{
"metadata": {
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 12 - Physical Equilibria"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1 - pg 295"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the no of degrees of freedom\n",
"#Initialization of variables\n",
"p=3\n",
"c=2\n",
"#calculations\n",
"f=2-p+c\n",
"#results\n",
"print '%s %d' %(\"no. of degrees of freedom =\",f)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"no. of degrees of freedom = 1\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2 - pg 301"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the change in temperature\n",
"#Initialization of variables\n",
"T=273.2 #K\n",
"vw=1.0001 #cm^3 /g\n",
"vi=1.0907 #cm^3 /g\n",
"hf=79.7 #cal/g\n",
"P1=76 #cm\n",
"P2=4.6 #cm\n",
"#calculations\n",
"dT=T*(vw-vi)*(P2-P1)*13.6*980.7/(hf*4.184*10**7)\n",
"#results\n",
"print '%s %.4f %s' %(\"change in temperature =\",dT,\"deg\")\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"change in temperature = 0.0071 deg\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3 - pg 302"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the value of dPbydT\n",
"#Initialization of variables\n",
"V=6.84 #cm^3 /g\n",
"#calculations\n",
"dPbydT=-1.7*4.184*10**7 /(2.19*V*0.06*1.01*10**6)\n",
"#results\n",
"print '%s %d %s' %(\"dPbydT =\",dPbydT,\" atm/deg\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"dPbydT = -78 atm/deg\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4 - pg 303"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Pressure\n",
"#Initialization of variables\n",
"P=6 #atm\n",
"T=273.2+25 #K\n",
"P=23.8 #mm\n",
"V=0.018 #lt/mol\n",
"R=0.08206 #lt am/deg mol\n",
"#calculations\n",
"dPa=V*P*4536/(R*T*760)\n",
"Pa=dPa+P\n",
"#results\n",
"print '%s %.1f %s' %(\"Pressure =\",Pa,\" mm\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Pressure = 23.9 mm\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5 - pg 305"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the mole fraction of methanol in vapor\n",
"#Initialization of variables\n",
"x=0.25\n",
"Ps1=96 #mm\n",
"Ps2=43.9 #mm\n",
"#calculations\n",
"P1=x*Ps1\n",
"P2=(1-x)*Ps2\n",
"P=P1+P2\n",
"Xdash=P1/P\n",
"#results\n",
"print '%s %.3f' %(\"mole fraction of methanol in vapor =\",Xdash)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"mole fraction of methanol in vapor = 0.422\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6 - pg 309"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Molal elevation constant\n",
"#Initialization of variables\n",
"Hv=539.6 #cal/g\n",
"T=273.2+100 #K\n",
"#calculations\n",
"Kb=1.987*T**2 /(1000*Hv)\n",
"#results\n",
"print '%s %.3f %s' %(\"Molal elevation constant =\",Kb,\" deg /mole /kg\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Molal elevation constant = 0.513 deg /mole /kg\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7 - pg 309"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Molecular weight of solute\n",
"#Initialization of variables\n",
"ms=0.5 #mol/kg\n",
"m=5 #g\n",
"mw=100 #g\n",
"Ws=1000 #g/kg\n",
"#calculations\n",
"Ma=m*Ws/(ms*mw)\n",
"#results\n",
"print '%s %d %s' %(\"Molecular weight of solute =\",Ma,\"g/mol \")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Molecular weight of solute = 100 g/mol \n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8 - pg 311"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the molality of the solution\n",
"#Initialization of variables\n",
"dT=0.23 #C\n",
"Kb=1.86 #deg/mol/kg\n",
"#calculations\n",
"m=dT/Kb\n",
"#results\n",
"print '%s %.2f %s' %(\"molality of solution =\",m,\"m\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"molality of solution = 0.12 m\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9 - pg 313"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Osmotic Pressure\n",
"#Initialization of variables\n",
"p=0.1 #m\n",
"T=30+273.2 #K\n",
"R=0.08206 #lt atm /deg/mol\n",
"P1=1 #atm\n",
"#calculations\n",
"w=1000/p\n",
"V=w/1000.\n",
"dP=R*T/V\n",
"P=dP+P1\n",
"#results\n",
"print '%s %.2f %s' %(\"Osmotic Pressure =\",P,\" atm \")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Osmotic Pressure = 3.49 atm \n"
]
}
],
"prompt_number": 10
}
],
"metadata": {}
}
]
}