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| author | hardythe1 | 2014-08-13 11:41:01 +0530 |
|---|---|---|
| committer | hardythe1 | 2014-08-13 11:41:01 +0530 |
| commit | 7e82f054d405211e1e8760524da8ad7c9fd75286 (patch) | |
| tree | 1790cf5a7460b48582da6c35417a85f3a1389a81 /Chemistry/Chapter_5.ipynb | |
| parent | 98bff1c301dd3b8b14983037a8a483e3eae1796d (diff) | |
| download | Python-Textbook-Companions-7e82f054d405211e1e8760524da8ad7c9fd75286.tar.gz Python-Textbook-Companions-7e82f054d405211e1e8760524da8ad7c9fd75286.tar.bz2 Python-Textbook-Companions-7e82f054d405211e1e8760524da8ad7c9fd75286.zip | |
adding book
Diffstat (limited to 'Chemistry/Chapter_5.ipynb')
| -rwxr-xr-x | Chemistry/Chapter_5.ipynb | 768 |
1 files changed, 768 insertions, 0 deletions
diff --git a/Chemistry/Chapter_5.ipynb b/Chemistry/Chapter_5.ipynb new file mode 100755 index 00000000..9649fe13 --- /dev/null +++ b/Chemistry/Chapter_5.ipynb @@ -0,0 +1,768 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5:Gases"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.1,Page no:177"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Pbaro=688 #pressure in mm Hg\n",
+ "\n",
+ "#Calculation\n",
+ "Patm=Pbaro/760.0 #pressure in atm\n",
+ "\n",
+ "#Result\n",
+ "print\"The presuure in atmospheres is :\",round(Patm,3),\"atm\\n\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The presuure in atmospheres is : 0.905 atm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.2,Page no:178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Pbaro=732 #pressure in mm Hg\n",
+ "\n",
+ "#Calculation\n",
+ "Patm=Pbaro/760.0 #pressure in atm\n",
+ "P=Patm*1.01325*10**2 #pressure in kilo Pascal\n",
+ "\n",
+ "#Result\n",
+ "print\"The presuure in kilo pascals is :\",round(P,1),\"kPa\\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The presuure in kilo pascals is : 97.6 kPa\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.3,Page no:186"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V=5.43 #volume, L\n",
+ "t=69.5 #temperature, C\n",
+ "\n",
+ "#Calculation\n",
+ "T=t+273 #temperature, K\n",
+ "n=1.82 #moles\n",
+ "R=0.0821 #universal gas constant, L.atm/(K.mol)\n",
+ "P=n*R*T/V #pressure, atm\n",
+ "\n",
+ "#Result\n",
+ "print\"The presuure in atmospheres is :\",round(P,2),\"atm\\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The presuure in atmospheres is : 9.42 atm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.4,Page no:187"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "m=7.4 #mass of NH3, g\n",
+ "#at STP for NH3 for 1mole of NH3\n",
+ "V1=22.41 # volume, L\n",
+ "NH3=17.03 #molar mass of NH3, g\n",
+ "\n",
+ "#Calculation\n",
+ "n=m/NH3 #moles of NH3\n",
+ "V=n*V1 #volume, L\n",
+ "\n",
+ "#Result\n",
+ "print\"The volume of NH3 under given conditions is :\",round(V,2),\"L\\n\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The volume of NH3 under given conditions is : 9.74 L\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.5,Page no:188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V1=0.55 #volume, L\n",
+ "P1=1 #pressure at sea level, atm\n",
+ "P2=0.4 #pressurea at 6.5km height, atm\n",
+ "\n",
+ "#n1=n2 and T1=T2 given hence P1V1=P2V2\n",
+ "#Calculation\n",
+ "V2=P1*V1/P2 \n",
+ "\n",
+ "#Result\n",
