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author | hardythe1 | 2015-05-05 14:21:39 +0530 |
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committer | hardythe1 | 2015-05-05 14:21:39 +0530 |
commit | fba055ce5aa0955e22bac2413c33493b10ae6532 (patch) | |
tree | be70ef4fccd07c9c88de778014219201b4ea971f /Physical_Chemsitry/Chapter13.ipynb | |
parent | 67068710030ddd6b6c809518c34af2e04e0bf7ca (diff) | |
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diff --git a/Physical_Chemsitry/Chapter13.ipynb b/Physical_Chemsitry/Chapter13.ipynb deleted file mode 100755 index 79973f5f..00000000 --- a/Physical_Chemsitry/Chapter13.ipynb +++ /dev/null @@ -1,401 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:b3b8e3a1df1ec221a5596f2fd615d0e7df4ba35a4012aeb734d5820e6983cd64"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 13 - Thermodynamic changes accompanying chemical reaction"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 320"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the dHp value\n",
- "#Initialization of variables\n",
- "n1=10 #mol\n",
- "n2=12 #mol\n",
- "#calculations\n",
- "dn=n1-n2\n",
- "#results\n",
- "print '%s %d %s' %(\"dHp = dEv-\",dn,\"*RT\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dHp = dEv- -2 *RT\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 322"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Enthalpy\n",
- "#Initialization of variables\n",
- "Ht1=-22063 #cal\n",
- "T=298.15 #K\n",
- "#calculations\n",
- "H=Ht1 +0.5293*T + 0.3398*10**-3 *T**2 - 2.039*10**-7 *T**3\n",
- "#results\n",
- "print '%s %d %s' %(\"Enthalpy =\",H,\"cal\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Enthalpy = -21880 cal\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - pg 326"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Entropy\n",
- "#Initialization of variables\n",
- "Cp=0.797 #cal/deg/mol\n",
- "#calculations\n",
- "S=Cp/3.\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Entropy =\",S,\" eu/mol\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy = 0.266 eu/mol\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - pg 328"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Change in entropy\n",
- "#Initialization of variables\n",
- "T1=77.32 #K\n",
- "P=1 #atm\n",
- "T2=126 #K\n",
- "Pc=33.5 #atm\n",
- "#calculations\n",
- "dS=27/32. *1.987*P/Pc *(T2/T1)**3\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Change in entropy =\",dS,\"eu/mol\")"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Change in entropy = 0.22 eu/mol\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - pg 330"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Change in entropy, enthalpy and free energy\n",
- "#Initialization of variables\n",
- "S1=57.47\n",
- "S2=50.34\n",
- "S3=49\n",
- "H1=8.09\n",
- "H2=21.06\n",
- "H3=0\n",
- "F1=12.39\n",
- "F2=20.72\n",
- "F3=0\n",
- "#calculations\n",
- "dS=S1-S2-0.5*S3\n",
- "dH=H1-H2-0.5*H3\n",
- "dF=F1-F2-0.5*F3\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Change in entropy =\",dS,\" eu\")\n",
- "print '%s %.2f %s' %(\"\\n Change in enthalpy =\",dH,\" kcal\")\n",
- "print '%s %.2f %s' %(\"\\n Change in free energy =\",dF,\"kcal\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Change in entropy = -17.37 eu\n",
- "\n",
- " Change in enthalpy = -12.97 kcal\n",
- "\n",
- " Change in free energy = -8.33 kcal\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - pg 334"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the change in free energy\n",
- "#Initialization of variables\n",
- "import math\n",
- "P1=0.01\n",
- "P2=0.1\n",
- "P3=0.01\n",
- "dF0=-54640 #cal\n",
- "T=298.15 #K\n",
- "R=1.987 #cal/deg\n",
- "#calculations\n",
- "Qp=P1/(P2*P3**0.5)\n",
- "dF=dF0+R*T*math.log(Qp)\n",
- "#results\n",
- "print '%s %d %s' %(\"change in free energy =\",dF,\"cal\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "change in free energy = -54640 cal\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - pg 335"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Equilibrium constant\n",
- "#Initialization of variables\n",
- "print \"From table 13.4 \"\n",
- "logKfwater=40.04724\n",
- "logKfH2=0\n",
- "logKfO2=0\n",
- "#calculations\n",
- "logK=logKfwater-logKfH2-0.5*logKfO2\n",
- "K=10**logK\n",
- "#results\n",
- "print '%s %.4e' %(\"Equilibrium constant = \",K)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table 13.4 \n",
- "Equilibrium constant = 1.1149e+40\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - pg 339"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the value of Kc\n",
- "#Initialization of variables\n",
- "Kp=1.1*10**40 #atm**-0.5\n",
- "dn=-0.5\n",
- "R=0.08206 #lt atm/deg mol\n",
- "T=298.15 #K\n",
- "#calculations\n",
- "Kc=Kp*(R*T)**(-dn)\n",
- "#results\n",
- "print '%s %.1e %s' %(\"Kc =\",Kc,\" (mol/lt)^-0.5\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Kc = 5.4e+40 (mol/lt)^-0.5\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9 - pg 339"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Density of equilibrium mixture\n",
- "#Initialization of variables\n",
- "import numpy\n",
- "Kp=0.141 #atm\n",
- "P=1 #atm\n",
- "nu=2\n",
- "R=0.08206 #lt atm/deg mol\n",
- "T=298.15 #K\n",
- "M=92.02 #g/mol\n",
- "#calculations\n",
- "p=([Kp+ 4*P,0, -Kp])\n",
- "z=numpy.roots(p)\n",
- "alpha=z[0]\n",
- "wbyV=P*M/(R*T*(1+(nu-1)*alpha))\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Density of the equilibrium mixture =\",wbyV,\" g/lt\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Density of the equilibrium mixture = 3.18 g/lt\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10 - pg 340"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Final pressure after equilibrium\n",
- "#Initialization of variables\n",
- "x=0.5\n",
- "P=0.468 #atm\n",
- "#calculations\n",
- "P1=x*P\n",
- "P2=x*P\n",
- "Kp=P1*P2\n",
- "#results\n",
- "print '%s %.4f %s' %(\"Final pressure after equilibrium =\",Kp,\" atm^2\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Final pressure after equilibrium = 0.0548 atm^2\n"
- ]
- }
- ],
- "prompt_number": 12
- }
- ],
- "metadata": {}
- }
- ]
-}
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