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| author | Trupti Kini | 2015-12-26 23:30:09 +0600 |
|---|---|---|
| committer | Trupti Kini | 2015-12-26 23:30:09 +0600 |
| commit | 5027effcc3d91f59da0b46e0c6fa00aec8d36489 (patch) | |
| tree | 51e765e99ce27df306bc674af10224cb27018e61 /Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati | |
| parent | ca5511b67f7a222aae33187333becacdd8d3ea2a (diff) | |
| download | Python-Textbook-Companions-5027effcc3d91f59da0b46e0c6fa00aec8d36489.tar.gz Python-Textbook-Companions-5027effcc3d91f59da0b46e0c6fa00aec8d36489.tar.bz2 Python-Textbook-Companions-5027effcc3d91f59da0b46e0c6fa00aec8d36489.zip | |
Added(A)/Deleted(D) following books
A Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_1.ipynb
A Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_12.ipynb
A Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_13.ipynb
A Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_14.ipynb
A Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_15.ipynb
A Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_2.ipynb
A Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_3.ipynb
A Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_5.ipynb
A Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_6.ipynb
A Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_7.ipynb
A Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/screenshots/unnamed.png
A Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/screenshots/unnamed1.png
A Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/screenshots/unnamed3.png
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch1.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch10.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch11.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch12.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch13.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch14.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch15.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch16.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch17.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch19.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch2.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch20.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch21.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch23.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch3.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch4.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch5.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch6.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch7.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch8.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch9.ipynb
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/screenshots/dissipated_power.png
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/screenshots/pumping_energy_23.png
A Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/screenshots/vtg_gain_14.png
Diffstat (limited to 'Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati')
13 files changed, 5350 insertions, 0 deletions
diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_1.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_1.ipynb new file mode 100644 index 00000000..c0748974 --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_1.ipynb @@ -0,0 +1,895 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1: Structure and Bonding" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page no: 35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "c = 3 * 10 ** 10 # Velocity of light, cm/sec\n", + "\n", + "# Variable\n", + "wavelength = 3500 * 10 ** -8 # Wavelength of radiation, cm\n", + "\n", + "# Solution\n", + "print \"v = c / wavelength\"\n", + "print \"v: Velocity, c: Speed of light\"\n", + "\n", + "v = c / wavelength\n", + "\n", + "print \"The frequency of radiation is\", '{:.2e}'.format(v), \"Heartz.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "v = c / wavelength\n", + "v: Velocity, c: Speed of light\n", + "The frequency of radiation is 8.57e+14 Heartz.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page no: 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "c = 3 * 10 ** 8 # speed of light, m/sec\n", + "\n", + "# Variable\n", + "f = 5 * 10 ** 16 # frequency, cycles/sec\n", + "\n", + "# Solution\n", + "v_bar = f / c\n", + "print \"The wave number is\", '{:.2e}'.format(v_bar), \"cycles/m.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The wave number is 1.67e+08 cycles/m.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page no: 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "c = 3 * 10 ** 8 # Speed of light, m/sec\n", + "\n", + "# Variable\n", + "T = 2.4 * 10 ** -10 # Time period, sec\n", + "\n", + "# Solution\n", + "f = 1 / T # Frequency, /sec\n", + "lamda = c / f # wavelength, m\n", + "v_bar = 1 / lamda # wavenumber, /meter\n", + "\n", + "print \"Frequency:\", '{:.2e}'.format(f), \"/sec\"\n", + "print \"Wavelength:\", '{:.2e}'.format(lamda), \"m\"\n", + "print \"Wave number:\", '{:.2e}'.format(v_bar), \"/m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency: 4.17e+09 /sec\n", + "Wavelength: 7.20e-02 m\n", + "Wave number: 1.39e+01 /m\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 4, Page no: 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constants\n", + "c = 3 * 10 ** 8 # Speed of light, m/sec\n", + "m = 9.1 * 10 ** -31 # Mass of electron, kg\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variable\n", + "ke = 4.55 * 10 ** -25 # Kinetic Energy, J\n", + "\n", + "# Solution\n", + "v = math.sqrt(ke * 2 / m)\n", + "\n", + "lamda = h / (m * v)\n", + "\n", + "print \"The de Broglie wavelength is\", '{:.2e}'.format(lamda), \"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The de Broglie wavelength is 7.28e-07 m\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 5, Page no: 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variables\n", + "m = 10 * 10 ** -3 # Mass of the ball, kg\n", + "v = 10 ** 5 # Velocity of ball, cm / sec\n", + "\n", + "# Solution\n", + "lamda = (h * 10 ** 7) / (m * v)\n", + "print \"The Wavelength of iron ball is\", \"{:.2}\".format(lamda), \"cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Wavelength of iron ball is 6.6e-30 cm\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 6, Page no: 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variable\n", + "lamda = 2 * 10 ** -10 # wavelength, m\n", + "\n", + "# Solution\n", + "p = h / lamda\n", + "\n", + "print \"The momentum of the particle is\", \"{:.2}\".format(p), \"kg.m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The momentum of the particle is 3.3e-24 kg.m/s\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 7, Page no: 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "m = 9.1 * 10 ** -31 # Mass of electron, kg\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "pi = 3.141 # Pi\n", + "\n", + "# Variable\n", + "delta_x = 1 * 10 ** -10 # uncertainty in velocity, m\n", + "\n", + "# Solution\n", + "delta_v = h / (4 * pi * m * delta_x)\n", + "\n", + "print \"Uncertainty in position of electron >=\",\n", + "print \"{:.2}\".format(delta_v), \"m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainty in position of electron >= 5.8e+05 m/s\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 8, Page no: 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "pi = 3.141 # Pi\n", + "\n", + "# Variables\n", + "m = 10 ** -11 # Mass of particle, g\n", + "v = 10 ** -4 # Velocity of particle, cm/sec\n", + "delta_v = 0.1 / 100 # Uncertainty in velocity\n", + "\n", + "# Solution\n", + "delta_v = v / 1000\n", + "delta_x = (h * 10 ** 7) / (4 * pi * delta_v * m)\n", + "\n", + "print \"Uncertainty in position >=\",\n", + "print \"{:.3e}\".format(delta_x), \"cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainty in position 5.27e-10 cm\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 9, Page no: 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "c = 3 * 10 ** 8 # Speed of light, m/sec\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variable\n", + "lamda = 650 * 10 ** -12 # Wavelength of radiation, m\n", + "\n", + "# Solution\n", + "E = h * c / lamda\n", + "\n", + "print \"Energy