+ "print\"The volume of He balloon at height 6.5km above sea level is :\",round(V2,1),\"L\\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The volume of He balloon at height 6.5km above sea level is : 1.4 L\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.6,Page no:189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P1=1.2 # pressure initial, atm\n",
+ "T1=18+273 #temperature initial, K\n",
+ "T2=85+273 #temperature final, K\n",
+ "#volume is constant\n",
+ "\n",
+ "#Calculation\n",
+ "P2=P1*T2/T1 # pressure final,atm\n",
+ "\n",
+ "#Result\n",
+ "print\"The final pressure is :\",round(P2,2),\"atm\\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The final pressure is : 1.48 atm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.7,Page no:189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P1=6.4 # pressure initial, atm\n",
+ "P2=1.0 # pressure final, atm\n",
+ "T1=8+273 #temperature initial, K\n",
+ "T2=25+273 #temperature final, K\n",
+ "V1=2.1 #volume initial, mL\n",
+ "\n",
+ "#Calculation\n",
+ "V2=P1*V1*T2/(T1*P2) # volume final, mL\n",
+ "\n",
+ "#Result\n",
+ "print\"The final volume is :\",round(V2),\"mL\\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The final volume is : 14.0 mL\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.8,Page no:191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "\n",
+ "#taking 1 mole of CO2\n",
+ "n=1 \n",
+ "R=0.0821 #universal gas constant, L. atm/K.mol\n",
+ "t=55 #temperature, C\n",
+ "T=t+273 #temperature, K\n",
+ "P=0.99 #.pressure, atm\n",
+ "M=44.01 #molar mass of CO2, g\n",
+ "\n",
+ "#Calculation\n",
+ "d1=P*M/(R*T) #density of CO2, g/L\n",
+ "#altenate method\n",
+ "#taking 1 mole of CO2\n",
+ "mass=M #mass of CO2 in g =mol mass since we are considering 1 mole of CO2\n",
+ "V=n*R*T/P #volume, L\n",
+ "d2=mass/V #density=mass/volume, g/L\n",
+ "\n",
+ "#Result\n",
+ "print\"The density of CO2 is :\",round(d1,2),\"g/L\\n\"\n",
+ "print\"(Alternate Method)the density of CO2 is :\",round(d2,2),\"g/L\\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The density of CO2 is : 1.62 g/L\n",
+ "\n",
+ "(Alternate Method)the density of CO2 is : 1.62 g/L\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.9,Page no:192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "d=7.71 # density, g/mL(given)\n",
+ "R=0.0821 #universal gas constant, L. atm/K.mol\n",
+ "T=36+273 # temp, K\n",
+ "P=2.88 #pressure, atm\n",
+ "\n",
+ "#Calculation\n",
+ "M1=d*R*T/P # mol. mass, g/mol\n",
+ "#alternate method\n",
+ "#considering 1 L of compound\n",
+ "V=1 #volume, L\n",
+ "n=P*V/(R*T) #no of moles\n",
+ "m=7.71 #mass per 1 L, g\n",
+ "M2=m/n # mol. mass, g/mol\n",
+ "\n",
+ "#Result\n",
+ "print\"The molecular mass of given compound is :\",round(M1,1),\"g/mol\\n\"\n",
+ "print\"{alternate method} the molecular mass of given compound is :\",round(M2,1),\" g/mol\\n\" \n",
+ "print\"The molecular formula can be only found by trial and error method as given in the book \\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The molecular mass of given compound is : 67.9 g/mol\n",
+ "\n",
+ "{alternate method} the molecular mass of given compound is : 67.9 g/mol\n",
+ "\n",
+ "The molecular formula can be only found by trial and error method as given in the book \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.10,Page no:193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "percentSi=33 #percent of Si in compound\n",
