{
"metadata": {
"name": "",
"signature": "sha256:e18fb187eca74984e6f3523731bde87a33ea02f1017bfbe68a1eb5fa826002ce"
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"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 3 - First law of Thermodynamics"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1 - pg 129"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the increase in energy\n",
"#initialisation of variables\n",
"P= 0.0060 #atm\n",
"M=18. #gm\n",
"L=80 #cal/gm\n",
"H=596.1 #cal/gm\n",
"#calculations\n",
"Hs=M*L+M*H\n",
"#results\n",
"print '%s %d %s' %('Net increase in energy = ',Hs,'cal')"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Net increase in energy = 12169 cal\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2 - pg 130"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the increase in energy\n",
"#initialisation of variables\n",
"P= 0.0060 #atm\n",
"V1= 0.0181 #l\n",
"H= -10730 #cal\n",
"V2= 22.4 #l\n",
"#CALCULATIONS\n",
"W= (V2-P*V1)*(1.987/.08205)\n",
"E= H+W\n",
"#RESULTS\n",
"print '%s %d %s' % (' increase in energy=',E,' cal ')\n",
"print 'The answer differs a bit from the textbook due to rounding off error'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" increase in energy= -10187 cal \n",
"The answer differs a bit from the textbook due to rounding off error\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3 - pg 132"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the increase in energy\n",
"#initialisation of variables\n",
"T1= 70 #C\n",
"T2= 10 #C\n",
"Cp= 18 #cal mole^-1 deg^-1\n",
"P= 1 #atm\n",
"m= 18. #g\n",
"d= 0.9778 #g/ml\n",
"d1= 0.9997 #g/ml\n",
"e= 1.987 #cal\n",
"x= 82.05 #ml atm\n",
"#CALCULATIONS\n",
"H= Cp*(T1-T2)\n",
"E= H-(e/x)*P*((m/d)-(m/d1))\n",
"#RESULTS\n",
"print '%s %.1f %s' % (' increase in energy=',E,'cal ')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" increase in energy= 1080.0 cal \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4 - pg 132"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the conversion factor\n",
"#initialisation of variables\n",
"i= 1 #amp\n",
"r= 2 #ohms\n",
"t= 10 #min\n",
"dT= 2.73 #C\n",
"x= 0.1 #cal/deg\n",
"x1= 100 #cal/deg\n",
"x2= 5 #cal/deg\n",
"#CALCULATIONS\n",
"w= i**2*r*t*60\n",
"H= (x+x1+x2)*dT\n",
"E= w/H\n",
"#RESULTS\n",
"print '%s %.2f %s' % ('conversion factor =',E,'cal ')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"conversion factor = 4.18 cal \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6 - pg 137"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the heat at constant pressure and volume\n",
"#initialisation of variables\n",
"Cp= 6.0954 #cal /mol deg\n",
"Cp1= 3.2533*10**-3 #cal /mol deg\n",
"Cp2= 1.071*10**-6 #cal /mol deg\n",
"T= 100 #C\n",
"T1= 0 #C\n",
"R= 1.987 #atml/cal K\n",
"#CALULATIONS\n",
"H= Cp*(T-T1)+(Cp1/2)*((T+273.2)**2-(T1+273.2)**2)-(Cp2/3)*((T+273.2)**3-(T1+273.2)**3)\n",
"q= H-R*(T-T1)\n",
"#RESULTS\n",
"print '%s %.1f %s' % (' Heat at constant pressure=',H,'cal ')\n",
"print '%s %.1f %s' % (' \\n Heat at constant volume=',q,'cal ')\n",
"print 'The answer differs a bit from the textbook due to rounding off error'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Heat at constant pressure= 703.4 cal \n",
" \n",
" Heat at constant volume= 504.7 cal \n",
