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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 20 - Statistical thermodynamics"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example I1 - Pg 477"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the relative populations of boat and chair conformations\n",
"#Initialization of variables\n",
"import math\n",
"E=22*1000. #kJ/mol\n",
"T=293. #K \n",
"#calculations\n",
"ratio=math.pow(math.e,(-E/(8.31451*T)))\n",
"#results\n",
"print '%s %.1e' %(\"Relative populations of boat and chair conformations is \",ratio)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Relative populations of boat and chair conformations is 1.2e-04\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example I2 - Pg 478"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the required ratio\n",
"#Initialization of variables\n",
"import math\n",
"g2=5.\n",
"g1=3.\n",
"E2=6.\n",
"E1=2.\n",
"k=1.38*math.pow(10,-23) #J/K\n",
"h=6.626*math.pow(10,-34) #J s\n",
"B=3.18*math.pow(10,11) #Hz\n",
"T =298 #K\n",
"#calculations\n",
"ratio=g2/g1 *(math.pow(math.e,((E1-E2)*h*B/(k*T))))\n",
"#results\n",
"print '%s %.2f' %(\"Ratio= \",ratio)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Ratio= 1.36\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example I3 - Pg 481"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the translational partition function\n",
"#Initialization of variables\n",
"import math\n",
"T=298 #K\n",
"m=32*1.66054*math.pow(10,-27) #kg\n",
"k=1.38066*math.pow(10,-23) #j/k\n",
"V=math.pow(10,-4) #m^3\n",
"h=6.62608*math.pow(10,-34) #J/s\n",
"#calculations\n",
"q=math.pow((2*math.pi*m*k*T),1.5) *V/h/h/h \n",
"#results\n",
"print '%s %.2e' %(\"Translational partition function = \",q)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Translational partition function = 1.75e+28\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E1 - Pg 479"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the partition function at 20 C\n",
"#Initialization of variables\n",
"import math\n",
"E=22 #kJ/mol\n",
"R=8.214 #J/K mol\n",
"T=293 #K\n",
"#Calculations\n",
"q=1+math.pow(math.e,(-E*1000. /(R*T)))\n",
"#results\n",
"print '%s %.4f' %(\"At 20 C, partition function = \",q)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"At 20 C, partition function = 1.0001\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E3 - Pg 485"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the contribution to rotational motion\n",
"#Initialization of variables\n",
"import math\n",
"k=1.38*math.pow(10,-23) #J/K\n",
"h=6.626*math.pow(10,-34) #J s\n",
"B=3.18*math.pow(10,11) #Hz\n",
"T=298 #K\n",
"R=8.314 #J/K mol\n",
"#calculations\n",
"Sm=R*(1+math.log(k*T/(h*B)))\n",
"#results\n",
"print '%s %.1f %s' %(\"Contribution to rotational motion=\",Sm,\"J/ K mol\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Contribution to rotational motion= 33.0 J/ K mol\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E5 - Pg 488"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Equilibrium constant\n",
"#Initialization of variables\n",
"import math\n",
"me=9.10939*math.pow(10,-31) #kg\n",
"k=1.38*math.pow(10,-23) #J/K\n",
"h=6.626*math.pow(10,-34) #J s\n",
"p=math.pow(10,5) #Pa\n",
"T=1000 #K\n",
"R=8.314 #J/K mol\n",
"I=376*1000. #J/mol\n",
"#calculations\n",
"K=math.pow((2*math.pi*me),1.5) *math.pow((k*T),2.5) /(p*h*h*h) *math.pow(math.e,(-I/(R*T)))\n",
"#results\n",
"print '%s %.2e' %(\"Equilibrium constant = \",K)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Equilibrium constant = 2.41e-19\n"
]
}
],
"prompt_number": 6
}
],
"metadata": {}
}
]
}
|
