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 "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": {}
  }
 ]
}