"The answer differs a bit from the textbook due to rounding off error\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7 - pg 140"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the work done in the process\n",
"#initialisation of variables\n",
"vl= 0.019 #l\n",
"vg= 16.07 #l\n",
"h= 1489. #mm of Hg\n",
"#CALCULATIONS\n",
"w= -(h/760)*(vl-vg)*(1.987/0.08206)\n",
"#RESULTS\n",
"print '%s %d %s' % (' Work done=',w,'cal ')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Work done= 761 cal \n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8 - pg 141"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the minimum work in both cases\n",
"#initialisation of variables\n",
"import math\n",
"n= 2 #moles\n",
"R= 0.08206 #J/mol K\n",
"T= 25 #C\n",
"b= 0.0428 #lmole^-1\n",
"a= 3.61 #atm l^2 mole^-1\n",
"V1= 20. #l\n",
"V2= 1. #l\n",
"#CALCULATIONS\n",
"w1= n*1.987*(273.2+T)*math.log10(V1/V2) *2.303\n",
"w= (n*R*(273.2+T)*2.303*math.log10((V1-n*b)/(V2-n*b))-a*n**2*((1/V2)-(1/V1)))*(1.987/0.08206)\n",
"#RESULTS\n",
"print '%s %.1f %s' % (' minimum work=',w1,'cal ')\n",
"print '%s %.1f %s' % (' \\n minimum work=',w,'cal ')\n",
"print 'The answer differs a bit from the textbook due to rounding off error'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" minimum work= 3550.7 cal \n",
" \n",
" minimum work= 3319.5 cal \n",
"The answer differs a bit from the textbook due to rounding off error\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9 - pg 144"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the final volume and temperature of the gas. Also calculate the work done in the process\n",
"#initialisation of variables\n",
"cv = 5.00 #cal mole^-1 deg^-1\n",
"R= 1.99 #cal mole^-1 deg^-1\n",
"p= 1 #atm\n",
"p1= 100. #atm\n",
"V= 75. #l\n",
"n= 3. #moles\n",
"R1= 0.08206 #cal/mol K\n",
"#CALCULATIONS\n",
"cp= cv+R\n",
"r= cp/cv\n",
"V1= V/(p1/p)**(1/r)\n",
"T2= p1*V1/(n*R1)\n",
"w= (p1*V1-p*V)*R/((r-1)*R1)\n",
"#RESULTS\n",
"print '%s %.2f %s' % (' final volume of gas =',V1,'l ')\n",
"print '%s %d %s' % (' \\n final temperature of gas =',T2,'K ')\n",
"print '%s %d %s' % (' \\n Work done =',w,'cal ')\n",
"print 'The answer differs a bit from the textbook due to rounding off error'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" final volume of gas = 2.78 l \n",
" \n",
" final temperature of gas = 1130 K \n",
" \n",
" Work done = 12384 cal \n",
"The answer differs a bit from the textbook due to rounding off error\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10 - pg 144"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the change in energy and enthalpy\n",
"#initialisation of variables\n",
"cv= 5 #cal mole^-1\n",
"P= 100 #atm\n",
"T= 1130 #K\n",
"T1= 812 #K\n",
"n= 3 #moles\n",
"R= 1.99 #cal/mole K\n",
"#CALCULTIONS\n",
"E= n*cv*(T1-T)\n",
"H= E+n*R*(T1-T)\n",
"#RESULTS\n",
"print '%s %.1f %s' % (' change in energy =',E,'cal ')\n",
"print '%s %.1f %s' % (' \\n change in enthalpy=',H,' cal ')\n",
"print 'The answer differs a bit from the textbook due to rounding off error'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" change in energy = -4770.0 cal \n",
" \n",
" change in enthalpy= -6668.5 cal \n",
"The answer differs a bit from the textbook due to rounding off error\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11 - pg 145"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the work done and final pressure\n",