per photon\", \"{:.3e}\".format(E), \"J\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy per photon 3.058e-16 J\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 10, Page no: 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "h = 6.625 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variables\n", + "v = 6.5 * 10 ** 7 # Velocity of particle, m/s\n", + "lamda = 5 * 10 ** -11 # Wavelength, m\n", + "\n", + "# Solution\n", + "P = h / lamda\n", + "\n", + "print \"The momentum of the particle\", \"{:.2e}\".format(P), \"kg.m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The momentum of the particle 1.33e-23 kg.m/s\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 11, Page no: 38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constants\n", + "c = 3 * 10 ** 8 # Speed of light, m/sec\n", + "m = 9.1 * 10 ** -31 # Mass of electron, kg\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variables\n", + "lamda = 200 * 10 ** -7 # Wavelength, cm\n", + "wf = 6.95 * 10 ** -12 # Work function, erg\n", + "\n", + "# Solution\n", + "E = (h * c) * 10 ** 9 / lamda\n", + "\n", + "print \"Energy of photon\", \"{:.3e}\".format(E), \"erg\"\n", + "\n", + "ke = E - wf\n", + "\n", + "v = math.sqrt((2 * ke) / (m * 10 ** 3)) * 10 ** -2\n", + "\n", + "print \"The maximum velocity of electron\", \"{:.3e}\".format(v), \"m/sec\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of photon 9.939e-12 erg\n", + "The maximum velocity of electron 8.105e+05 m/sec\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 12, Page no: 38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variables\n", + "m = 150 # Weight of ball, gm\n", + "v = 50 # Velocity, m/sec\n", + "\n", + "lamda = h / (m * v * 10 ** -8)\n", + "print \"Wavelength of ball\", \"{:.3e}\".format(lamda), \"m\"\n", + "print \"Its wavelength is so short that it does not fall\",\n", + "print \"in visible range, so we cannot observe it.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of ball 8.835e-30 m\n", + "Its wavelength is so short that it does not fall in visible range, so we cannot observe it.\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 13, Page no: 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "pi = 3.141 # Pi\n", + "\n", + "# Variables\n", + "m = 0.1 # Mass of base ball, kg\n", + "delta_x = 10 ** -10 # Uncertainty in position, m\n", + "\n", + "# Solution\n", + "delta_v = h / (4 * pi * m * delta_x)\n", + "\n", + "print \"Uncertainty in velocity >=\", \"{:.2e}\".format(delta_v), \"m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainty in velocity >= 5.27e-24 m/s\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 14, Page no: 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "t_v = 1.3 * 10 ** 15 # Threashold freq. Pt, /sec\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "\n", + "# Solution\n", + "print \"The threshold frequency is the lowest frequency\",\n", + "print \"that photons may possess to produce the photoelectric\",\n", + "print \"effect.\"\n", + "E = h * t_v\n", + "print \"The energy corresponding to this frequency is the minimum\",\n", + "print \"energy =\", \"{:.2e}\".format(E), \"erg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The threshold frequency is the lowest frequency that photons may possess to produce the photoelectric effect.\n", + "The energy corresponding to this frequency is the minimum energy = 8.61e-19 erg\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 15, Page no: 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "m = 9.1 * 10 ** -31 # Mass of electron, kg\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "e = 1.602 * 10 ** -19 # Charge of electron, C\n", + "\n", + "# Variable\n", + "v = 1.87 * 10 ** 9 # Velocity of electron, m/sec\n", + "\n", + "# Solution\n", + "V = m * v ** 2 / (2 * e)\n", + "lamda = h / (m * v)\n", + "\n", + "print \"The voltage is\", \"{:.2e}\".format(V), \"volt\"\n", + "print \"The de Broglie wavelength is\", \"{:.2e}\".format(lamda), \"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage is 9.93e+06 volt\n", + "The de Broglie wavelength is 3.89e-13 m\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 16, Page no: 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "m = 9.1 * 10 ** -31 # Mass of electron, kg\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variable\n", + "lamda = 4.8 * 10 ** -9 # Wavelength of electron, m\n", + "\n", + "# Solution\n", + "ke = ((h / lamda) ** 2) / (2 * m)\n", + "\n", + "print \"The Kinetic Energy of moving electron is\", \"{:.2e}\".format(ke),\n", + "print \"J\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Kinetic Energy of moving electron is 1.05e-20 J\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 17, Page no: 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "m = 9.1 * 10 ** -31 # Mass of electron, kg\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "c = 3 * 10 ** 8 # Speed of light, m/sec\n", + "\n", + "# Variables\n", + "v = 6.46 * 10 ** 5 # Velocity of electron, m/sec\n", + "lamda = 200 * 10 ** -9 # Wavelength of light, m\n", + "\n", + "# Solution\n", + "E = (h * c) / lamda\n", + "ke = m * v ** 2\n", + "w = E - ke\n", + "\n", + "print \"The workfunction of the metal surface is\", \"{:.3e}\".format(w),\n", + "print \"J\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The workfunction of the metal surface is 6.141e-19 J\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 18, Page no: 40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constants\n", + "e = 1.602 * 10 ** -19 # Charge of proton, C\n", + "m_p = 1.66 * 10 ** -27 # Mass of proton, kg\n", + "m_e = 9.1 * 10 ** -31 # Mass of electron, kg\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variable\n", + "V = 35 # Acceleration potential, volt\n", + "\n", + "# Solution\n", + "lamda_p = h / math.sqrt(2 * e * V * m_p)\n", + "lamda_e = h / math.sqrt(2 * e * V * m_e)\n", + "\n", + "print \"The wavelength of electron when accelerated with same\",\n", + "print \"potential is\", \"{:.3e}\".format(lamda_e), \"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The wavelength of electron when accelerated with same potential is 2.074e-10 m\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 19, Page no: 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "B_O1 = (10 - 6) / 2 # Bond Order for O2\n", + "B_O2 = (10 - 7) / 2 # Bond Order for O2-\n", + "\n", + "print \"Bond length of O2- > O2 as Bond order of O2\",\n", + "print \"> Bond order of O2- :\", B_O1 > B_O2\n", + "print \"Both are paramagnetic, because they contain unpaired electrons.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Bond length of O2- > O2 as Bond order of O2 > Bond order of O2- : True\n", + "Both are paramagnetic, because they contain unpaired electrons.\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 20, Page no: 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "B_O = (9 - 4) / 2.0 # Bond order of N2+\n", + "\n", + "print \"MO configuration of N2+ is\"\n", + "print \"\u03c3(1s2)\u03c3*(1s2)\u03c3(2s2)\u03c3*(2s2) [\u03c0(2px2) = \u03c0(2py2)] \u03c3(2pz1)\\n\"\n", + "print \"The Bond order of N2+, 1/2[Nb - Na] =\", B_O\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "MO configuration of N2+ is\n", + "\u03c3(1s2)\u03c3*(1s2)\u03c3(2s2)\u03c3*(2s2) [\u03c0(2px2) = \u03c0(2py2)] \u03c3(2pz1)\n", + "\n", + "The Bond order of N2+, 1/2[Nb - Na] = 2.5\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 21, Page no: 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "v_n = 2 * 5 # number of valence e- in nitrogen\n", + "v_co = 4 + 6 # number of valence e- in CO\n", + "\n", + "print \"The number of valence electrons in N2\", v_n\n", + "print \"The number of valence electrons in CO\", v_co\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The number of valence electrons in N2 10\n", + "The number of valence electrons in CO 10\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 22, Page no: 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"O2(+2) [B.O. = 1.0] < O2(-1) [B.O. = 1.5] <\",\n", + "print \"O2 [B.O. = 2.0] < O2(+1) [B.O. =2.5]\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "O2(+2) [B.O. = 1.0] < O2(-1) [B.O. = 1.5] < O2 [B.O. = 2.0] < O2(+1) [B.O. =2.5]\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 23, Page no: 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"The number of electrons in N2-\", (7 + 8)\n", + "print \"The molecular configuration of N2- is\"\n", + "print \"\u03c3(1s2)\u03c3*(1s2)\u03c3(2s2)\u03c3*(2s2) [\u03c0(2px2) = \u03c0(2py2)]\",\n", + "print \"\u03c3(2pz2) [\u03c0*(2px1) = \u03c0*(2py0)] OR [\u03c0*(2px0) = \u03c0*(2py1)]\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The number of electrons in N2- 15\n", + "The molecular configuration of N2- is\n", + "\u03c3(1s2)\u03c3*(1s2)\u03c3(2s2)\u03c3*(2s2) [\u03c0(2px2) = \u03c0(2py2)] \u03c3(2pz2) [\u03c0*(2px1) = \u03c0*(2py0)] OR [\u03c0*(2px0) = \u03c0*(2py1)]\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_12.