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+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 :  Partial Molal Properties"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 6.1  Page: 108\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''\n",
+      "calculate\n",
+      "Partial molal volume of ethanol in water at zero molality\n",
+      "Partial molal volume of ethanol in water at unity molality \n",
+      "'''\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "T = 20.             #[C]\n",
+      "m_1 = 0.            #[molal]\n",
+      "m_2 = 1.            #[molal]\n",
+      "\n",
+      "# Calculations\n",
+      "# The data given in the figure 6.2 , as reported in book, can be repersented with excellent accuracy by a simple data fitting equation\n",
+      "#V = 1.0019+0.054668*m-0.000418*m**(2)\n",
+      "# Where 'V' is( solution volume, liters per 1000g of water ) and 'm' is the molality of ethanol in water\n",
+      "#The partial molal volume is obtained by differentiating the expression of the 'V' with respect to 'm'\n",
+      "# v_ethanol = dV/dm = 0.054668-2*0.000418*m\n",
+      "# So that at zero molality \n",
+      "m = 0               #[molal]\n",
+      "# the partial molal volume is \n",
+      "v_1 = 0.054668-2*0.000418*m         #[L/mol]\n",
+      "# and at\n",
+      "m = 1.              #[molal]\n",
+      "v_2 = 0.054668-2*0.000418*m         #[L/mol]\n",
+      "v_1 = v_1*1000                      #[cm**(3)/mol]\n",
+      "v_2 = v_2*1000                      #[cm**(3)/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \"Partial molal volume of ethanol in water at zero molality is  %f cm**3/mol\"%(v_1)\n",
+      "print \" Partial molal volume of ethanol in water at unity molality is %f cm**3/mol\"%(v_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Partial molal volume of ethanol in water at zero molality is  54.668000 cm**3/mol\n",
+        " Partial molal volume of ethanol in water at unity molality is 53.832000 cm**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 6.2  Page: 109\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculate Volume change on mixing etanol and water\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "n_eth = 1.          #[mol]\n",
+      "W_water = 1.        #[kg]\n",
+      "Temp = 20.          #[C]\n",
+      "\n",
+      "# For pure ethanol at 20C\n",
+      "v_ethanol = 58.4            #[cm**(3)/mol]\n",
+      "v_ethanol = v_ethanol/1000  # [L/mol]\n",
+      "v_water = 1.0019            #[L/1000g]\n",
+      "\n",
+      "# Calculations\n",
+      "# Molality of ethanol in water is\n",
+      "m = n_eth/W_water           #[molal]\n",
+      "# We have the equation used in the previous example as\n",
+      "V_final_mix = 1.0019+0.054668*m-0.000418*m**(2)\n",
+      "\n",
+      "# Where 'V' is( solution volume, liters per 1000g of water ) and 'm' is the molality of ethanol in water\n",
+      "# V is the final volume of the solution \n",
+      "# The volume expansion on moxing is \n",
+      "V_exp = V_final_mix-v_ethanol-v_water       #[L]\n",
+      "V_exp = V_exp*1000                          #[cm**(3)]\n",
+      "\n",
+      "# Results\n",
+      "print \"Volume change on mixing emath.tanol and water is %0.3f cubic cm\"%(V_exp)\n",
+      "# We see that there is a net contraction on mixing of the volume of the ethanol added.\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volume change on mixing emath.tanol and water is -4.150 cubic cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 6.3  Page: 109\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# find Volume change on mixing etanol and water \n",
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "# All the data are same as in the previous example \n",
+      "# The equation 6.5 reported in the book is \n",
+      "v_i_average = 0.05425       #[L/mol]\n",
+      "# and\n",
+      "v_i_0 = 0.0584              #[L/mol]\n",
+      "delta_n = 1.00              #[mol]\n",
+      "\n",
+      "# Calculations\n",
+      "delta_V_mixing = (v_i_average-v_i_0)*delta_n        #[L]\n",
+      "delta_V_mixing = delta_V_mixing*1000                #[cm**(3)]\n",
+      "\n",
+      "# Results\n",
