removing problem statements

1 files changed, 0 insertions, 132 deletions

diff --git a/Introduction_To_Chemical_Engineering/ch2.ipynb b/Introduction_To_Chemical_Engineering/ch2.ipynb
index 7f49d9ae..845b485b 100644
--- a/Introduction_To_Chemical_Engineering/ch2.ipynb
+++ b/Introduction_To_Chemical_Engineering/ch2.ipynb
@@ -27,21 +27,17 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the volume of oxygen that can be obtained\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "p1=15.     #in bar\n",
       "p2=1.013  #in bar\n",
       "t1=283.    #in K\n",
       "t2=273.    #in K\n",
       "v1=10.     #in l\n",
       "\n",
-      "# Calculations\n",
       "v2=p1*v1*t2/(t1*p2);\n",
       "\n",
-      "# Results\n",
       "print \"volume of oxygen = %f liters\"%(v2)\n"
      ],
      "language": "python",
@@ -69,16 +65,13 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find volumetric composition,partial pressue of each gas and total pressure of mixture\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "nCO2 = 2./44;     #moles of CO2\n",
       "nO2 = 4./32;     #moles of O2\n",
       "nCH4 = 1.5/16;   #moles of CH4\n",
       "\n",
-      "# Calculations and Results\n",
       "total_moles = nCO2+nO2+nCH4;\n",
       "yCO2 = nCO2/total_moles;\n",
       "yO2 = nO2/total_moles;\n",
@@ -122,21 +115,17 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find equivalent mass of metal\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "P=104.3     #total pressure in KPa\n",
       "pH2O=2.3    #in KPa\n",
       "pH2=P-pH2O; #in KPa\n",
       "\n",
-      "# Calculations and Results\n",
       "VH2=209*pH2*273/(293*101.3)\n",
       "\n",
       "print \"volume of hydrogen obtained = %f ml\"%(VH2)\n",
       "\n",
-      "#calculating amount of metal having 11.2l of hydrogen\n",
       "\n",
       "m=350/196.08*11.2   #mass of metal in grams\n",
       "print \"mass of metal equivalent to 11.2 litre/mol of hydrogen = %f gm\"%(m)\n"
@@ -167,15 +156,11 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find NaCl content in NaOH solution\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "w=2   #in gm\n",
       "m=0.287  #in gm\n",
       "\n",
-      "# Calculations and Results\n",
-      "#precipitate from 58.5gm of NaCl=143.4gm\n",
       "mNaCl=58.5/143.4*m;\n",
       "\n",
       "print \"mass of NaCl = %f gm\"%(mNaCl )\n",
@@ -209,20 +194,16 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the carbon content in sample\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "w=4.73    #in gm5\n",
       "VCO2=5.30    #in liters\n",
       "\n",
-      "# Calculations\n",
       "weight_CO2=44/22.4*VCO2;\n",
       "carbon_content=12./44*weight_CO2;\n",
       "percentage_content=(carbon_content/w)*100;\n",
       "\n",
-      "# Results\n",
       "print \"percentage amount of carbon in sample = %f\"%(percentage_content)\n"
      ],
      "language": "python",
@@ -250,19 +231,15 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the volume of air\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "volume_H2=0.5    #in m3\n",
       "volume_CH4=0.35  #in m3\n",
       "volume_CO=0.08   #in m3\n",
       "volume_C2H4=0.02 #in m3\n",
       "volume_oxygen=0.21  #in m3 in air\n",
       "\n",
-      "# Calculations\n",
-      "#required oxygen for various gases\n",
       "H2=0.5*volume_H2;\n",
       "CH4=2*volume_CH4;\n",
       "CO=0.5*volume_CO;\n",
@@ -271,7 +248,6 @@
       "total_O2=H2+CH4+CO+C2H4;\n",
       "oxygen_required=total_O2/volume_oxygen;\n",
       "\n",
-      "# Results\n",
       "print \"amount of oxygen required = %f cubic meter\"%(oxygen_required)\n"
      ],
      "language": "python",
@@ -299,23 +275,19 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the volume of sulphuric acid and mass of water consumed\n",
       "\n",
       "import math \n",
       "\n",
       "\n",
-      "# Variables\n",
       "density_H2SO4 = 1.10   #in g/ml\n",
       "mass_1 = 100*density_H2SO4;   #mass of 100ml of 15% solution\n",
       "mass_H2SO4 = 0.15*mass_1;\n",
       "density_std = 1.84   #density of 96% sulphuric acid\n",
       "mass_std = 0.96*density_std;   #mass of H2SO4 in 1ml 96% H2SO4\n",