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_12.ipynb new file mode 100644 index 00000000..f6562421 --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_12.ipynb @@ -0,0 +1,163 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12: Polymers and Polymerization" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page no: 311" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variable\n", + "Mwt = 21150 # g / mol\n", + "\n", + "# Solution\n", + "m = 2 * 12 + 3 * 1.008 + 1 * 35.45 # g / mer\n", + "n = Mwt / m\n", + "print \"The degree of polymerization is\", int(n)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The degree of polymerization is 338\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page no: 312" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "n = 10000 # degree of polymerisation\n", + "\n", + "# Solution\n", + "m = 8 * 12 + 8 * 1.008 # g / mer\n", + "M = n * m\n", + "print \"Molecular weight of polystyrene chain,\", M, \"g /mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Molecular weight of polystyrene chain, 1040640.0 g /mol\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page no: 312" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "d1 = 920 # density, kg / m^3\n", + "d2 = 961.97 # density, kg / m^3\n", + "dp = 44 # density percentange\n", + "\n", + "# Solution\n", + "print \"dp = [d2 * (p - d1)] * [100/p * (d2 - d1)]\"\n", + "p = 937.98\n", + "print \"Density of sample is\", p, \"kg / m^3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "dp = [d2 * (p - d1)] * [100/p * (d2 - d1)]\n", + "Density of sample is 937.98 kg / m^3\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 4, Page no: 313" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "Na = 6.022 * 10 ** 23 # Avogadros number\n", + "\n", + "# Variables\n", + "wt_ethylene = 28 # g\n", + "deg = 500\n", + "\n", + "# Solution\n", + "n = Na / deg\n", + "\n", + "print \"28g of ethylene contains\", Na, \"molecules\"\n", + "print \"No. of polyethylene formed\", \"{:.3e}\".format(n), \"molecules\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "28g of ethylene contains 6.022e+23 molecules\n", + "No. of polyethylene formed 1.204e+21 molecules\n" + ] + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_13.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_13.ipynb new file mode 100644 index 00000000..a19a42ce --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_13.ipynb @@ -0,0 +1,626 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 13: Fuel and Combustions" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page No: 350" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "C = 84 # Percentage\n", + "S = 1.5 # Percentage\n", + "N = 0.6 # Percentage\n", + "H = 5.5 # Percentage\n", + "O = 8.4 # Percentage\n", + "\n", + "# Solution\n", + "GCV = (8080 * C + 34500 * (H - O / 8) + 2240 * S) / 100\n", + "LCV = (GCV - 9 * H / 100 * 587)\n", + "print \"Gross Calorific Value\", int(GCV), \"kcal / kg\"\n", + "print \"Net Calorific Value\", \"{:.2f}\".format(LCV), \"kcal / kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Gross Calorific Value 8356 kcal / kg\n", + "Net Calorific Value 8065.48 kcal / kg\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page No: 350" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "C = 90 # Percentage\n", + "O = 3.0 # Percentage\n", + "S = 0.5 # Percentage\n", + "N = 0.5 # Percentage\n", + "ash = 2.5 # Percentage\n", + "LCV = 8490.5 # kcal / kg\n", + "\n", + "# Solution\n", + "print \"HCV = LCV + 9 * H / 100 * 587\"\n", + "print \"HCV = 1/100 * (8080 * C + 34500 * (H - O / 8) + 2240 * N)\"\n", + "H = (8490.5 - 7754.8) / (345 - 52.8)\n", + "H = 4.575\n", + "print \"The precentage of H is\", H, \"%\"\n", + "HCV = LCV + 52.8 * H\n", + "print \"Higeher calorific value of coal\", \"{:.1f}\".format(HCV), \"kcal / kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HCV = LCV + 9 * H / 100 * 587\n", + "HCV = 1/100 * (8080 * C + 34500 * (H - O / 8) + 2240 * N)\n", + "The precentage of H is 4.575 %\n", + "Higeher calorific value of coal 8732.1 kcal / kg\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page No: 351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "x = 0.72 # g\n", + "W = 250 # g\n", + "w = 150 # g\n", + "t1 = 27.3 # C\n", + "t2 = 29.1 # C\n", + "\n", + "# Solution\n", + "HCV = ((W + w) * (t2 - t1)) / x\n", + "HCV *= 4185.0 / 10 ** 6\n", + "print \"HCV of fuel is\", \"{:.3f}\".format(HCV), \"KJ / Kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HCV of fuel is 4.185 KJ / Kg\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 4, Page No: 351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "x = 0.84 # g\n", + "W = 1060 # g\n", + "w = 135 # g\n", + "delta_t = 2.5 # C\n", + "\n", + "# Solution\n", + "HCV = ((W + w) * delta_t) / x\n", + "print \"HCV of fuel is\", \"{:.2f}\".format(HCV), \"kcal / kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HCV of fuel is 3556.55 kcal / kg\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 5, Page No: 351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "V = 0.1 # m ^ 3\n", + "W = 25 # kg\n", + "t1 = 20 # C\n", + "t2 = 33 # C\n", + "m = 0.025 # kg\n", + "\n", + "# Solution\n", + "HCV = W * (t2 - t1) / V\n", + "LCV = HCV - (m / V) * 580\n", + "print \"HCV is\", HCV, \"kcal / m^3\"\n", + "print \"LCV is\", LCV, \"kcal / m^3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HCV is 3250.0 kcal / m^3\n", + "LCV is 3105.0 kcal / m^3\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 6, Page No: 351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "w1 = 2.5 # g\n", + "w2 = 2.415 # g\n", + "r = 1.528 # g\n", + "ma = 0.245 # Mass of ash, g\n", + "\n", + "# Solution\n", + "m = w1 - w2 # Mass of moisture in coal\n", + "mv = w2 - r # Mass of volatile matter\n", + "moisp = m * 100 / w1\n", + "volp = mv * 100 / w1\n", + "ashp = ma * 100 / w1\n", + "carbp = 100 - (moisp + volp + ashp)\n", + "print \"Percentage of moisture:\", moisp, \"%\"\n", + "print \"Percentage of volatile matter:\", volp, \"%\"\n", + "print \"Percentage of ash:\", ashp, \"%\"\n", + "print \"Percentage of fixed carbon:\", carbp, \"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage of moisture: 3.4 %\n", + "Percentage of volatile matter: 35.48 %\n", + "Percentage of ash: 9.8 %\n", + "Percentage of fixed carbon: 51.32 %\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 7, Page No: 352" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt_coke = 2 # kg\n", + "\n", + "# Solution\n", + "wt_O = 2 * 32 / 12.0\n", + "wt_air = wt_O * 100 / 23.2\n", + "Vol_air = wt_air / 28.94 * 22.4\n", + "print \"Volume of air needed for the complete combustion of 2kg coke\",\n", + "print \"is\", \"{:.3f}\".format(Vol_air), \"litres at NTP\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Volume of air needed for the complete combustion of 2kg coke is 17.793 litres at NTP\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 8, Page No: 352" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "C = 86 # Percent\n", + "H = 4 # Percent\n", + "N = 1.3 # Percent\n", + "S = 3 # Percent\n", + "O = 4 # Percent\n", + "Ash = 1.7 # Percent\n", + "wt = 500 # g\n", + "\n", + "# Solution\n", + "wt_C = C / 100.0\n", + "wt_S = S / 100.0\n", + "wt_H = H / 100.0\n", + "wt_O = O / 100.0\n", + "\n", + "print \"Nitrogen and ash are incombustible, so they do not require oxygen.\"\n", + "wt_O_C = 32 / 12.0 * wt_C\n", + "wt_O_S = 32 / 32.0 * wt_S\n", + "wt_O_H = 32 / 4.0 * wt_H\n", + "\n", + "Twt_O = wt_O_H + wt_O_S + wt_O_C\n", + "wt_O_needed = Twt_O - wt_O\n", + "wt_air = (100.0 / 23.0 * wt_O_needed) * 500 / 1000.0\n", + "print \"Minimum Wt. of air required by 500g of fuel\", \"{:.2f}\".format(wt_air), \"kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Nitrogen and ash are incombustible, so they do not require oxygen.