"#initialisation of variables\n",
"import math\n",
"k= 1.435 \n",
"k1= 17.845*10**-3 #K**-1\n",
"k2= -4.165*10**-6 #K**-2\n",
"T= 200. #C\n",
"T1= 0. #C\n",
"P= 10. #atm\n",
"R= 1.987 #cal/mol K\n",
"k3= 3.422\n",
"#CALCULATIONS\n",
"W= k*(T-T1)+(k1/2)*((273+T)**2-(273+T1)**2)+(k2/3)*((273+T)**3-(273+T1)**3)\n",
"P2= (P/math.e**((k*math.log((273+T1)/(273+T))+k1*(T1-T)+(k2/2)*((273+T1)**2-(273+T)**2))/R))/100.\n",
"#RESULTS\n",
"print '%s %d %s' % (' work done by methane =',W,'cal ')\n",
"print '%s %.2f %s' % (' \\n final pressure=',P2,'atm ')"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" work done by methane = 1499 cal \n",
" \n",
" final pressure= 0.77 atm \n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 12 - pg 150"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the fraction of liquid\n",
"#initialisation of variables\n",
"P= 100 #atm\n",
"P1= 1 #atm\n",
"R= 1.99 #cal/mol**-1 K**-1\n",
"k= 0.3 #atm**-1\n",
"E= 1600 #cal\n",
"T= -183 #C\n",
"T1= 0 #C\n",
"#CALCULATIONS\n",
"X= (k*3.5*R*(P-P1))/(3.5*R*(T1-T)+E)\n",
"#RESULTS\n",
"print '%s %.3f' % (' fraction of liquid = ',X)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" fraction of liquid = 0.072\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 13 - pg 152"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the enthalpy change of the reaction\n",
"#initialisation of variables\n",
"H= -21.8 #kcal\n",
"H1= 3.3 #kcal\n",
"#CALCULATIONS\n",
"H2= H-H1\n",
"#RESULTS\n",
"print '%s %.1f %s' % (' Enthalpy =',H2,'kcal ')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Enthalpy = -25.1 kcal \n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14 - pg 153"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the heat of hydrogenation\n",
"#initialisation of variables\n",
"H= -68.317 #kcal\n",
"H1= -310.615 #kcal\n",
"H2= -337.234 #kcal\n",
"R= 1.987 #cal/mol^-1 K^-1\n",
"T= 298.2 #K\n",
"n= 1 #mole\n",
"n1= 1 #mole\n",
"n2= 1 #mole\n",
"#CALCULATIONS\n",
"E= H+H1-H2-(n-n1-n2)*R*T*10**-3\n",
"#RESULTS\n",
"print '%s %.3f %s' % (' Heat of hydrogenation =',E,'kcal ')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Heat of hydrogenation = -41.105 kcal \n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15 - pg 155"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the enthalpy of the process\n",
"#initialisation of variables\n",
"Hf= -196.5 #kcal\n",
"H= -399.14 #kcal\n",
"#CALCULATIONS\n",
"H1= (H-Hf)*1000\n",
"#RESULTS\n",
"print '%s %d %s' % (' Enthalpy =',H1,' kcal ')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Enthalpy = -202640 kcal \n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 16 - pg 157"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Enthalpy change\n",
"#initialisation of variables\n",
"H= -350.2 #kcal\n",
"H1= -128.67 #kcal\n",
"H2= -216.90 #kcal\n",
"#CALCULATIONS\n",
"H3= H-(H1+H2)\n",
"#RESULTS\n",
"print '%s %.1f %s' % (' Enthalpy =',H3,'kcal ')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Enthalpy = -4.6 kcal \n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17 - pg 158"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the enthalpy of the process\n",
"#initialisation of variables\n",
"H= -40.023 #kcal\n",
"H1= -22.063 #kcal\n",
"#CALCULATIONS\n",
"H2= H-H1\n",
"#RESULTS\n",
