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| author | hardythe1 | 2015-01-28 14:31:21 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-01-28 14:31:21 +0530 |
| commit | 53f72e6790ff23b43c8f6a0b69d6386940671429 (patch) | |
| tree | 7745af6dbf2f5b2972b23f9f5a7a19c695a27321 /Elements_of_Physical_Chemistry/Chapter15.ipynb | |
| parent | 7b78be04fe05bf240417e22f74b3fc22e7a77d19 (diff) | |
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added books
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| -rwxr-xr-x | Elements_of_Physical_Chemistry/Chapter15.ipynb | 140 |
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diff --git a/Elements_of_Physical_Chemistry/Chapter15.ipynb b/Elements_of_Physical_Chemistry/Chapter15.ipynb new file mode 100755 index 00000000..ada2a951 --- /dev/null +++ b/Elements_of_Physical_Chemistry/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": {}
+ }
+ ]
+}
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