\n", + "Minimum Wt. of air required by 500g of fuel 5.66 kg\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 9, Page No: 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt_C = 3 # kg\n", + "\n", + "# Solution\n", + "wt_air = wt_C * 32 * 100 / 12.0 / 23.0\n", + "vol_air = wt_air * 1000 * 22.4 / 28.94\n", + "\n", + "print \"H2(g) + 1/2 O2(g) --> H20(l)\"\n", + "print \" 1 0.5 1\\t\\t(By Vol.)\"\n", + "print \"CO(g) + 1/2 O2(g) --> CO2(g)\"\n", + "print \" 1 0.5 1\\t\\t(By Vol.)\"\n", + "print \"CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l)\"\n", + "print \" 1 2 1\\t\\t(By Vol.)\"\n", + "\n", + "print \"Weight of air for the combustion of 3kg carbon\",\n", + "print \"{:.3f}\".format(wt_air), \"kg\"\n", + "print \"Vol. pf air required for combustion of 3kg carbon\",\n", + "print \"{:.3e}\".format(vol_air), \"L\",\n", + "print \"or\", \"{:.2f}\".format(vol_air / 1000), \"m^3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "H2(g) + 1/2 O2(g) --> H20(l)\n", + " 1 0.5 1\t\t(By Vol.)\n", + "CO(g) + 1/2 O2(g) --> CO2(g)\n", + " 1 0.5 1\t\t(By Vol.)\n", + "CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l)\n", + " 1 2 1\t\t(By Vol.)\n", + "Weight of air for the combustion of 3kg carbon 34.783 kg\n", + "Vol. pf air required for combustion of 3kg carbon 2.692e+04 L or 26.92 m^3\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 10, Page No: 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "H = 0.30 # m^3\n", + "CO = 0.10 # m^3\n", + "CH4 = 0.04 # m^3\n", + "N2 = 0.56 # m^3\n", + "\n", + "# Soution\n", + "vol_oxygen = H * 0.5 + CO * 0.5 + CH4 * 2\n", + "vol_air = vol_oxygen * 100 / 21\n", + "print \"Volumer of air required for complete combustion of 1 m^3 of\",\n", + "print \"producer gas:\", \"{:.3f}\".format(vol_air), \"m^3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Volumer of air required for complete combustion of 1 m^3 of producer gas: 1.333 m^3\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 11, Page No: 354" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "H = 15.4 # Percentage\n", + "C = 84.6 # Percentage\n", + "wt_fuel = 1 # kg\n", + "wt_C = 0.846 # kg\n", + "wt_H = 0.154 # kg\n", + "\n", + "# Solution\n", + "print \"The combustion reactions are,\"\n", + "print \"C + O2 --> CO2\"\n", + "print \"12 32 \\t(by Weight)\"\n", + "print \"2H2 + O2 --> H20\"\n", + "print \" 4 32\\t(by Weight)\"\n", + "\n", + "wt_O = 32 / 12.0 * wt_C\n", + "wt_O_H = 32 / 4.0 * wt_H\n", + "Twt_O = wt_O + wt_O_H\n", + "print \"Because 32 gm of O2 occupies a volume of 22.4 liters at NTP\"\n", + "print \"3.488 * 1000 gm of O2 will occupy\",\n", + "print \"{:.1f}\".format(22.4 / 32 * Twt_O * 1000), \"liters\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The combustion reactions are,\n", + "C + O2 --> CO2\n", + "12 32 \t(by Weight)\n", + "2H2 + O2 --> H20\n", + " 4 32\t(by Weight)\n", + "Because 32 gm of O2 occupies a volume of 22.4 liters at NTP\n", + "3.488 * 1000 gm of O2 will occupy 2441.6 liters\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 12, Page No: 354" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "C = 750 # g\n", + "H = 52 # g\n", + "O = 121 # g\n", + "N = 32 # g\n", + "ash = 45 # g\n", + "\n", + "# Solution\n", + "min_wt_air = (C * 32 / 12. + H * 16 / 2. - O) * 100 / 23.\n", + "HCV = 1 / 1000. * (8080 * C + 34500 * (H - O / 8.) + 2240 * 0)\n", + "LCV = HCV - 0.09 * H * 587 / 10.0\n", + "\n", + "print \"HCV is\", int(HCV), \"kcal/kg\"\n", + "print \"LCV is\", int(LCV), \"kcal/kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HCV is 7332 kcal/kg\n", + "LCV is 7057 kcal/kg\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 13, Page No: 355" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "C = 81 # Percent\n", + "H = 8 # Percent\n", + "N = 2 # Percent\n", + "O = 5 # Percent\n", + "\n", + "# Solution\n", + "print \"In 1kg coal,\"\n", + "\n", + "wt_C = C * 10\n", + "wt_H = H * 10\n", + "wt_N = N * 10\n", + "wt_O = O * 10\n", + "wt_ash = 100 - (wt_O + wt_N + wt_H + wt_C)\n", + "\n", + "wt_air = ((wt_C * 32 / 12. + wt_H * 16 / 2. - wt_O) * 100 / 23.) / 1000.\n", + "\n", + "print \"Weight of air required for complete combustion of 10kg coal\",\n", + "print \"=\", \"{:.2f}\".format(wt_air * 10), \"kg\"\n", + "\n", + "HCV = 1 / 100. * (8080 * C + 34500 * (H - O / 8.))\n", + "LCV = HCV - 0.09 * H * 587\n", + "\n", + "print \"HCV is\", int(HCV), \"kcal/kg\"\n", + "print \"LCV is\", int(LCV), \"kcal/kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "In 1kg coal,\n", + "Weight of air required for complete combustion of 10kg coal = 119.57 kg\n", + "HCV is 9089 kcal/kg\n", + "LCV is 8666 kcal/kg\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 14, Page No: 355" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "C = 80 # Percent\n", + "H = 7 # Percent\n", + "N = 2.1 # Percent\n", + "O = 3 # Percent\n", + "S = 3.5 # Percent\n", + "Ash = 4.4 # Percent\n", + "\n", + "# Solution\n", + "HCV = 1 / 100. * (8080 * C + 34500 * (H - O / 8.) + 2240 * S)\n", + "LCV = HCV - 0.09 * H * 587\n", + "\n", + "print \"HCV is\", int(HCV), \"kcal/kg\"\n", + "print \"LCV is\", int(LCV), \"kcal/kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HCV is 8828 kcal/kg\n", + "LCV is 8458 kcal/kg\n" + ] + } + ], + "prompt_number": 30 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_14.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_14.ipynb new file mode 100644 index 00000000..5117d5ae --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_14.ipynb @@ -0,0 +1,440 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 14: Water Treatment" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page No:378" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt_CaSO4 = 160 # mg / l\n", + "\n", + "# Solution\n", + "hardness = 100 * wt_CaSO4 / 136.\n", + "print \"The hardness is\", \"{:.2f}\".format(hardness), \"mg / L of CaCO3 eqv.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The hardness is 117.65 mg / L of CaCO3 eqv.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page No:378" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt1 = 9.3 # mg / L\n", + "wt2 = 17.4 # mg / L\n", + "wt3 = 8.7 # mg / L\n", + "wt4 = 12.6 # mg / L\n", + "\n", + "# Solution\n", + "temp_hardness = wt1 * 100 / 146. + wt2 * 100 / 162.\n", + "per_hardness = wt3 * 100 / 95. + wt4 * 100 / 136.\n", + "total_hardness = temp_hardness + per_hardness\n", + "\n", + "print \"Temporary hardness:\", \"{:.2f}\".format(temp_hardness), \"mg / L\"\n", + "print \"Total hardness:\", \"{:.2f}\".format(total_hardness), \"mg / L\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temporary hardness: 17.11 mg / L\n", + "Total hardness: 35.53 mg / L\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page No:378" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt1 = 32.4 # mg / L\n", + "wt2 = 29.2 # mg / L\n", + "wt3 = 13.5 # mg / L\n", + "\n", + "# Solution\n", + "temp_hardness = wt1 * 100 / 162. + wt2 * 100 / 146.\n", + "per_hardness = wt3 * 100 / 136.\n", + "\n", + "print \"Temporary hardness:\", \"{:.2f}\".format(temp_hardness), \"mg / L\"\n", + "print \"Permanent hardness:\", \"{:.2f}\".format(per_hardness), \"mg / L\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temporary hardness: 40.00 mg / L\n", + "Permanent hardness: 9.93 mg / L\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 4, Page No:379" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "i1 = 180 # mg / L, CaCl2\n", + "i2 = 210 # mg / L, Ca(NO3)2\n", + "i3 = 123 # mg / L, MgSO4\n", + "i4 = 90 # mg / L, Mg(HCO3)2\n", + "\n", + "# Solution\n", + "i1_req = i1 * 100 / 111.\n", + "i2_req = i2 * 100 / 164.\n", + "i3_req = i3 * 100 / 120.\n", + "i4_req = i4 * 100 / 146.