"print '%s %.3f %s' % (' Enthalpy =',H2,' kcal ')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Enthalpy = -17.960 kcal \n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 18 - pg 162"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the enthalpy change in the process\n",
"#initialisation of variables\n",
"H= -112.148 #k cal\n",
"H1= 101.99 #k cal\n",
"Hx=-112.148 #kcal\n",
"Hy=-111.015 #kcal\n",
"Hz=-.64\n",
"Hsol=-9.02\n",
"#CALCULATIONS\n",
"H2= H+H1\n",
"H3=2*Hx-2*Hy\n",
"H4=-10*Hz\n",
"H5=Hsol-5*Hz\n",
"#RESULTS\n",
"print '%s %.2f %s' % (' Enthalpy in case 1=',H2,'k cal ')\n",
"print '%s %.3f %s' % (' Enthalpy in case 2=',H3,'k cal ')\n",
"print '%s %.1f %s' % (' Enthalpy in case 3=',H4,'k cal ')\n",
"print '%s %.2f %s' % (' Enthalpy in case 4=',H5,'k cal ')"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Enthalpy in case 1= -10.16 k cal \n",
" Enthalpy in case 2= -2.266 k cal \n",
" Enthalpy in case 3= 6.4 k cal \n",
" Enthalpy in case 4= -5.82 k cal \n"
]
}
],
"prompt_number": 34
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19 - pg 167"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the dE and dH in the process\n",
"#initialisation of variables\n",
"cp=18.\n",
"T2=373 #K\n",
"T1=298 #K\n",
"T3=403.2 #K\n",
"hvap=9713 #cal\n",
"H4= 0 #cal\n",
"E4= 0 #cal\n",
"a=7.1873\n",
"b=2.3733e-3\n",
"c=.2084e-6\n",
"R=1.987\n",
"#RESULTS\n",
"H1=cp*(T2-T1)\n",
"H2=hvap\n",
"H3=a*(T3-T2) + b/2 *(T3**2-T2**2) + c/3 *(T3**3-T2**3)\n",
"E1=H1\n",
"E2=H2-R*T2\n",
"E3=H3-R*(T3-T2)\n",
"H= H1+H2+H3+H4\n",
"E= E1+E2+E3+E4\n",
"#RESULTS\n",
"print '%s %d %s' % (' Enthalpy=',H,'cal ')\n",
"print '%s %d %s' % (' \\n Energy=',E,' cal ')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Enthalpy= 11308 cal \n",
" \n",
" Energy= 10507 cal \n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 20 - pg 171"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the enthalpy change \n",
"#initialisation of variables\n",
"H= -114009.8 #cal\n",
"x= -5.6146 #K**-1\n",
"y= 0.9466*10**-3 #K**-2\n",
"z= 0.1578*10**-6 #K**-3\n",
"T= 1000\n",
"#CALCULATIONS\n",
"H1= H+x*T+y*T**2+z*T**3\n",
"#RESULTS\n",
"print '%s %.1f %s' %(' Enthalpy =',H1,'cal ')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Enthalpy = -118520.0 cal \n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 21 - pg 173"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#caculate the temperature achieved\n",
"#Initialization of variables\n",
"import numpy\n",
"a=72.3639\n",
"b=36.2399e-3\n",
"c=3.7621e-6\n",
"H=214920\n",
"#calculations\n",
"vec=([-c/3,b/2,a,-H])\n",
"vec2=numpy.roots(vec)\n",
"vec22=(vec2[2])\n",
"#results\n",
"print '%s %.1f %s' %('The required temperature observed is', vec22,'K')"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required temperature observed is 2059.4 K\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 22 - pg 175"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the change in enthalpy\n",
"#initialisation of variables\n",
"T= 298 #K\n",
"R= 1.987 #atmcc/mol K\n",
"x= 128.16\n",
"y= 0.9241\n",
"H= -8739 #cal\n",
"n1= 10 #mol\n",
"n2= 12 #mol\n",
"#CALCULATIONS\n",
"E= (x/y)*H\n",
"H= (E+R*T*(n1-n2))/1000\n",
"#RESULTS\n",
"print '%s %.1f %s' %(' Enthalpy =',H,'kcal mole^-1 ')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Enthalpy = -1213.2 kcal mole^-1 \n"
]
}
],
"prompt_number": 21
}
],
"metadata": {}
}
]
}