+      "print \"Volume change on mixing etanol and water is %f cm**3\"%(delta_V_mixing)\n",
+      "# Which is same as the solution in example 6.2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volume change on mixing etanol and water is -4.150000 cm**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 6.4  Page: 113\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''\n",
+      "calculate\n",
+      "Partial molar volume of the ethanol \n",
+      "Partial molar volume of the water \n",
+      "'''\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "m = 1.              #[molal] Molality of the solution with respect to ethanol\n",
+      "M_water = 18.       #[g/mol] molecular weight of water\n",
+      "\n",
+      "# Calculations\n",
+      "# First we convert molality to mole fraction\n",
+      "x_ethanol = m/(m + 1000/M_water)\n",
+      "\n",
+      "# For the low range of data point on figure 6.5(page 112), we can fit an equation\n",
+      "# (Specific volume ) = 0.018032 + 0.037002*x_ethanol - 0.039593*x_ethanol**(2) + 0.21787*x_ethanol**(3)\n",
+      "# This is applicable for (0 < x_ethanol < 0.04 ), which is the case we have\n",
+      "\n",
+      "# So\n",
+      "v_math_tan = 0.018032 + 0.037002*x_ethanol - \\\n",
+      "0.039593*x_ethanol**(2) + 0.21787*x_ethanol**(3)            #[L/mol]\n",
+      "\n",
+      "# Now we will find the derivative of the specific volume with respect to x_ethanol at the known point x_ethanol\n",
+      "# (dv/dx_ethanol) =  0.037002 - 2*0.039593*x_ethanol + 3*0.21787*x_ethanol**(2)\n",
+      "# Hence\n",
+      "v_derv_math_tan = 0.037002 - 2*0.039593*x_ethanol + 3*0.21787*x_ethanol**(2)        #[L/mol]\n",
+      "\n",
+      "# By simple geometry from the figure 6.6(page 113) of the book we find\n",
+      "# a = v_math_tan + (1-x_math_tan)*(dv/dx_1)_math_tan\n",
+      "# b = v_math_tan - x_math_tan*(dv/dx_1)_math_tan\n",
+      "\n",
+      "# We have a = v_ethanol and b = v_water\n",
+      "x_math_tan = x_ethanol\n",
+      "# So\n",
+      "v_ethanol = v_math_tan + (1-x_math_tan)*(v_derv_math_tan)       #[L/mol]\n",
+      "v_water = v_math_tan - x_math_tan*(v_derv_math_tan)             #[L/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \" Partial molar volume of the ethanol in the given solution is %f L/mol\"%(v_ethanol)\n",
+      "print \" Partial molar volume of the water in the given solution is %f L/mol\"%(v_water)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Partial molar volume of the ethanol in the given solution is 0.053848 L/mol\n",
+        " Partial molar volume of the water in the given solution is 0.018042 L/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 6.7  Page: 117\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''\n",
+      "calculate\n",
+      "Partial molar enthalpy of water in the mixture\n",
+      "Partial molar enthalpy of H2SO4 in the mixture\n",
+      "'''\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "x_sulph = 0.6\n",
+      "x_water = 0.4\n",
+      "Temp = 200.             #[F]\n",
+      "# In the given figure 6.8 in the book, drawing the math.tangent to the 200F curve at 60 wt% H2SO4, we find that it intersects the 0%(pure water) axis at      25 Btu/lbm, and the 100% H2SO4 axis at -100Btu/lbm. i.e.\n",
+      "h_water_per_pound = 25.         #[Btu/lbm]\n",
+      "h_sulph_per_pound = -100.       #[Btu/lbm]\n",
+      "# also molecular weight of water and sulphuric acid are\n",
+      "M_water = 18.                   #[lbm/lbmol]\n",
+      "M_sulph = 98.                   #[lbm/lbmol]\n",
+      "\n",
+      "# Calculations\n",
+      "# Using equation 6.20 given in the book we have\n",
+      "h_water = h_water_per_pound*M_water     #[Btu/lbmol]\n",
+      "h_sulph = h_sulph_per_pound*M_sulph     #[Btu/lbmol]\n",
+      "\n",
+      "# Results\n",
+      "print \"Partial molar enthalpy of water in the mixture is  %f Btu/lbmol\"%(h_water)\n",
+      "print \" Partial molar enthalpy of H2SO4 in the mixture is %f Btu/lbmol\"%(h_sulph)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Partial molar enthalpy of water in the mixture is  450.000000 Btu/lbmol\n",