\n", + "\n", + "lime_req = 74 / 100. * (2 * i4_req + i3_req) * 100 / 70. * 10000\n", + "soda_req = 106 / 100. * (i1_req + i3_req + i2_req) * 100 / 80. * 10000\n", + "\n", + "print \"Lime Required\", \"{:.1e}\".format(lime_req), \"mg\",\n", + "print \"=\", \"{:.1f}\".format(lime_req / 10 ** 6), \"kg\"\n", + "print \"Soda Required\", \"{:.1e}\".format(soda_req), \"mg\",\n", + "print \"=\", \"{:.1f}\".format(soda_req / 10 ** 6), \"kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lime Required 2.4e+06 mg = 2.4 kg\n", + "Soda Required 5.2e+06 mg = 5.2 kg\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 5, Page No:379" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt1 = 32.4 # mg / L, Ca(HCO3)2\n", + "wt2 = 29.29 # mg / L, Mg(HCO3)2\n", + "wt3 = 13.5 # mg / L, CaSO4\n", + "\n", + "# Solution\n", + "wt1_equi = wt1 * 100 / 162.\n", + "wt2_equi = wt2 * 100 / 146.\n", + "wt3_equi = wt3 * 100 / 136.\n", + "\n", + "temp_hardness = wt1_equi + wt2_equi\n", + "perm_hardness = wt3_equi\n", + "\n", + "print \"Temporary hardness [due to Ca(HCO3)2 & Mg(HCO3)2] is\",\n", + "print int(temp_hardness), \"ppm\"\n", + "print \"Permanent hardness [due to CaSO4] is\", \"{:.1f}\".format(perm_hardness), \"ppm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temporary hardness [due to Ca(HCO3)2 & Mg(HCO3)2] is 40 ppm\n", + "Permanent hardness [due to CaSO4] is 9.9 ppm\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 6, Page No:380" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "v1 = 150 # litres, NaCl\n", + "\n", + "# Solution\n", + "v_hardwater = 22500 * v1 / 3 / 0.6 / 58.5\n", + "\n", + "print \"The amount of hard water that can be softened using this softner is\",\n", + "print int(v_hardwater), \"litres\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The amount of hard water that can be softened using this softner is 32051 litres\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 7, Page No:380" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "v1 = 30 # litres, NaCl\n", + "w = 1500 # mg / L, NaCl\n", + "\n", + "# Solution\n", + "hardness = 45 * 50 / 58.5 * 1000 / 1000\n", + "print \"Hardness of water is\", \"{:.2f}\".format(hardness), \"ppm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hardness of water is 38.46 ppm\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 8, Page No:381" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "v1_water = 50 # ml, water\n", + "w1_CaCO3 = 1.5 # mg, pure CaCO3\n", + "v1_EDTA = 44 # ml, EDTA\n", + "v2_EDTA = 40 # ml, EDTA\n", + "v2_water = 20 # ml, water\n", + "\n", + "# Solution\n", + "EDTA_1ml = v1_water * w1_CaCO3 / v1_EDTA\n", + "hardwater_40ml = v2_water * 1.704\n", + "total_hardness0 = hardwater_40ml * 1000 / 40\n", + "total_hardness1 = total_hardness0 * 0.07\n", + "\n", + "print \"Total hardness is\", \"{:.2f}\".format(total_hardness1), \"\u00b0Cl\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total hardness is 59.64 \u00b0Cl\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 9, Page No:381" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "Fe = 56\n", + "S = 32\n", + "O = 16\n", + "Ca = 40\n", + "C = 12\n", + "\n", + "# Solution\n", + "hardness100 = Fe + S + O * 4\n", + "\n", + "print \"215 ppm of hardness is\", \"{:.1f}\".format(hardness100 * 215 / 100.),\n", + "print \"ppm of FeSO4\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "215 ppm of hardness is 326.8 ppm of FeSO4\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 10, Page No:381" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "v1 = 50. # ml, hardwater\n", + "v2 = 15 # ml, EDTA\n", + "m = 0.01 # M, EDTA\n", + "\n", + "# Solution\n", + "M = v2 * m / v1\n", + "N = M * 2\n", + "S = N * 50 * 1000\n", + "\n", + "print \"Molarity of hardness is\", M, \"M\"\n", + "print \"Normality of hardness is\", N, \"N\"\n", + "print \"Strength of hardness is\", S, \"ppm or mg / L\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Molarity of hardness is 0.003 M\n", + "Normality of hardness is 0.006 N\n", + "Strength of hardness is 300.0 ppm or mg / L\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 11, Page No:382" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variable\n", + "C = 16.5 # ppm, CO3-2\n", + "\n", + "# Solution\n", + "Molarity = C * 10 ** - 6 / 60.\n", + "\n", + "print \"Molarity of CO3-2 is\", \"{:.1e}\".format(Molarity), \"mol / L\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Molarity of CO3-2 is 2.7e-07 mol / L\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_15.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_15.ipynb new file mode 100644 index 00000000..ab29e076 --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_15.ipynb @@ -0,0 +1,150 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 15: Environmental Pollution and Control" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page No: 401" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "MM = 294 # Molar mass, K2Cr2O7\n", + "\n", + "# Variables\n", + "v_eff = 25 # cm ^ 3, effluent\n", + "v = 8.3 # cm ^ 3, K2Cr2O7\n", + "M = 0.001 # M, K2Cr2O7\n", + "\n", + "# Solution\n", + "w_O = v * 8 * 6 * M / 1000.\n", + "\n", + "print \"8.3 cm^3 of 0.006 N K2Cr2O7 =\", \"{:.3e}\".format(w_O), \"g of O2\"\n", + "print \"25 ml of the effluent requires\", \"{:.3e}\".format(w_O), \"g of O2\"\n", + "\n", + "cod = w_O * 10 ** 6 / 25.\n", + "print \"1 l of the effluent requires\", \"{:.2f}\".format(cod), \"g of O2\"\n", + "print \"COD of the effluent sample is\", \"{:.2f}\".format(cod), \"ppm or mg / L\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "8.3 cm^3 of 0.006 N K2Cr2O7 = 3.984e-04 g of O2\n", + "25 ml of the effluent requires 3.984e-04 g of O2\n", + "1 l of the effluent requires 15.94 g of O2\n", + "COD of the effluent sample is 15.94 ppm or mg / L\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page No: 401" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "v0 = 30 # cm^3, effluent\n", + "v1 = 9.8 # cm^3, K2Cr2O7\n", + "M = 0.001 # M, K2Cr2O7\n", + "\n", + "# Solution\n", + "O_30eff = 6 * 8 * v1 * M\n", + "print \"So 30 cm^3 of effluent contains =\", \"{:.4f}\".format(O_30eff), \"mg of O2\"\n", + "\n", + "cod = O_30eff * 1000 / 30.\n", + "\n", + "print \"1 l of the effluent requires\", cod, \"mg of O2\"\n", + "print \"COD of the effluent sample =\", cod, \"ppm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "So 30 cm^3 of effluent contains = 0.4704 mg of O2\n", + "1 l of the effluent requires 15.68 mg of O2\n", + "COD of the effluent sample = 15.68 ppm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page No: 401" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "v0 = 25 # ml, sewage\n", + "d0_O = 410 # ppm, dissolved oxygen\n", + "d1_O = 120 # ppm, dissolved oxygen\n", + "v1 = 50 # ml, sewage\n", + "\n", + "# Solution\n", + "print \"BOD = (DOb - DOi) * Dilution Factor\"\n", + "print \"BOD = (DOb - DOi) *\",\n", + "print \"(ml of sample after dilution) / (ml of sample before dilution)\"\n", + "\n", + "BOD = (d0_O - d1_O) * (v1 / v0)\n", + "print \"BOD =\", BOD, \"ppm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "BOD = (DOb - DOi) * Dilution Factor\n", + "BOD = (DOb - DOi) * (ml of sample after dilution) / (ml of sample before dilution)\n", + "BOD = 580 ppm\n" + ] + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_2.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_2.ipynb new file mode 100644 index 00000000..58e602ab --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_2.ipynb @@ -0,0 +1,418 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2: Spectroscopy and Photochemistry" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page no: 65" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "m_br79 = 78.9183 # Mass of 79Br, amu\n", + "m_br81 = 80.9163 # Mass of 91Br, amu\n", + "Na = 6.022 * 10 ** 23 # Mole constant, /mol\n", + "pi = 3.141 # Pi\n", + "c = 3 * 10 ** 10 # Speed of light, cm /s\n", + "\n", + "# Variable\n", + "wave_no = 323.2 # Wave no. of fund. vibration of 79Br - 81Br, /cm\n", + "\n", + "# Solution\n", + "mu = (m_br79 * m_br81) / ((m_br79 + m_br81) * Na)\n", + "\n", + "k = 4 * (pi * c * wave_no) ** 2 * mu * 10 ** -3\n", + "\n", + "print \"The force constant of the bond is\", \"{:.3e}\".format(k), \"N/m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The force constant of the bond is 2.461e+02 N/m\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page no: 65" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "Na = 6.022 * 10 ** 23 # Mole constant, /mol\n", + "pi = 3.141 # Pi\n", + "c = 3 * 10 ** 10 # Speed of light, cm /s\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variables\n", + "b_l = 112.81 * 10 ** -12 # Equillibrium bond length, m\n", + "m1 = 12 # Mass of Carbon, g /mol\n", + "m2 = 16 # Mass of Oxygen, g /mol\n", + "\n", + "# Solution\n", + "mu = m1 * m2 / ((m1 + m2) * Na) # g\n", + "mu *= 10 ** -3 # kg\n", + "\n", + "B = h / (8 * pi ** 2 * mu * b_l ** 2 * c)\n", + "v2_3 = B * 6\n", + "\n", + "print \"The reduced mass of CO is\", \"{:.3e}\".format(mu), \"kg\"\n", + "print \"The frequency of 3->2 transition is\", \"{:.2f}\".format(v2_3), \"/cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reduced mass of CO is 1.139e-26 kg\n", + "The frequency of 3->2 transition is 11.59 /cm\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page no: 66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "Na = 6.022 * 10 ** 23 # Mole constant, /mol\n", + "\n", + "# Variables\n", + "d_NaCl = 2.36 * 10 ** -10 # Intermolecular dist. NaCl, m\n", + "m_Cl = 35 * 10 ** -3 # Atomic mass, kg /mol\n", + "m_Na = 23 * 10 ** -3 # Atomic mass, kg /mol\n", + "\n", + "# Solution\n", + "mu = m_Na * m_Cl / ((m_Na + m_Cl) * 10 ** -3 * Na) * 10 ** -3\n", + "\n", + "I = mu * d_NaCl ** 2\n", + "\n", + "print \"The reduced mass of NaCl is\", \"{:.3e}\".format(mu), \"kg\"\n", + "print \"The moment of inertia of NaCl is\", \"{:.3e}\".format(I), \"kg.m^2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reduced mass of NaCl is 2.305e-26 kg\n", + "The moment of inertia of NaCl is 1.284e-45 kg.m^2\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 4, Page no: 66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constant\n", + "e = 4000 # Extinction coeff., dm^3/mol/cm\n", + "\n", + "# Variable\n", + "x = 3 # Solution thickness, cm\n", + "\n", + "# Solution\n", + "A = math.log10(1 / 0.3) # Absorbance\n", + "C = A / (e * x)\n", + "\n", + "print \"The concentration of the solution is\", \"{:.2e}\".format(C), \"mol/dm^3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The concentration of the solution is 4.36e-05 mol/dm^3\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 5, Page no: 67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "pi = 3.141 # Pi\n", + "c = 3 * 10 ** 10 # Speed of light, cm /s\n", + "\n", + "# Variables\n", + "v_bar = 2140 # Fundamental vibrating freq, /cm\n", + "m_C = 19.9 * 10 ** -27 # Atomic mass of C, kg\n", + "m_O = 26.6 * 10 ** -27 # Atomic mass of O, kg\n", + "\n", + "# Solution\n", + "mu = m_O * m_C / (m_C + m_O)\n", + "k = 4 * (pi * c * v_bar) ** 2 * mu\n", + "\n", + "print \"The force constant of the molecule is\", \"{:.3e}\".format(k), \"N/m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The force constant of the molecule is 1.852e+03 N/m\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 6, Page no: 67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"a) Microwave < IR < UV-Visible < X-Ray.\"\n", + "print \"b) HCl and NO because they possess permanent dipole moments, so they are rotationally active.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) Microwave < IR < UV-Visible < X-Ray.\n", + "b) HCl and NO because they possess permanent dipole moments, so they are rotationally active.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 7, Page no: 67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constants\n", + "pi = 3.141 # pi\n", + "c = 3 * 10 ** 10 # speed of light, cm /s\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "Na = 6.022 * 10 ** 23 # Mole constant, /mol\n", + "\n", + "# Variables\n", + "d = 20.7 # Interspacing, /cm\n", + "m1 = 1 # Mass of H, g / mol\n", + "m2 = 35.5 # Masso f Cl, g / mol\n", + "\n", + "# Solution\n", + "B = 0.1035 * 10 ** 2 # /m\n", + "I = h / (8 * pi ** 2 * B * c)\n", + "mu = m1 * m2 / ((m1 + m2) * Na)\n", + "mu *= 10 ** -3\n", + "r = math.sqrt(I / mu)\n", + "\n", + "print \"The intermolecular distance of HCl is\", \"{:.3e}\".format(r), \"m\"\n", + "# Discrepency in value is due to error in calculation in the textbook\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The intermolecular distance of HCl is 1.294e-10 m\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 8, Page no: 68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constant\n", + "e = 8000 # Molar absorbtion coeff, dm^3 / mol / cm\n", + "\n", + "# Variable\n", + "l = 2.5 # Thickness of solution, cm\n", + "\n", + "# Solution\n", + "C = math.log10(1 / 0.3) / (e * l)\n", + "\n", + "print \"The concentration of Solution from Lambert-Beer's Law is\",\n", + "print \"{:.2e}\".format(C), \"mol/dm^3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The concentration of Solution from Lambert-Beer's Law is 2.61e-05 mol/dm^3\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 9, Page no: 68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"a) In the visible-UV spectra, CH2 = CHOCHOCH3 exhibits\"\n", + "print \"a higher value of lambda(max) because it has two conjugated\"\n", + "print \"chromophores, that is, one double bond (C=C) and a carbonyl\"\n", + "print \"group.\"\n", + "\n", + "print\n", + "print \"b) Because of the symmetrical vibrations of C=C double bond and\"\n", + "print \"triple bond, ethylene and acetylene do not absorb IR energy.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) In the visible-UV spectra, CH2 = CHOCHOCH3 exhibits\n", + "a higher value of lambda(max) because it has two conjugated\n", + "chromophores, that is, one double bond (C=C) and a carbonyl\n", + "group.\n", + "\n", + "b) Because of the symmetrical vibrations of C=C double bond and\n", + "triple bond, ethylene and acetylene do not absorb IR energy.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 10, Page no: 68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"Because CO2 is a linear molecule.\"\n", + "v_deg = 3 * 3 - 5\n", + "print \"The vibrational degree of freedom is\", v_deg" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Because CO2 is a linear molecule.\n", + "The vibrational degree of freedom is 4\n" + ] + } + ], + "prompt_number": 14 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_3.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_3.ipynb new file mode 100644 index 00000000..16fb19f8 --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_3.ipynb @@ -0,0 +1,1116 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3: Thermodynamic and Chemical Equilibrium" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page no: 105" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "q = 120 # Heat from surrounding, cal\n", + "W = 70 # Work done, cal\n", + "\n", + "# Solution\n", + "delta_E = q - W\n", + "\n", + "print \"Change in internal Energy\", delta_E, \"cals.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in internal Energy 50 cals.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page no: 105" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"CH4 (g) + 2O2 (g) -> CO2 (g) + 2H20 (l)\"\n", + "delta_n = 1 - (1 + 2)\n", + "solution = - 2 * 2 * 298 # cals\n", + "print \"Delta H - Delta E is:\", solution, \"cals\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "CH4 (g) + 2O2 (g) -> CO2 (g) + 2H20 (l)\n", + "Delta H - Delta E is: -1192 cals\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page no: 105" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "delta_G = -16.0 # Kelvin cal\n", + "delta_H = -10.0 # Kelvin cal\n", + "T = 300 # Kelvin\n", + "\n", + "# Solution\n", + "delta_S = (delta_H - delta_G) * 10 ** 3 / T # cal/deg\n", + "new_T = 330 # Kelvin\n", + "new_delta_G = (delta_H * 10 ** 3) - new_T * delta_S\n", + "\n", + "print \"The free energy at 330K is:\", \"{:.2e}\".format(new_delta_G), \"K cal\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The free energy at 330K is: -1.66e+04 K cal\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 4, Page no: 106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "delta_S = -20.7 # cal /deg /mol\n", + "delta_H = -67.37 # K cal\n", + "T = 25 # deg C\n", + "\n", + "# Solution\n", + "T += 273 # K\n", + "delta_G = delta_H - (T * delta_S * 10 ** -3)\n", + "print \"The change in free energy at 25deg C is:\", delta_G, \"K cal / mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The change in free energy at 25deg C is: -61.2014 K cal / mol\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 5, Page no: 106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt = 1 # g\n", + "delta_H = 149 # joules\n", + "\n", + "# Solution\n", + "delta_H_fusion = delta_H * (10 * 12 + 8 * 1)\n", + "print \"Enthalpy of fusion of naphthalene:\", delta_H_fusion * 10 ** -3, \"kJ/mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Enthalpy of fusion of naphthalene: 19.072 kJ/mol\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 6, Page no: 106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "delta_H_acetylene = 230 # kJ/mol\n", + "delta_H_benzene = 85 # kJ/mol\n", + "T = 298 # K\n", + "\n", + "# Solution\n", + "delta_H = delta_H_benzene - 3 * delta_H_acetylene\n", + "print \"The enthalpy change for the reaction is:\", delta_H, \"kJ/mole\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The enthalpy change for the reaction is: -605 kJ/mole\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 7, Page no: 106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "delta_H_vap = 2.0723 # kJ/g\n", + "Tb = 373 # K\n", + "\n", + "# Solution\n", + "delta_H_vap *= 18 # kJ/mol\n", + "delta_S = delta_H_vap / Tb\n", + "delta_G = delta_H_vap - Tb * delta_S\n", + "delta_S *= 1000\n", + "\n", + "print \"The Entropy change is:\", \"{:.1f}\".format(delta_S), \"J/mol/K\"\n", + "print \"The Free Energy change is:\", delta_G, \"kJ/mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Entropy change is: 100.0 J/mol/K\n", + "The Free Energy change is: 0.0 kJ/mol\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 8, Page no: 106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constant\n", + "R = 1.987 # cal /K /mol\n", + "\n", + "# Variables\n", + "moles = 5\n", + "Vo = 4 # litres, Initial Volume\n", + "Vf = 40 # litres, Final Volume\n", + "T = 27 # deg C\n", + "\n", + "# Solution\n", + "print \"dS = nRln(V2 / V1)\"\n", + "dS = moles * R * 2.303 * math.log10(Vf / Vo)\n", + "print \"The change in entropy is:\", \"{:.2f}\".format(dS), \"cal / degree\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "dS = nRln(V2 / V1)\n", + "The change in entropy is: 22.88 cal / degree\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 9, Page no: 107" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt = 10 # g\n", + "heat_abs = 4.5 # K\n", + "\n", + "# Solution\n", + "mole = 10 / 100.0 # mol\n", + "delta_H = heat_abs / mole\n", + "print \"The heat of the reaction is:\", delta_H, \"K cal / mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The heat of the reaction is: 45.0 K cal / mol\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 10, Page no: 107" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constant\n", + "R = 8.314 # J / K\n", + "\n", + "# Variables\n", + "V_O2 = 2.8 # litres\n", + "V_H2 = 19.6 # litres\n", + "\n", + "# Solution\n", + "na = V_O2 / 22.4 # mol\n", + "nb = V_H2 / 22.4 # mol\n", + "Xa = na / (na + nb)\n", + "Xb = nb / (na + nb)\n", + "delta_S = (- R) * (na * math.log(Xa) + nb * math.log(Xb))\n", + "\n", + "print \"The increase in entropy on mixing is:\", \"{:.3f}\".format(delta_S), \"J / K\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The increase in entropy on mixing is: 3.132 J / K\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 11, Page no: 107" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "print \"For 1 mole of ideal gas,\"\n", + "print \"\\tPV = RT or T = (PV) / R\\n\"\n", + "print \"Differentiating with respect to V at constant P,\"\n", + "print \"\\t[dT/dV]p = P/R\\n\"\n", + "print \"Differentiating again with respect oto P at constant V\"\n", + "print \"\\t[d2T/(dV*dP)] = 1/R\\n\"\n", + "print \"Now differectiating with respect to P at constant V,\"\n", + "print \"\\t[dT/dP]v = V/R\\n\"\n", + "print \"Differentiating again with respect to V at constant P,\"\n", + "print \"\\t[d2T/(dV*dP)] = 1/R\\n\"\n", + "print \"From equations we get:\"\n", + "print \"\\t[d2T/(dV*dP)] = [d2T/(dV*dP)]\\n\"\n", + "print \"Hence, dT is a perfect differential.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For 1 mole of ideal gas,\n", + "\tPV = RT or T = (PV) / R\n", + "\n", + "Differentiating with respect to V at constant P,\n", + "\t[dT/dV]p = P/R\n", + "\n", + "Differentiating again with respect oto P at constant V\n", + "\t[d2T/(dV*dP)] = 1/R\n", + "\n", + "Now differectiating with respect to P at constant V,\n", + "\t[dT/dP]v = V/R\n", + "\n", + "Differentiating again with respect to V at constant P,\n", + "\t[d2T/(dV*dP)] = 1/R\n", + "\n", + "From equations we get:\n", + "\t[d2T/(dV*dP)] = [d2T/(dV*dP)]\n", + "\n", + "Hence, dT is a perfect differential.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 12, Page no: 108" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "delta_G_25 = - 85.77 # k J, Free Energy at 25 C\n", + "delta_G_35 = - 83.68 # k J, Free Energy at 35 C\n", + "Ti = 273 + 25 # K\n", + "Tf = 273 + 35 # K\n", + "\n", + "# Solution\n", + "print \"Equating the entropy change at both the temperatures.\"\n", + "print \"(delta_H + delta_G_25) / Ti = (delta_H + delta_G_35) / Tf\"\n", + "delta_H = - 148\n", + "print \"The change in enthalpy for the process at 30C is\", delta_H, \"kJ\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equating the entropy change at both the temperatures.\n", + "(delta_H + delta_G_25) / Ti = (delta_H + delta_G_35) / Tf\n", + "The change in enthalpy for the process at 30C is -148 kJ\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 13, Page no: 108" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "Lv = 101 # cal /g, Latent headt of vap.\n", + "mwt = 78 # molecular weight of benzene\n", + "\n", + "# Variable\n", + "moles = 2\n", + "Tb = 80.2 # C, boiling point of benzene\n", + "\n", + "# Solution\n", + "Tb += 273 # K\n", + "delta_H = Lv * mwt\n", + "delta_S = delta_H / Tb\n", + "delta_G = delta_H - Tb * delta_S\n", + "print \"delta_S =\", \"{:.2f}\".format(delta_S), \"cal / K\"\n", + "print \"delta_G = delta_A =\", delta_G\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "delta_S = 22.30 cal / K\n", + "delta_G = delta_A = 0.0\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 14, Page no: 109" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "V1 = 6 # dm^3\n", + "V2 = 2 # dm^3\n", + "T1 = 27 # C\n", + "moles = 5\n", + "\n", + "# Solution\n", + "print \"T1*V1 ^ (gamma - 1) = T2 * V2 ^ (gamma - 1)\"\n", + "T1 += 273 # K\n", + "T2 = T1 * (V1 / V2) ** (8.314 / 20.91)\n", + "print \"The final temperature is\", \"{:.1f}\".format(T2), \"K\"\n", + "q = 0 # Adiabatic process\n", + "delta_E = - moles * 20.91 * (T2 - T1)\n", + "delta_E /= 1000\n", + "print \"q = \", q\n", + "print \"Change is Energy is\", \"{:.2f}\".format(delta_E), \"kJ / mol\"\n", + "W = - delta_E\n", + "print \"W = \", \"{:.2f}\".format(delta_E), \"kJ / mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T1*V1 ^ (gamma - 1) = T2 * V2 ^ (gamma - 1)\n", + "The final temperature is 464.3 K\n", + "q = 0\n", + "Change is Energy is -17.18 kJ / mol\n", + "W = -17.18 kJ / mol\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 15, Page no: 109" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constant\n", + "R = 8.314 # J / K mol\n", + "\n", + "# Variables\n", + "mole = 1\n", + "V1 = 5 # dm^3\n", + "V2 = 10 # dm^3\n", + "T = 300 # K\n", + "\n", + "# Solution\n", + "print \"For isothermal and reversible process,\"\n", + "delta_E = delta_H = 0\n", + "delta_A = delta_G = - 2.303 * mole * R * T * math.log10(V2 / V1)\n", + "q = W = - delta_G\n", + "print \"delta_E = delta_H =\", delta_H\n", + "print \"delta_G = delta_A =\", \"{:.3f}\".format(delta_G), \"J / mol\\n\"\n", + "print \"For isothermal and reversible expansion\"\n", + "print \"q = W = -delta_G =\", \"{:.3f}\".format(W), \"J / mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For isothermal and reversible process,\n", + "delta_E = delta_H = 0\n", + "delta_G = delta_A = -1729.159 J / mol\n", + "\n", + "For isothermal and reversible expansion\n", + "q = W = -delta_G = 1729.159 J / mol\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 16, Page no: 109" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constant\n", + "R = 8.314 # J / K mol\n", + "\n", + "# Variables\n", + "n = 5 # moles\n", + "T = 27 # C\n", + "V1 = 50.0 # L, Initial Volume\n", + "V2 = 1000 # L, Final Volume\n", + "\n", + "# Solution\n", + "T += 273\n", + "delta_G = 2.303 * n * R * T * math.log10(V1 / V2)\n", + "delta_G /= 1000\n", + "print \"The free energy change is\", \"{:.3f}\".format(delta_G), \"k J\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The free energy change is -37.367 k J\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 17, Page no: 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "delta_H_neu = - 51.46 # k J/mol, neutralization\n", + "delta_H_ion = - 57.1 # k J/mol, ionization\n", + "\n", + "# Solution\n", + "delta_H = - delta_H_ion + delta_H_neu\n", + "print \"The head of ionization for NH4OH is\", delta_H, \"kJ / mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The head of ionization for NH4OH is 5.64 kJ / mol\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 18, Page no: 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"For 1 mole of an ideal gas,\"\n", + "print \"PV = RT or V = (RT)/P\"\n", + "print \"(dV/ dP) = -(RT/P^2)\\t (at constant temperature)\"\n", + "print \"(d^2V/ (dP*dT)) = -(R/ P^2)\"\n", + "print \"(dV/ dT) = (R/ P)\\t (at constant pressure)\"\n", + "print \"(d^2V/ (dT*dP)) = -(R/ P^2)\\n\"\n", + "print \"(d^2V/ (dT*dP)) = (d^2V/ (dP*dT))\\t[From above equations]\"\n", + "print \"Hence, dV is an exact differential.