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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 13 : Dilute solution laws"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 13.1 Page No : 478"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"weight = 10. \t\t\t #weight of NaCl in grams\n",
"volume = 1. \t \t\t #volume of water in litres\n",
"weight_water = 1000. \t\t\t # weight of water in grams (Weight = Volume*Density, density of water = 1g/cc = 1g/ml = 1000g/l)\n",
"molwt_NaCl = 58.5 \t\t\t #molecular weight of NaCl in grams\n",
"molwt_water = 18. \t\t\t #molecular weight of water in grams\n",
"hf = 6.002; \t\t\t #enthalpy change of fusion in kJ/mol at 0 degree celsius\n",
"P = 101.325; \t\t\t #pressure in kPa\n",
"T = 273.15; \t\t\t # freezing point temperature of water at the given pressure in K\n",
"R = 8.314; \t\t\t #universal gas constant in J/molK;\n",
"\n",
"# Calculations\n",
"x2 = (weight/molwt_NaCl)/((weight/molwt_NaCl)+(weight_water/molwt_water))\n",
"delt = (R*T**2*x2)/(hf*10**3)\n",
"\n",
"# Results\n",
"print ' The depression in freezing point of water when 10g of NaCl solute is added = %0.2f K'%(delt);\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The depression in freezing point of water when 10g of NaCl solute is added = 0.32 K\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 13.2 Page No : 480"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"weight = 10.; \t\t \t #weight of NaCl in grams\n",
"volume = 1.; \t\t\t #volume of water in litres\n",
"weight_water = 1000.; \t\t\t # weight of water in grams (Weight = Volume*Density, density of water = 1g/cc = 1g/ml = 1000g/l)\n",
"molwt_NaCl = 58.5; \t\t\t #molecular weight of NaCl in grams\n",
"molwt_water = 18; \t\t\t #molecular weight of water in grams\n",
"lat_ht = 2256.94; \t\t\t #latent heat of vaporization in kJ/kg at 100 degree celsius (obtained from steam tables)\n",
"P = 101.325; \t\t\t #pressure in kPa\n",
"T = 373.15; \t\t\t #boiling point temperature of water at the given pressure in K\n",
"R = 8.314; \t\t\t #universal gas constant in J/molK\n",
"\n",
"# Calculations\n",
"x2 = 0.0031\n",
"hv = (lat_ht*molwt_water)/1000\n",
"delt = (R*T**2*x2)/(hv*10**3)\n",
"\n",
"# Results\n",
"print ' The elevation in boiling point of water when 10g of NaCl solute is added = %0.2f K'%(delt);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The elevation in boiling point of water when 10g of NaCl solute is added = 0.09 K\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 13.3 Page No : 481"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"weight = 10.; \t\t\t #weight of NaCl in grams\n",
"weight_water = 1000.; \t\t\t # weight of water in grams\n",
"molwt_NaCl = 58.5; \t\t\t #molecular weight of NaCl in grams\n",
"molwt_water = 18.; \t\t\t #molecular weight of water in grams\n",
"T = 300.; \t\t\t #prevailing temperature of water in K\n",
"R = 8.314; \t\t\t #universal gas constant in (Pa m**3)/(mol K);\n",
"v = 18*10**-6;\t\t\t #molar volume in m**3/mol\n",
"\n",
"# Calculations\n",
"x2 = 0.0031\n",
"pi = ((R*T*x2)/v)*10**-3\n",
"\n",
"# Results\n",
"print ' The osmotic pressure of a solution conatining 10g of NaCl in 1000g of water at 300K = %0.2f kPa'%(pi);\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The osmotic pressure of a solution conatining 10g of NaCl in 1000g of water at 300K = 429.56 kPa\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 13.4 Page No : 483"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"# Variables\n",
"temp = 20. \t\t\t # prevailing tempearture in degree celsius\n",
"melt_temp = 80.05; \t\t\t # melting point of naphthalene in degree celsius\n",
"hf = 18.574 \t\t\t # enthalpy of fusion in kJ/mol\n",
"R = 8.314 \t\t\t # universal gas constant in J/molK\n",
"\n",
"# Calculations\n",
"t = temp+273.15\n",
"melt_t = melt_temp+273.15\n",
"x2 = math.exp(((hf*10**3)/R)*((1./melt_t)-(1./t)))\n",
"\n",
"# Results\n",
"print ' The ideal solubility of naphthalene at 20 degree celsius = %0.4f'%(x2);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The ideal solubility of naphthalene at 20 degree celsius = 0.2737\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 13.5 Page No : 483"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"t = 295.43; \t\t\t #prevailing temperature in K\n",
"sat_p = 6.05; \t\t\t #Sasturation pressure of carbon dioxide at the prevailing temperature in MPa\n",
"p = 0.1; \t\t\t #pressure at which solubility has to be determined in MPa\n",
"\n",
"# Calculations\n",
"x2 = p/sat_p\n",
"\n",
"# Results\n",
"print ' The solubility of carbon dioxide expressed in mole fraction of carbon dioxide in solution\\\n",
" at 0.1MPa = %0.4f'%(x2);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The solubility of carbon dioxide expressed in mole fraction of carbon dioxide in solution at 0.1MPa = 0.0165\n"
]
}
],
"prompt_number": 6
}
],
"metadata": {}
}
]
}
|
