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+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 15 : The Phase Rule"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 15.2  Page: 401"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate independent relations among these four species\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "\n",
+      "# The system contains four species\n",
+      "print \" In this system there are four identifiable chemical species%(which are C,O2,CO2 and CO. The balanced equations we can write among them are\"\n",
+      "\n",
+      "print \"    C + 0.5O2 = CO\"\n",
+      "print \"    C + O2 = CO2\"\n",
+      "print \"    CO + 0.5O2 = CO2\"\n",
+      "print \"    CO2 + C = 2CO\"\n",
+      "\n",
+      "# Let we call these equations A, B, C and D respectively\n",
+      "# These relations are not independent.\n",
+      "# If we add A and C and cancel like terms, we obtain B.\n",
+      "# So, If we want independent chemical equilibria we must remove equation C\n",
+      "\n",
+      "# Now, if we reverse the direction of B and add it to A, we see that D is also not independent.\n",
+      "# Thus, there are only two independent relations among these four species and \n",
+      "print \" There are only two independent relations among these four species and\"\n",
+      "\n",
+      "# V = C + 2 - P\n",
+      "# and we have\n",
+      "V = 2# No of the variable\n",
+      "P = 2# No of the phases\n",
+      "# So\n",
+      "C = V + P - 2\n",
+      "print \"    C = V + P - 2\"\n",
+      "print \"    C = 4 - 2 = 2\"\n",
+      "print \" Thus, this is a two-component system\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " In this system there are four identifiable chemical species%(which are C,O2,CO2 and CO. The balanced equations we can write among them are\n",
+        "    C + 0.5O2 = CO\n",
+        "    C + O2 = CO2\n",
+        "    CO + 0.5O2 = CO2\n",
+        "    CO2 + C = 2CO\n",
+        " There are only two independent relations among these four species and\n",
+        "    C = V + P - 2\n",
+        "    C = 4 - 2 = 2\n",
+        " Thus, this is a two-component system\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 15.3   Page: 402\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Define Phase rule\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "#***Data***#\n",
+      "# This contains three species.\n",
+      "print \" The three species in this system are H2 N2 and NH3\"\n",
+      "N = 3\n",
+      "print \" There is only one balanced chemical reaction among these species\"\n",
+      "Q = 1\n",
+      "\n",
+      "# 2NH3 = N2 + 3H2\n",
+      "C = N - Q\n",
+      "print \"    C = N - Q = %0.0f\"%(C)\n",
+      "# Now let us we made the system by starting with pure ammonia.\n",
+      "# Assuming that all the species are in the gas phase, ammonia dissociates in H2 and N2 in the ratio of 3:1.\n",
+      "print \" Let we start with pure ammonia in the system then ammonia will dissociate in H2 and N2 in the ratio of 3:1.\"\n",
+      "\n",
+      "# We can write an equation among their mole fractions, viz\n",
+      "# y_H2 = 3*y_N2\n",
+      "print \" And the relation between their mole fraction is    y_H2 = 3*y_N2\"\n",
+      "\n",
+      "# We might modify the phase rule to put in another symbol for stoichiometric restrictions, but the common usage is to write that \n",
+      "# Components = species - (independent reactions) - (stoichiometric restriction)\n",
+      "# and stoichiometric restriction SR is \n",
+      "SR = 1\n",
+      "# so\n",
+      "c = N-Q-SR\n",
+      "print \" We have the modified phase rule as    Components = species - independent reactions - stoichiometric restriction\"\n",
+      "print \"    C = N - Q - SR = %0.0f\"%(c)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The three species in this system are H2 N2 and NH3\n",
+        " There is only one balanced chemical reaction among these species\n",
+        "    C = N - Q = 2\n",
+        " Let we start with pure ammonia in the system then ammonia will dissociate in H2 and N2 in the ratio of 3:1.\n",
+        " And the relation between their mole fraction is    y_H2 = 3*y_N2\n",
+        " We have the modified phase rule as    Components = species - independent reactions - stoichiometric restriction\n",
+        "    C = N - Q - SR = 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 15.4   Page: 403\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Find number of the components present in the test tube\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "#***Data***#\n",