       "\n",
-      "# Calculations\n",
       "volume_std = mass_H2SO4/mass_std;    #volume of 96%H2SO4\n",
       "mass_water = mass_1 - mass_H2SO4;\n",
       "\n",
-      "# Results\n",
       "print \"volume of 0.96 H2SO4 required = %f ml\"%(volume_std)\n",
       "print \"mass of water required = %f g\"%(mass_water)\n"
      ],
@@ -345,18 +317,15 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find molarity,molality and normality\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "w_H2SO4=0.15    #in gm/1gm solution\n",
       "density=1.10   #in gm/ml\n",
       "m=density*1000;   #mass per liter\n",
       "weight=m*w_H2SO4;   #H2SO4 per liter solution\n",
       "molar_mass=98;\n",
       "\n",
-      "# Calculations and Results\n",
       "Molarity=weight/molar_mass;\n",
       "print \"Molarity = %f mol/l\"%(Molarity)\n",
       "\n",
@@ -394,18 +363,14 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find normality\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "molar_mass_BaCl2=208.3;      #in gm\n",
       "equivalent_H2SO4=0.144;\n",
       "\n",
-      "# Calculations\n",
       "normality=equivalent_H2SO4*1000/28.8;\n",
       "\n",
-      "# Results\n",
       "print \"Normality = %f N\"%(normality)\n"
      ],
      "language": "python",
@@ -433,20 +398,16 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find amount of KClO3 precipitated\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "solubility_70=30.2     #in gm/100gm\n",
       "w_solute=solubility_70*350/130.2;     #in gm\n",
       "\n",
-      "# Calculations\n",
       "w_water=350-w_solute;\n",
       "solubility_30=10.1    #in gm/100gm\n",
       "precipitate=(solubility_70-solubility_30)*w_water/100\n",
       "\n",
-      "# Results\n",
       "print \"amount precipitated = %f gm\"%(precipitate)\n"
      ],
      "language": "python",
@@ -474,19 +435,15 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the pressure for solubility of CO2\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "absorbtion_coefficient=1.71    #in liters\n",
       "molar_mass=44;\n",
       "\n",
-      "# Calculations\n",
       "solubility=absorbtion_coefficient*molar_mass/22.4;   #in gm\n",
       "pressure=8/solubility*101.3;\n",
       "\n",
-      "# Results\n",
       "print \"pressure required = %f kPa\"%(pressure)\n"
      ],
      "language": "python",
@@ -514,23 +471,19 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the vapor pressure of water\n",
       "\n",
       "import math \n",
       "\n",
       "\n",
-      "# Variables\n",
       "w_water=540.   #in gm\n",
       "w_glucose=36.  #in gm\n",
       "m_water=18.;      #molar mass of water\n",
       "m_glucose=180.;   #molar mass of glucose\n",
       "\n",
-      "# Calculations\n",
       "x=(w_water/m_water)/(w_water/m_water+w_glucose/m_glucose);\n",
       "p=8.2*x;\n",
       "depression=8.2-p;\n",
       "\n",
-      "# Results\n",
       "print \"depression in vapor pressure = %f Pa\"%(depression*1000)\n"
      ],
      "language": "python",
@@ -558,21 +511,17 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the boiling point of solution\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "w_glucose=9.    #in gm\n",
       "w_water=100.    #in gm\n",
       "E=0.52;\n",
       "m=90/180.;    #moles/1000gm water\n",
       "\n",
-      "# Calculations\n",
       "delta_t=E*m;\n",
       "boiling_point=100+delta_t;\n",
       "\n",
-      "# Results\n",
       "print \"boiling_point of water = %f degreeC\"%(boiling_point)\n"
      ],
      "language": "python",
@@ -600,16 +549,13 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the molar mass and osmotic pressure\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "K=1.86;\n",
       "c=15   #concentration of alcohol\n",
       "delta_t=10.26;\n",
       "\n",
-      "# Calculations and Results\n",
       "m=delta_t/K;   #molality\n",
       "M=c/(m*85);    #molar mass\n",
       "print \"molar mass = %f gm\"%(M*1000)\n",
@@ -648,22 +594,18 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find u_in, M_v, k'\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "u_in = 0.575   #from the graph\n",
