +Year of publication: 2001 +Isbn: 0199608113 +Edition: 3
\ No newline at end of file diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/chapter0.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/chapter0.ipynb new file mode 100755 index 00000000..4cc1774a --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/chapter0.ipynb @@ -0,0 +1,198 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 0 - Introduction"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 4"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the final pressure of the gas\n",
+ "#Initialization of variables\n",
+ "P=1.115 #bar\n",
+ "#Calculations\n",
+ "Conv_fac=1/1.01325\n",
+ "FinalP=Conv_fac*P #Final pressure\n",
+ "#Results\n",
+ "print '%s %.3f' %('Final pressure in atmospheres (atm)= ',FinalP)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Final pressure in atmospheres (atm)= 1.100\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I1 - Pg 5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the pressure at the foot of the column\n",
+ "#Initialization of variables\n",
+ "h=0.760 #m\n",
+ "d=1.36*10000 #kg/m^3\n",
+ "#Calculations\n",
+ "P=9.81*d*h\n",
+ "#Results\n",
+ "print '%s %.3e' %('Pressure at the foot of the column (Pa)= ',P)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pressure at the foot of the column (Pa)= 1.014e+05\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 6"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the hydrostatic pressure and pressure in apparatus\n",
+ "#Initialization of variables\n",
+ "h=0.1 #m\n",
+ "d=1000 #Kg/m^3\n",
+ "Patm=100021. #Pa\n",
+ "#Calculations\n",
+ "P=9.81*h*d\n",
+ "Papp=(Patm-P)/1000.\n",
+ "#Results\n",
+ "print '%s %.3f' %('Hydrostatic pressure(Pa) = ',P )\n",
+ "print '%s %.3f' %('\\n Pressure in apparatus(kPa) = ',Papp)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Hydrostatic pressure(Pa) = 981.000\n",
+ "\n",
+ " Pressure in apparatus(kPa) = 99.040\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I3 - Pg 8"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the no.of moles of copper required\n",
+ "import math\n",
+ "#Initialization of variables\n",
+ "N=8.8*math.pow(10,22)\n",
+ "NA=6.023*math.pow(10,23) #mol^-1\n",
+ "#Calculations\n",
+ "n=N/NA\n",
+ "#Results\n",
+ "print '%s %.2f' %('No. of moles of Cu ( mol Cu)= ',n)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No. of moles of Cu ( mol Cu)= 0.15\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I4 - Pg 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the amount of Carbon atoms \n",
+ "#Initialization of variables\n",
+ "m=21.5 #g\n",
+ "Mc=12.01 #g/mol\n",
+ "#Calculations\n",
+ "nc=m/Mc\n",
+ "#Results\n",
+ "print '%s %.2f %s' %(\"Amount of C atoms=\",nc,\"mol C\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Amount of C atoms= 1.79 mol C\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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