+        " Partial molar enthalpy of H2SO4 in the mixture is -9800.000000 Btu/lbmol\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 6.8  Page: 119\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# find The amount of heat removed to keep the temperature constant\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "x_sulph = 0.6\n",
+      "x_water = 0.4\n",
+      "M_i = 18.           #[lbm/lbmol]\n",
+      "Temp = 200.         #[F]\n",
+      "# From Equation 6.11 as given in the book, we have \n",
+      "# dQ/dm_in = h_i-h_in\n",
+      "# where h_i is partial molal enthalpy which is taken from the example 6.7 and h_in is the pure species molar enthalpy which is read from the figure 6.8.\n",
+      "# So at 200F we have \n",
+      "h_i = 25.           #[Btu/lbm]\n",
+      "h_in = 168.         #[Btu/lbm]\n",
+      "\n",
+      "# Calculations\n",
+      "# hence\n",
+      "dQ_by_dm_in = h_i-h_in          #[Btu/lbm]\n",
+      "# Now \n",
+      "dQ_by_dn_in = M_i*dQ_by_dm_in       #[Btu/lbmol]\n",
+      "\n",
+      "# Results\n",
+      "print \"The amount of heat removed to keep the temperature consmath.tant is %f Btu/lbm of water added\"%(dQ_by_dm_in)\n",
+      "# The negative sign shows that this mixing is exothermic we must remove 143 Btu/lbm of water added.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The amount of heat removed to keep the temperature consmath.tant is -143.000000 Btu/lbm of water added\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 6.9  Page: 119\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# find The amount of heat added or removed \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "m_sulph = 0.4\n",
+      "m_water = 0.6\n",
+      "m = m_sulph+m_water\n",
+      "Temp = 200.             #[F]\n",
+      "# Here at 200F we can read the solution enthalpy h_solution and pure H2SO4 enthalpy h_sulph such that\n",
+      "h_solution = -43.       #[Btu/lbm]\n",
+      "h_sulph = 168.           #[Btu/lbm]\n",
+      "# By energy balance, umath.sing h_0_water from example 6.7 in the book i.e.\n",
+      "h_0_water = 53.        #[Btu/lbm]\n",
+      "\n",
+      "# We find \n",
+      "# Calculations\n",
+      "delta_Q = m*h_solution-(m_sulph*h_sulph+m_water*h_0_water)      #[Btu]\n",
+      "\n",
+      "# Results\n",
+      "print \"The amount of heat added or removed is %f Btu\"%(delta_Q)\n",
+      "# We must remove the given amount of to hold the temperature consmath.tant.\n",
+      "# Note However the book has some mistake in calculation and reporting -172 Btu\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The amount of heat added or removed is -142.000000 Btu\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 6.10  Page: 120\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# calculate enthalpy of the solution\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "x_sulph = 0.6\n",
+      "x_water = 0.4\n",
+      "Temp = 200.         #[F]\n",
+      "# At the 200F we have\n",
+      "h_water = 25.       #[Btu/lbm]\n",
+      "h_sulph = -100.     #[Btu/lbm]\n",
+      "\n",
+      "# Calculations\n",
+      "# From equation 6.16 (as reporated in the book), rewritten for masses instead of moles we have \n",
+      "h_solution = h_water*x_water+h_sulph*x_sulph        # [Btu/lbm]\n",
+      "\n",
+      "# Results\n",
+      "print \"Enthalpy of the solution is %f Btu/lbm\"%(h_solution)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enthalpy of the solution is -50.000000 Btu/lbm\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 6.11  Page: 121\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# find Value of the dv_b/dx_a at x_b =0\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "x_b = 0\n",
+      "x_a = 1\n",
+      "# We have\n",
+      "#dv_a/dx_a = 3*x_b**(2)+2*x_b\n",
+      "# We have the equation \n",
+      "# dv_b/dx_a = -(dv_a/dx_a)/(x_b/x_a)\n",
+      "# So\n",
+      "# dv_b/dx_a = -(x_a/x_b)*(3*x_b**(2)+2*x_b) \n",
+      "# Calculations\n",
+      "dv_b_by_dx_a = x_a*(-3*x_b-2)\n",
+      "\n",
+      "# Results\n",
+      "print \"Value of the dv_b/dx_a at x_b =0 is %0.0f\"%(dv_b_by_dx_a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of the dv_b/dx_a at x_b =0 is -2\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file