       "u_s = 0.295   #in mPa-s\n",
       "\n",
-      "# Calculations\n",
       "M_v = (u_in/(5.80*10**-5))**(1/0.72);\n",
       "u_red = 0.628;   #in dl/g\n",
       "\n",
       "c = 0.40   #in g/dl\n",
       "k = (u_red-u_in)/((u_in**2)*c);\n",
       "\n",
-      "# Results\n",
       "print \"k = %f Mv = %fu_in = %f dl/gm\"%(k,M_v,u_in)\n"
      ],
      "language": "python",
@@ -691,11 +633,9 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the molecular formula\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "C=54.5     #% of carbon\n",
       "H2=9.1     #% of hydrogen\n",
       "O2=36.4    #% of oxygen\n",
@@ -705,13 +645,11 @@
       "molar_mass=88.;\n",
       "density=44.;\n",
       "\n",
-      "# Calculations\n",
       "ratio=molar_mass/density;\n",
       "x=ratio*2;\n",
       "y=ratio*1;\n",
       "z=ratio*4;\n",
       "\n",
-      "# Results\n",
       "print \"x = %f y = %f z = %f\"%(x,y,z)\n",
       "print \"formula of butyric acid is = C4H8O2\"\n"
      ],
@@ -741,11 +679,9 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find molecular foemula \n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "C=93.75    #% of carbon\n",
       "H2=6.25    #% of hydrogen\n",
       "x=C/12     #number of carbon atoms\n",
@@ -753,12 +689,10 @@
       "molar_mass=64\n",
       "density=4.41*29;\n",
       "\n",
-      "# Calculations\n",
       "ratio=density/molar_mass;\n",
       "x=round(ratio*5);\n",
       "y=round(ratio*4);\n",
       "\n",
-      "# Results\n",
       "print \"x = %f y = %f\"%(x,y)\n",
       "print \"formula of butyric acid is = C10H8\"\n",
       "\n"
@@ -789,11 +723,9 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find molecular formula\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "C=50.69     #% of carbon\n",
       "H2=4.23     #% of hydrogen\n",
       "O2=45.08    #% of oxygen\n",
@@ -802,7 +734,6 @@
       "b=H2/2;    #number of hydrogen molecules\n",
       "molar_mass=71;\n",
       "\n",
-      "# Calculations and Results\n",
       "def f(m):\n",
       "    return (2.09*1000)/(60*m);\n",
       "\n",
@@ -849,18 +780,15 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the molecular formula\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "C=64.6     #% of carbon\n",
       "H2=5.2     #% of hydrogen\n",
       "O2=12.6    #% of oxygen\n",
       "N2=8.8     #% of nitrogen\n",
       "Fe=8.8     #% of iron\n",
       "\n",
-      "# Calculations\n",
       "a=C/12;    #number of carbon molecules\n",
       "c=8.8/14;   #number of nitrogen molecules\n",
       "b=H2;    #number of hydrogen molecules\n",
@@ -871,7 +799,6 @@
       "\n",
       "molar_mass=63.3/cm;\n",
       " \n",
-      "# Results \n",
       "print \"a = %d, b = %d, c = %d, d = %d, e = %d\"%(a*6.5,b*6.5,c*6.5,d*6.5,e*6.5)\n",
       "print \"formula of butyric acid is = C34H33N4O5Fe\"\n",
       "\n"
@@ -902,20 +829,16 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find sequence of deposition\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "E1=-0.25;\n",
       "E2=0.80;\n",
       "E3=0.34;\n",
       "\n",
-      "# Calculations\n",
       "a=[E1,E2,E3];\n",
       "sorted(a)\n",
       "\n",
-      "# Results\n",
       "print \"sorted potential in volts =\"\n",
       "print  (a)\n",
       "print  (\"E2>E3>E1\")\n",
@@ -951,22 +874,18 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the emf of cell\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "E0_Zn=-0.76;\n",
       "E0_Pb=-0.13;\n",
       "c_Zn=0.1;\n",
       "c_Pb=0.02;\n",
       "\n",
-      "# Calculations\n",
       "E_Zn=E0_Zn+(0.059/2)*math.log10(c_Zn);\n",
       "E_Pb=E0_Pb+(0.059/2)*math.log10(c_Pb);\n",
       "E=E_Pb-E_Zn;\n",
       "\n",
-      "# Results\n",
       "print \"emf of cell = %f V\"%(E)\n",
       "print \"Since potential of lead is greater than that of zinc thus reduction will occur at\\\n",
       " lead electrode and oxidation will occur at zinc electrode\"\n"
@@ -997,22 +916,18 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the emf of cell\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "E0_Ag=0.80;\n",