+      "# We have been given the reaction \n",
+      "# CaCO3(s) = CaO(s) + CO2(g)\n",
+      "\n",
+      "# Here we have three species and one balanced chemical reaction between them\n",
+      "# So\n",
+      "N = 3# No of species\n",
+      "Q = 1 # no of reaction\n",
+      "\n",
+      "# Since CO2 will mostly be in the gas phase and CaCO3 and CaO will each form separate solid phases, \n",
+      "# there is no equation we can write among the mole fractions in any of the phases.\n",
+      "# Hence, there is no stoichiometric restriction i.e.\n",
+      "SR = 0\n",
+      "# and the number of the components is\n",
+      "C = N - Q - SR\n",
+      "\n",
+      "print \"Number of the components presents in the test tube are %0.0f\"%(C)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of the components presents in the test tube are 2\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 15.5   Page: 403\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''\n",
+      "find\n",
+      "(a) How many phases are present?\n",
+      "(b) How many degrees of freedom are there?\n",
+      "(c) If we place a sample of pure CaCO 3 in an evacuated container and heat it, will we find a unique P-T curve?\n",
+      "'''\n",
+      "import math \n",
+      "\n",
+      "#***Data***#\n",
+      "# We have been given the reaction \n",
+      "# CaCO3(s) = CaO(s) + CO2(g)\n",
+      "# The CaCO3 and CaO form separate solid phases, so we have three phases, two solid and one gas. \n",
+      "# So\n",
+      "P = 3\n",
+      "# This is a two component system, so\n",
+      "C = 2\n",
+      "\n",
+      "# From the phase rule\n",
+      "V = C + 2 - P\n",
+      "\n",
+      "# If there is only one degree of freedom, then the system should have a unique P-T curve.\n",
+      "# Reference [ 2, page 214 ] as reported in the book, shows the data to draw such a curve, which can be well represented by\n",
+      "# math.log(p/torr) = 23.6193 - 19827/T\n",
+      "\n",
+      "print \" The no. of phases present in the system are %0.0f \"%(P)\n",
+      "print \" Total no of degrees of freedom is %0.0f \"%(V)\n",
+      "print \" Since there is only one degree of freedom so the system has a unique P-T curve\"\n",
+      "print \" which can be well represented by     logp/torr = 23.6193 - 19827/T\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The no. of phases present in the system are 3 \n",
+        " Total no of degrees of freedom is 1 \n",
+        " Since there is only one degree of freedom so the system has a unique P-T curve\n",
+        " which can be well represented by     logp/torr = 23.6193 - 19827/T\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 15.6   Page: 404\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# find number of components\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "#***Data***#\n",
+      "# The system consists of five species.\n",
+      "print \" The five species present in the system are H2O%( HCl%( H+%( OH- and Cl-. \"\n",
+      "# So\n",
+      "N = 5               # Number of the species \n",
+      "print \" Here we have two chemical relations:\"\n",
+      "print \"    H2O = H+  +  OH- \"\n",
+      "print \"    HCl = H+  +  Cl- \"\n",
+      "\n",
+      "# so\n",
+      "Q = 2               # No of the reactions\n",
+      "\n",
+      "# In addition we have electroneutrality, which says that at equilibrium the total no of positive ions in the solution must be the same as the total no of nagative ions,or\n",
+      "# [H+] = [OH-] + [Cl-]\n",
+      "# To maintain electroneutrality number of positive and negative ion should be same.\n",
+      "# Here [H+] smath.radians(numpy.arcmath.tan(s for the molality of hydrogen ion. This is convertible to a relation among the 'mu's' hence,\n",
+      "# it is an additional restriction, so\n",
+      "SR = 1 \n",
+      "# So\n",
+      "# The number of components is\n",
+      "C = N - Q - SR\n",
+      "\n",
+      "print \" Number of the components present in the system are     C = N - Q - SR = %0.0f\"%(C)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The five species present in the system are H2O%( HCl%( H+%( OH- and Cl-. \n",
+        " Here we have two chemical relations:\n",
+        "    H2O = H+  +  OH- \n",
+        "    HCl = H+  +  Cl- \n",
+        " Number of the components present in the system are     C = N - Q - SR = 2\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
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