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Diffstat (limited to 'Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch1.ipynb')
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diff --git a/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch1.ipynb b/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch1.ipynb index 48ad85eb..23e275b1 100644 --- a/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch1.ipynb +++ b/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch1.ipynb @@ -27,38 +27,26 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Calculate Mass Fraction, Mole fraction, Molality and PPM.\n", "\n", "import math \n", "\n", - "# Variables\n", "m_i = 10. #[g]\n", "m_w = 990. #[g]\n", "M_i = 342.3 #[g]\n", "M_w = 18. #[g]\n", "\n", - "# Calculations\n", - "# The mass fraction is \n", - "# ( mass fraction of sucrose ) = x_i (by mass) = m_i/(sum of all subsmath.tances)\n", "\n", "x_i = m_i/(m_i+m_w)\n", "x_i = x_i*100. # [in percentage]\n", "\n", - "# This is also the weight fraction.\n", - "# The mole fraction is\n", - "# ( mole fraction of sucrose ) = x_j (by mole) = n_i/(sum of number moles of all the subsmath.tances)\n", "n_i = m_i/M_i # number of moles of sucrose\n", "n_w = m_w/M_w # number of moles of water\n", "x_j = n_i/(n_i+n_w)\n", "x_j = x_j*100 # [in percentage]\n", "\n", - "# The molality, a concentration unit is widely used in equilibrium calculations, is defined as \n", - "# m (molality) = (moles of solute)/(kg of solvent)\n", "m = n_i/m_w*1000 #[molal]\n", - "# For solutions of solids and liquids (but not gases) ppm almost always means ppm by mass, so \n", "x_ppm = x_i*10**(6)/100. #[ppm]\n", "\n", - "# Results\n", "print \" sucrose concentration in terms of the mass fraction is %f%%\"%(x_i)\n", "print \" sucrose concentration in terms of the mole fraction is %f%%\"%(x_j)\n", "print \" sucrose concentration in terms of the molality is %f molal\"%(m)\n", @@ -92,36 +80,25 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Calculate Mass concentration , Mole concentration and Molarity\n", "\n", "import math \n", "\n", - "# Variables\n", "T = 20. #[C]\n", "d = 1.038143/1000*10.**(6) #[kg/m**(3)]\n", "m_i = 10. #[g] mass of sucrose\n", "M_i = 342.3 #[g/mol] molecular weight of sucrose\n", "\n", - "# Calculations\n", - "# In the previous example i.e. example 1.1 the mass was chosen to be 1.00 kg, so that\n", "m = 1.00 #[kg]\n", "V = m/d*1000 #[L]\n", "\n", - "# The mass concentration is\n", - "# m_1 ( mass concentration of sucrose ) = (mass of sucrose)/(volume of solution)\n", "m_1 = m_i/V #[g/L]\n", "\n", - "# The mole concentration is \n", - "# m_2 ( mole concentration of sucrose ) = (moles of sucrose)/(volume of solution)\n", "\n", "m_2 = (m_i/M_i)/V #[mol/L]\n", "\n", - "# Results\n", "print \" Mass concentration of the solution is %f g/L\"%(m_1)\n", "print \" Mole concentration of the solution is %f mol/L\"%(m_2)\n", "\n", - "# By the definition of the molarity, molarity is mole concentration of the solute\n", - "# so molarity \n", "print \" Molarity of the solution is %f mol/L\"%(m_2)\n" ], "language": "python", |