       "E0_AgNO3=0.80;\n",
       "c_Ag=0.001;\n",
       "c_AgNO3=0.1;\n",
       "\n",
-      "# Calculations\n",
       "E_Ag=E0_Ag+(0.059)*math.log10(c_Ag);\n",
       "E_AgNO3=E0_AgNO3+(0.059)*math.log10(c_AgNO3);\n",
       "E=E_AgNO3-E_Ag;\n",
       "\n",
-      "# Results\n",
       "print \"emf of cell = %f V\" %(E)\n",
       "print \"since E is positive, the left hand electrode will be anode and\\\n",
       " the electron will travel in the external circuit from the left hand to the right hand electrode\"\n"
@@ -1043,18 +958,14 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find emf of cell\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "pH=12;    #pH of solution\n",
       "E_H2=0;\n",
       "\n",
-      "# Calculations\n",
       "E2=-0.059*pH;\n",
       "E=E_H2-E2;\n",
       "\n",
-      "# Results\n",
       "print \"EMF of cell = %f V\"%(E)\n"
      ],
      "language": "python",
@@ -1082,20 +993,16 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find amount of silver deposited\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "I=3   #in Ampere\n",
       "t=900   #in s\n",
       "m_eq=107.9    #in gm/mol\n",
       "F=96500;\n",
       "\n",
-      "# Calculations\n",
       "m=(I*t*m_eq)/F;\n",
       "\n",
-      "# Results\n",
       "print \"mass = %f gm\"%(m)\n"
      ],
      "language": "python",
@@ -1123,24 +1030,20 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the time for electroplating\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "volume=10*10*0.005;    #in cm3\n",
       "mass=volume*8.9;\n",
       "F=96500;\n",
       "atomic_mass=58.7    #in amu\n",
       "current=2.5     #in Ampere\n",
       "\n",
-      "# Calculations\n",
       "charge=(8.9*F*2)/atomic_mass;\n",
       "yield_=0.95;\n",
       "actual_charge=charge/(yield_*3600);\n",
       "t=actual_charge/current;\n",
       "\n",
-      "# Results\n",
       "print \"time required = %f hours\"%(t)\n"
      ],
      "language": "python",
@@ -1168,17 +1071,13 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find hardness of water\n",
       "\n",
-      "# Variables\n",
       "m_MgSO4=90.   #in ppm\n",
       "MgSO4_parts=120.;\n",
       "CaCO3_parts=100.;\n",
       "\n",
-      "# Calculations\n",
       "hardness=(CaCO3_parts/MgSO4_parts)*m_MgSO4;\n",
       "\n",
-      "# Results\n",
       "print \"hardness of water = %f mg/l\"%(hardness)\n"
      ],
      "language": "python",
@@ -1214,7 +1113,6 @@
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "m1 = 162.   #mass of calcium bi carbonate in mg\n",
       "m2 = 73.   #mass of magnesium bi carbonate in mg\n",
       "m3 = 136.  # mass of calsium sulfate in mg\n",
@@ -1222,19 +1120,16 @@
       "m5 = 500.  #mass of sodium cloride in mg\n",
       "m6 = 50.   # mass of potassium cloride in mg\n",
       "\n",
-      "# Calculations and Results\n",
       "content_1 = m1*100/m1;    #content of calcium bi carbonate in mg\n",
       "content_2 = m2*100/(2*m2);   #content of magnesium bi carbonate in mg\n",
       "content_3 = m3*100/m3;  # content of calsium sufate in mg\n",
       "content_4 = m4*100/m4;   # content of magnesium cloride\n",
       "\n",
-      "#part_1\n",
       "\n",
       "temp_hardness = content_1 + content_2;   #depends on bicarbonate only\n",
       "total_hardness = content_1+content_2+content_3+content_4;\n",
       "print \"total hardness = %.0f mg/l temporary hardness = %.0f mg/l\"%(temp_hardness,total_hardness)\n",
       "\n",
-      "#part 2\n",
       "wt_lime = (74./100)*(content_1+2*content_2+content_4);\n",
       "actual_lime = wt_lime/0.85;\n",
       "print \"amount of lime required = %.1f mg/l\"%(actual_lime)\n",
@@ -1270,19 +1165,15 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find hardness of water\n",
       "\n",
-      "# Variables\n",
       "volume_NaCl=50.    #in l\n",
       "c_NaCl=5000.       #in mg/l\n",
       "\n",
-      "# Calculations\n",
       "m=volume_NaCl*c_NaCl;\n",
       "equivalent_NaCl=50/58.5;\n",
       "\n",
       "hardness=equivalent_NaCl*m;\n",
       "\n",
-      "# Results\n",
       "print \"hardness of water = %f mg/l\"%(hardness/1000.)\n"