+ "percentF=67 #percent of F in compound\n",
+ "P=1.7 #pressure, atm\n",
+ "T=35+273 #temp. in K\n",
+ "m=2.38 #mass, g\n",
+ "V=0.21 #volume, L\n",
+ "R=0.0821 #universal Gas constant, L.atm/K.mol\n",
+ "m_sif3=85.09\n",
+ "\n",
+ "#Calculation\n",
+ "nSi=percentSi/28.01 #moles of Si in 100g compound\n",
+ "nF=percentF/19 #moles of F in 100g compound\n",
+ "n=P*V/(R*T) #moles\n",
+ "M=m/n #mol. mass=mass/moles, g/mol\n",
+ "e=M/m_sif3\n",
+ "\n",
+ "#Result\n",
+ "print\"The molecular mass of given compound is :\",round(M),\"g/mol\\n\"\n",
+ "print\"Therefore,molecular formula is (SiF3)%.f\"%round(e),\"OR Si2F6\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The molecular mass of given compound is : 169.0 g/mol\n",
+ "\n",
+ "Therefore,molecular formula is (SiF3)2 OR Si2F6\n"
+ ]
+ }
+ ],
+ "prompt_number": 57
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.11,Page no:194"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "VC2H2=7.64 #volume of acetylene, L\n",
+ "\n",
+ "#Calculation\n",
+ "VO2=VC2H2*5/2.0 #volume of O2 required for complete combustion as 5mol O2 react with 2mol acetylene for complete combustion\n",
+ "\n",
+ "#Result\n",
+ "print\"The volume of O2 required for complete combustion of acetylene is :\",VO2,\"L\\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The volume of O2 required for complete combustion of acetylene is : 19.1 L\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.12,Page no:195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "R=0.0821 #universal Gas constant, L.atm/K.mol\n",
+ "T=80+273 #temp in K\n",
+ "P=823/760.0 #pressure in atm\n",
+ "m=60.0 #mass of NaN3, g\n",
+ "NaN3=65.02 #mol. mass of NaN3, g\n",
+ "\n",
+ "#Calculation\n",
+ "nN2=m*3.0/(2.0*NaN3) #moles of N2\n",
+ "nN2=round(nN2,2)\n",
+ "VN2=nN2*R*T/P #from ideal gas law\n",
+ "#Result\n",
+ "print\"\\t the volume of N2 generated is :\",round(VN2,1),\"L\\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\t the volume of N2 generated is : 36.9 L\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.13,Page no:196"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "R=0.0821 #universal Gas constant, L.atm/K.mol\n",
+ "T=312 #temp in K\n",
+ "V=2.4*10**5 #volume, L\n",
+ "P1=7.9*10**-3 #pressure initial in atm\n",
+ "P2=1.2*10**-4 #pressure final in atm\n",
+ "\n",
+ "#Calculation\n",
+ "Pdrop=P1-P2 #pressure drop, atm\n",
+ "n=Pdrop*V/(R*T) #moles of Co2 reacted\n",
+ "Li2CO3=73.89 #mol. mass of Li2CO3, g\n",
+ "mLi2CO3=n*Li2CO3 #mass of Li2CO3, g\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"The mass of Li2CO3 formed is :%.1e\"%mLi2CO3,\"g Li2CO3\\n\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mass of Li2CO3 formed is :5.4e+03 g Li2CO3\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.14,Page no:199"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "nNe=4.46 #moles of Ne\n",
+ "nXe=2.15 #moles of Xe\n",
+ "nAr=0.74 #moles of Ar\n",
+ "PT=2 #total pressure in atm\n",
+ "\n",
+ "#Calculation\n",
+ "XNe=nNe/(nNe+nAr+nXe) #mole fraction of Ne\n",
+ "XAr=nAr/(nNe+nAr+nXe) #mole fraction of Ar\n",
+ "XXe=nXe/(nNe+nAr+nXe) #mole fraction of Xe\n",
+ "PNe=XNe*PT #partial pressure of Ne\n",
+ "PAr=XAr*PT #partial pressure of Ar\n",
+ "PXe=XXe*PT #partial pressure of Xe\n",
+ "\n",
+ "#Result\n",
