{
"metadata": {
"name": "",
"signature": "sha256:33e389243f3edbfa798e02801dff08e24da71a0ad10959698b7564f561b867d4"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter17-Ideal solutions"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1-pg 480"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#intilization variables\n",
"x=0.25\n",
"pa=40\n",
"pb=50\n",
"ya=0.25\n",
"alpha=1.25\n",
"#calculation\n",
"P=x *pa+(1-x)*pb\n",
"y=x*pa/P\n",
"yb=(1-y)\n",
"xa=alpha*y/(1+(alpha-1)*y)\n",
"xb=(1-x)\n",
"#results\n",
"print'%s %.2f %s'%('total pressure of an ideal solution',P,'kpa')\n",
"print'%s %.2f %s %.2f %s '%('composition of the gaseous phase',y,'' and ' ',yb,'')\n",
"print'%s %.2f %s %.2f %s '%('the composition of last drop',xa,' ' and ' ',xb,'') "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"total pressure of an ideal solution 47.50 kpa\n",
"composition of the gaseous phase 0.21 0.79 \n",
"the composition of last drop 0.25 0.75 \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example2-pg484"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"##initialisation of variables\n",
"T= 290 ##K\n",
"xa= 0.4\n",
"xb= 0.6\n",
"P= 600 ##kPa\n",
"V= 60 ##L\n",
"R= 8.314 ##J/mol K\n",
"Mp= 44 ##kg/kmol\n",
"Mb= 58.12 ##kg/kmol\n",
"vp= 0.00171 ##m**3/kg\n",
"vb= 0.00166 ##m**3/kg\n",
"na= 0.1 ##kmol\n",
"nb= 0.15 ##kmol\n",
"V1= 0.04000 ##m**3\n",
"xa= 0.4 \n",
"np= 2\n",
"Vc= 0.1 ##m**3\n",
"##CALCULATIONS\n",
"Pasat= math.e**(14.435-(2255/T))\n",
"Pbsat= math.e**(14.795-(2770/T))\n",
"P1= xa*Pasat+xb*Pbsat\n",
"Na1= P*V/(100*R*T)\n",
"Vp= vp*Mp\n",
"Vb= vb*Mb\n",
"V= na*Vp+nb*Vb\n",
"Vv= V1-V\n",
"nv= P1*Vv/(R*T)\n",
"ya= xa*Pasat/P\n",
"yb=1-ya\n",
"Na= na+ya*nv\n",
"Nb= nb+yb*nv\n",
"##RESULTS\n",
"print'%s %.2f %s'% (' initial pressure= ',P1,' kPa')\n",
"print'%s %.2f %s'% (' moles of propane= ',Na1,' kmol')\n",
"print'%s %.2f %s'% (' initial mole of propane= ',Na,' kmol')\n",
"print'%s %.2f %s'% (' initial mole of butane= ',Nb,' kmol')\n",
"print'%s %.2f %s'% (' numbar of phases= ',np,'')\n",
"print'%s %.2f %s'% (' volume in final state=',Vc,' m^3')"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" initial pressure= 874.89 kPa\n",
" moles of propane= 0.15 kmol\n",
" initial mole of propane= 0.11 kmol\n",
" initial mole of butane= 0.15 kmol\n",
" numbar of phases= 2.00 \n",
" volume in final state= 0.10 m^3\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example3-pg489"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\t\n",
"#calculate pressure of the phase of pure A\n",
"##initialisation of variables\n",
"p0= 10. ##Mpa\n",
"R= 8.314 ##J/mol K\n",
"T= 30. ##C\n",
"va= 0.02 ##m^3/kmol\n",
"xa= 0.98\n",
"##CALCULATIONS\n",
"p= p0+(R*(273.15+T)*math.log(xa)/(va*1000.))\n",
"##RESULTS\n",
"print'%s %.2f %s'%('Pressure of the phase of pure A=',p,'Mpa')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Pressure of the phase of pure A= 7.45 Mpa\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example4-pg491"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate the boiling point elevation\n",
"##initialisation of variables\n",
"hfg= 2257.0 ##kJ/kg\n",
"Tb= 100 ##C\n",
"R= 8.314 ##J/mol K\n",
"m2= 10 ##gms\n",
"M2= 58.5 ##gms\n",
"m1= 90. ##gms\n",
"M1= 18. ##gms\n",
"##CALCULATIONS\n",
"x2= (m2/M2)/((m2/M2)+(m1/M1))\n",
"dT= R*math.pow(273.15+Tb,2)*x2/(M1*hfg)\n",
"##RESULTS\n",
"print'%s %.3f %s'%(' Boiling point elevation=',dT,'C')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Boiling point elevation= 0.942 C\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example5-pg494"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate Osomatic pressures\n",
"##initialisation of variables\n",
"M1= 18.02 ##gms\n",
"m1= 0.965 ##gms\n",
"m2= 0.035 ##gms\n",
"M2= 58.5 ##gms\n",
"R= 8.314 ##J/mol K\n",
"M= 18.02 ##kg\n",
"T= 20. ##C\n",
"vf= 0.001002 ##m^3\n",
"x21= 0.021856 ##m^3\n",
"##CALCULATIONS\n",
"n1= m1/M1\n",
"n2= m2/M2\n",
"x1= n1/(n1+n2)\n",
"x2= n2/(n2+n1)\n",
"P= R*(273.15+T)*x2/(M*vf)\n",
"P1= R*(273.15+T)*x21/(M*vf)\n",
"##RESULTS\n",
"print'%s %.1f %s'%('Osmotic pressure=',P,'kpa')\n",
"print'%s %.1f %s'%('Osmotic pressure=',P1,'kpa')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Osmotic pressure= 1491.4 kpa\n",
"Osmotic pressure= 2950.2 kpa\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example6-pg495"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#what is useful work in the process and heat interaction and maximum work and irreversibility\n",
"##initialisation of variables\n",
"W= 0.\n",
"Q= 0.\n",
"R= 8.314 ##J/mol K\n",
"T0= 300. ##K\n",
"x= 5./13.\n",
"n1= 0.5 ##kmol/s\n",
"n2= 0.8 ##kmol/s\n",
"##CALCULATIONS\n",
"W1= (n1+n2)*R*T0*(x*math.log(1/x)+(1-x)*math.log(1./(1.-x)))+470\n",
"I= W1\n",
"##RESULTS\n",
"print'%s %.f %s'%('useful work of the process=',W,'kW') \n",
"print'%s %.f %s'%('heat interaction=',Q,'kW') \n",
"print'%s %.1f %s'%('maximum work=',W1,'kW') \n",
"print'%s %.1f %s'%('irreversibility=',I,'kW')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"useful work of the process= 0 kW\n",
"heat interaction= 0 kW\n",
"maximum work= 2630.4 kW\n",
"irreversibility= 2630.4 kW\n"
]
}
],
"prompt_number": 8
}
],
"metadata": {}
}
]
}