      ],
      "language": "python",
@@ -1310,16 +1201,13 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the total vapor pressure and molar compositions\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "m_benzene = 55.   #in kg\n",
       "m_toluene = 28.   #in kg\n",
       "m_xylene = 17.    # in kg\n",
       "\n",
-      "# Calculations and Results\n",
       "mole_benzene = m_benzene/78.;\n",
       "mole_toluene = m_toluene/92.;\n",
       "mole_xylene = m_xylene/106.;\n",
@@ -1364,22 +1252,17 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the mixture composition\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "vapor_pressure=8.    #in kPa\n",
       "pressure=100.       #in kPa\n",
       "\n",
-      "# Calculations and Results\n",
-      "#part 1\n",
       "volume=1     #in m3\n",
       "volume_ethanol=volume*(vapor_pressure/pressure);\n",
       "volume_air=1-volume_ethanol;\n",
       "print \"volumetric composition:- air composition = %f ethanol compostion = %f\"%(volume_air*100,volume_ethanol*100)\n",
       "\n",
-      "#part 2\n",
       "molar_mass_ethanol=46;\n",
       "molar_mass_air=28.9;\n",
       "mass_ethanol=0.08*molar_mass_ethanol;    #in kg\n",
@@ -1388,16 +1271,13 @@
       "fraction_air=(mass_air*100)/(mass_air+mass_ethanol);\n",
       "print \"composition by weight:-Air = %f Ethanol vapor = %f\"%(fraction_air,fraction_ethanol)\n",
       "\n",
-      "#part 3\n",
       "mixture_volume=22.3*(101.3/100)*(299./273);   #in m3\n",
       "weight_ethanol=mass_ethanol/mixture_volume;\n",
       "print \"weight of ethanol/cubic meter = %f Kg\"%(weight_ethanol)\n",
       "\n",
-      "#part 4\n",
       "w_ethanol=mass_ethanol/mass_air;\n",
       "print \"weight of ethanol/kg vapor free air = %f Kg\"%(w_ethanol)\n",
       "\n",
-      "#part 5\n",
       "moles_ethanol=0.08/0.92;\n",
       "print \"kmol of ethanol per kmol of vapor free air = %f\"%(moles_ethanol)\n"
      ],
@@ -1430,23 +1310,18 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find relative saturation and dew point\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "vapor_pressure=8.   #in kPa\n",
       "volume_ethanol=0.05;\n",
       "\n",
       "\n",
-      "# Calculations and Results\n",
-      "#basis 1kmol of mixture\n",
       "partial_pressure=volume_ethanol*100;\n",
       "relative_saturation=partial_pressure/vapor_pressure;\n",
       "mole_ratio=volume_ethanol/(1-volume_ethanol);\n",
       "print \"mole ratio = %f \\nrelative saturation = %f %%\"%(mole_ratio,relative_saturation*100)\n",
       "\n",
-      "#basis 1kmol saturated gas mixture at 100kPa\n",
       "volume_vapor=(8./100)*100;\n",
       "ethanol_vapor=volume_vapor/100.;\n",
       "air_vapor=1-ethanol_vapor;\n",
@@ -1455,7 +1330,6 @@
       "\n",
       "print \"percentage saturation = %f %%\"%(percentage_saturation*100)\n",
       "\n",
-      "#dew point\n",
       "print \"corresponding to partial pressure of 5kPa we get a dew point of 17.3 degree celcius\"\n"
      ],
      "language": "python",
@@ -1486,26 +1360,21 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the properties of humid air\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "p = 4.24   #in kPa\n",
       "H_rel = 0.8;\n",
       "\n",
-      "# Calculations and Results\n",
       "p_partial = p*H_rel;\n",
       "molal_H = p_partial/(100-p_partial);\n",
       "print \"initial molal humidity = %.3f\"%(molal_H)\n",
       "\n",
-      "#part 2\n",
       "P = 200.   #in kPa\n",
       "p_partial = 1.70   #in kPa\n",
       "final_H = p_partial/(P-p_partial);\n",
       "print \"final molal humidity = %.4f\"%(final_H)\n",
       "\n",
-      "#part 3\n",
       "p_dryair = 100 - 3.39;\n",
       "v = 100*(p_dryair/101.3)*(273./303);\n",
       "moles_dryair = v/22.4;\n",
@@ -1514,7 +1383,6 @@
       "water_condensed = (vapor_initial-vapor_final)*18;\n",
       "print \"amount of water condensed = %f kg\"%(water_condensed)\n",
       "\n",
-      "#part 4\n",
       "total_air = moles_dryair+vapor_final;\n",
       "final_v = 22.4*(101.3/200)*(288./273)*total_air;\n",
       "print \"final volume of wety air = %f m**3\"%(final_v)\n"