+ "print\"The partial pressures of Ne, Ar and Xe are:\"\n",
+ "print\"PNe=\",round(PNe,2),\"atm\\n\",\"PAr=\",round(PAr,2),\"atm\\nPXe=\",round(PXe,3),\"atm \"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The partial pressures of Ne, Ar and Xe are:\n",
+ "PNe= 1.21 atm\n",
+ "PAr= 0.2 atm\n",
+ "PXe= 0.585 atm \n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.15,Page no:200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "PT=762 #pressure total, mmHg\n",
+ "PH2O=22.4 #pressure of water vapor, mmHg\n",
+ "PO2=PT-PH2O #pressure of O2, frm Dalton's law, mmHg\n",
+ "M=32 #mol mass of O2, g\n",
+ "R=0.0821 #universal Gas constant, L.atm/K.mol\n",
+ "T=24+273 #temp in K\n",
+ "V=0.128 #volume in L\n",
+ "\n",
+ "#Calculation\n",
+ "m=(PO2/760)*V*M/(R*T) #mass of mass of O2 collected, g\n",
+ "\n",
+ "#Result\n",
+ "print\"The mass of O2 collected is :\",round(m,3),\"g\\n\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mass of O2 collected is : 0.163 g\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.16,Page no:207"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "R=8.314 #universal Gas constant, J/K mol\n",
+ "T=25+273 #temp in K\n",
+ "import math\n",
+ "#Calculation\n",
+ "#for O2\n",
+ "M=4.003*10**-3 #mol mass in kg\n",
+ "Urms=math.sqrt(3*R*T/M) #rms velocity, m/s\n",
+ "print\"\\t the rms velocity of O2 collected is :%.2e\"%Urms,\"m/s\\n\"\n",
+ "#for N2\n",
+ "M=28.02*10**-3 #mol mass in kg\n",
+ "Urms=math.sqrt(3*R*T/M) #rms velocity, m/s\n",
+ "\n",
+ "#Result\n",
+ "print\"\\t the rms velocity of N2 collected is :\",round(Urms),\" m/s\\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\t the rms velocity of O2 collected is :1.36e+03 m/s\n",
+ "\n",
+ "\t the rms velocity of N2 collected is : 515.0 m/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.17,Page no:209"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t2=1.5 #diffusion time of compound, min\n",
+ "t1=4.73 #diffusion time of Br, min\n",
+ "M2=159.8 #mol mass of Br gas, g\n",
+ "\n",
+ "#Calculation\n",
+ "M=(t2/t1)**2*M2 #molar gas of unknown gas, g(from Graham's Law of Diffusion)\n",
+ "\n",
+ "#Result\n",
+ "print\"\\t the molar mass of unknown gas is :\",round(M,1),\"g/mol\\n\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\t the molar mass of unknown gas is : 16.1 g/mol\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 76
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5.18,Page no:213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "#(a)\n",
+ "V=5.2 #volume, L\n",
+ "T=47+273 \n",
+ "n=3.5 \n",
+ "R=0.0821 #universal Gas constant, L.atm/K.mol\n",
+ "\n",
+ "#Calculation\n",
+ "P=n*R*T/V \n",
+ "print\"The pressure of NH3 gas from ideal gas equation is :\",round(P,1),\"atm\\n\"\n",
+ "#(b)\n",
+ "a=4.17 #constant, atm.L2/mol2\n",
+ "b=0.0371 #constant, L/mol\n",
+ "Pc=a*n**2/V**2 #pressure correction term, atm\n",
+ "Vc=n*b #volume correction term, L\n",
+ "P=n*R*T/(V-Vc)-Pc #from van der waals equation, pressure, atm\n",
+ "\n",
+ "#Result\n",
+ "print\"The pressure of NH3 gas from van der waals equation is :\",round(P,1),\"atm\\n\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The pressure of NH3 gas from ideal gas equation is : 17.7 atm\n",
+ "\n",
+ "The pressure of NH3 gas from van der waals equation is : 16.2 atm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |