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| author | hardythe1 | 2015-05-05 14:21:39 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-05-05 14:21:39 +0530 |
| commit | fba055ce5aa0955e22bac2413c33493b10ae6532 (patch) | |
| tree | be70ef4fccd07c9c88de778014219201b4ea971f /Physical_Chemsitry/Chapter12.ipynb | |
| parent | 67068710030ddd6b6c809518c34af2e04e0bf7ca (diff) | |
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Diffstat (limited to 'Physical_Chemsitry/Chapter12.ipynb')
| -rwxr-xr-x | Physical_Chemsitry/Chapter12.ipynb | 348 |
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diff --git a/Physical_Chemsitry/Chapter12.ipynb b/Physical_Chemsitry/Chapter12.ipynb deleted file mode 100755 index 92f1a058..00000000 --- a/Physical_Chemsitry/Chapter12.ipynb +++ /dev/null @@ -1,348 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:65a756a76ade1ec4b9387e0db33894444f541be4a142adaa7f77f2e89d036580"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 12 - Physical Equilibria"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 295"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the no of degrees of freedom\n",
- "#Initialization of variables\n",
- "p=3\n",
- "c=2\n",
- "#calculations\n",
- "f=2-p+c\n",
- "#results\n",
- "print '%s %d' %(\"no. of degrees of freedom =\",f)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "no. of degrees of freedom = 1\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 301"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the change in temperature\n",
- "#Initialization of variables\n",
- "T=273.2 #K\n",
- "vw=1.0001 #cm^3 /g\n",
- "vi=1.0907 #cm^3 /g\n",
- "hf=79.7 #cal/g\n",
- "P1=76 #cm\n",
- "P2=4.6 #cm\n",
- "#calculations\n",
- "dT=T*(vw-vi)*(P2-P1)*13.6*980.7/(hf*4.184*10**7)\n",
- "#results\n",
- "print '%s %.4f %s' %(\"change in temperature =\",dT,\"deg\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "change in temperature = 0.0071 deg\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - pg 302"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the value of dPbydT\n",
- "#Initialization of variables\n",
- "V=6.84 #cm^3 /g\n",
- "#calculations\n",
- "dPbydT=-1.7*4.184*10**7 /(2.19*V*0.06*1.01*10**6)\n",
- "#results\n",
- "print '%s %d %s' %(\"dPbydT =\",dPbydT,\" atm/deg\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dPbydT = -78 atm/deg\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - pg 303"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Pressure\n",
- "#Initialization of variables\n",
- "P=6 #atm\n",
- "T=273.2+25 #K\n",
- "P=23.8 #mm\n",
- "V=0.018 #lt/mol\n",
- "R=0.08206 #lt am/deg mol\n",
- "#calculations\n",
- "dPa=V*P*4536/(R*T*760)\n",
- "Pa=dPa+P\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Pressure =\",Pa,\" mm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure = 23.9 mm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - pg 305"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the mole fraction of methanol in vapor\n",
- "#Initialization of variables\n",
- "x=0.25\n",
- "Ps1=96 #mm\n",
- "Ps2=43.9 #mm\n",
- "#calculations\n",
- "P1=x*Ps1\n",
- "P2=(1-x)*Ps2\n",
- "P=P1+P2\n",
- "Xdash=P1/P\n",
- "#results\n",
- "print '%s %.3f' %(\"mole fraction of methanol in vapor =\",Xdash)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "mole fraction of methanol in vapor = 0.422\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - pg 309"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Molal elevation constant\n",
- "#Initialization of variables\n",
- "Hv=539.6 #cal/g\n",
- "T=273.2+100 #K\n",
- "#calculations\n",
- "Kb=1.987*T**2 /(1000*Hv)\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Molal elevation constant =\",Kb,\" deg /mole /kg\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Molal elevation constant = 0.513 deg /mole /kg\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - pg 309"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Molecular weight of solute\n",
- "#Initialization of variables\n",
- "ms=0.5 #mol/kg\n",
- "m=5 #g\n",
- "mw=100 #g\n",
- "Ws=1000 #g/kg\n",
- "#calculations\n",
- "Ma=m*Ws/(ms*mw)\n",
- "#results\n",
- "print '%s %d %s' %(\"Molecular weight of solute =\",Ma,\"g/mol \")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Molecular weight of solute = 100 g/mol \n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - pg 311"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the molality of the solution\n",
- "#Initialization of variables\n",
- "dT=0.23 #C\n",
- "Kb=1.86 #deg/mol/kg\n",
- "#calculations\n",
- "m=dT/Kb\n",
- "#results\n",
- "print '%s %.2f %s' %(\"molality of solution =\",m,\"m\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "molality of solution = 0.12 m\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9 - pg 313"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Osmotic Pressure\n",
- "#Initialization of variables\n",
- "p=0.1 #m\n",
- "T=30+273.2 #K\n",
- "R=0.08206 #lt atm /deg/mol\n",
- "P1=1 #atm\n",
- "#calculations\n",
- "w=1000/p\n",
- "V=w/1000.\n",
- "dP=R*T/V\n",
- "P=dP+P1\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Osmotic Pressure =\",P,\" atm \")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Osmotic Pressure = 3.49 atm \n"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
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