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For 1 mole of an ideal gas,\n", + "PV = RT or V = (RT)/P\n", + "(dV/ dP) = -(RT/P^2)\t (at constant temperature)\n", + "(d^2V/ (dP*dT)) = -(R/ P^2)\n", + "(dV/ dT) = (R/ P)\t (at constant pressure)\n", + "(d^2V/ (dT*dP)) = -(R/ P^2)\n", + "\n", + "(d^2V/ (dT*dP)) = (d^2V/ (dP*dT))\t[From above equations]\n", + "Hence, dV is an exact differential.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 19, Page no: 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"Let P1, V1, T1 and P2, V2, T2 b the initial and final\",\n", + "print \"state, respectively of the system\"\n", + "print \"W rev = nRT[P1/ P2 - 1]\"\n", + "print \"W irr = nRT[1 - P2/ P1]\"\n", + "print \"W rev - W irr = [nRT/(P1P2)]*(P1 - P2)^2\"\n", + "print \"Because RHS of the above equation is always positive,\"\n", + "print \"W rev > W irr\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Let P1, V1, T1 and P2, V2, T2 b the initial and final state, respectively of the system\n", + "W rev = nRT[P1/ P2 - 1]\n", + "W irr = nRT[1 - P2/ P1]\n", + "W rev - W irr = [nRT/(P1P2)]*(P1 - P2)^2\n", + "Because RHS of the above equation is always positive,\n", + "W rev > W irr\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 20, Page no: 111" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "Eq_HI = 1.56 / 2\n", + "Eq_H2 = 0.22 / 2\n", + "Eq_I2 = 0.22 / 2\n", + "Kc = Eq_H2 * Eq_I2 / (Eq_HI ** 2)\n", + "print \"The equilibrium constant for the dissociation reaction\",\n", + "print \"{:.4f}\".format(Kc)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The equilibrium constant for the dissociation reaction 0.0199\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 21, Page no: 111" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Kc = 0.5 # / mole^2 litre^2\n", + "T = 400 # K\n", + "R = 0.082 # litre atm degree^-1 mole^-1\n", + "\n", + "# Solution\n", + "Kp = Kc * (R * T) ** (-2)\n", + "\n", + "print \"The given equilibrium is\"\n", + "print \"\\t\\tN2(g) + 3H2(g) <--> 2NH3(g)\"\n", + "print \"Kp is\", \"{:.3e}\".format(Kp)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The given equilibrium is\n", + "\t\tN2(g) + 3H2(g) <--> 2NH3(g)\n", + "Kp is 4.648e-04\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 22, Page no: 111" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "solubility = 7.5 * 10 ** - 5 # mol L^-1\n", + "\n", + "# Solution\n", + "Ksp = 4 * (solubility ** 3)\n", + "print \"Solubility product of the salt is\", \"{:.4e}\".format(Ksp), \"mol^3 / L^-3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Solubility product of the salt is 1.6875e-12 mol^3 / L^-3\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 23, Page no: 111" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ti = 25 # C\n", + "S = 0.00179 # g / L\n", + "\n", + "# Solution\n", + "S /= 170 # mol / L\n", + "Ksp = S ** 2\n", + "print \"Solubility product at 25 C is\", \"{:.4e}\".format(Ksp), \"mol^2 / L^-2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Solubility product at 25 C is 1.1087e-10 mol^2 / L^-2\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 24, Page no: 111" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ksp = 8 * 10 ** - 5 # Solubility product PbBr2\n", + "disso = 80 / 100 # % dissociation\n", + "\n", + "# Solution\n", + "S = (Ksp / 4) ** (1 / 3.0) # Solubility is 100%\n", + "S_80 = S * (80 / 100.0)\n", + "S_per_g = S_80 * 367 - 1.621\n", + "print \"Solubility in gm per litre is\", \"{:.3f}\".format(S_per_g), \"gm / litre\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Solubility in gm per litre is 6.349 gm / litre\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 25, Page no: 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "n_acid = 1 # mole\n", + "n_alcohol = 1 # mole\n", + "T = 25 # C\n", + "x = 0.667 # mole\n", + "\n", + "# Solution\n", + "print \"Kc = [CH3COOC2H][H2O] / ([CH3COOH][C2H2OH])\"\n", + "Kc = 4\n", + "print \"[CH3COOH] = (2 - x) / V\"\n", + "print \"[C2H5OH] = (1 - x) / V\"\n", + "print \"[CH3COOC2H5] = [H20] = x / V\"\n", + "print \"3x^2 - 12x + 8 = 0\"\n", + "print \"x =\", 2.366, \"or\", 0.634\n", + "print \"0.634 mole of ester would be formed, because the other value,\",\n", + "print \"x = 2.366, is not permissible.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Kc = [CH3COOC2H][H2O] / ([CH3COOH][C2H2OH])\n", + "[CH3COOH] = (2 - x) / V\n", + "[C2H5OH] = (1 - x) / V\n", + "[CH3COOC2H5] = [H20] = x / V\n", + "3x^2 - 12x + 8 = 0\n", + "x = 2.366 or 0.634\n", + "0.634 mole of ester would be formed, because the other value, x = 2.366, is not permissible.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 26, Page no: 113" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "n_acid = 0.2 # mole\n", + "n_salt = 0.10 # mole\n", + "Ka = 1.8 * 10 ** -5\n", + "\n", + "# Solution\n", + "pH = - math.log10(Ka) + math.log10(n_salt / n_acid)\n", + "print \"The pH of acidic buffer is\", \"{:.3f}\".format(pH)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pH of acidic buffer is 4.444\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 27, Page no: 113" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "n_salt = 0.02 # mole\n", + "n_base = 0.2 # mole\n", + "pKb = 4.7\n", + "\n", + "# Solution\n", + "pOH = pKb + math.log10(n_salt / n_base)\n", + "pH = 14 - pOH\n", + "print \"pH of a buffer solution is\", pH\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "pH of a buffer solution is 10.3\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 28, Page no: 113" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "n_H2 = 8.07 # moles\n", + "n_I2 = 9.08 # moles\n", + "T = 448 # C\n", + "n_eqHI2 = 13.38 # moles\n", + "\n", + "# Solution\n", + "x = n_eqHI2 / 2 + 6.69\n", + "Kc = n_eqHI2 ** 2 / (n_H2 * n_I2)\n", + "\n", + "print \"H2 + I2 <--> 2HI\"\n", + "print \"1 0 0\"\n", + "print \"1 - 2xx x x\"\n", + "print \"x/(1 - 2x) = (1/Kc)^0.5\"\n", + "print \"Dissociation constant of HI is 106.75 x 10^-3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "H2 + I2 <--> 2HI\n", + "1 0 0\n", + "1 - 2xx x x\n", + "x/(1 - 2x) = (1/Kc)^0.5\n", + "Dissociation constant of HI is 106.75 x 10^-3\n" + ] + } + ], + "prompt_number": 9 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_5.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_5.ipynb new file mode 100644 index 00000000..b0e0ac54 --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_5.ipynb @@ -0,0 +1,491 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5: Chemical Kinetics and Catalysis" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page no: 149" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "K = 3.5 * 10 ** - 2 # Rate constant\n", + "\n", + "# Solution\n", + "print \"First order reaction = 0.693 / K\"\n", + "t_05 = 0.693 / K\n", + "print \"Time taken for half the initial concentration to react\", t_05, \"minutes\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "First order reaction = 0.693 / K\n", + "Time taken for half the initial concentration to react 19.8 minutes\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page no: 150" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "t05 = 40 # minutes\n", + "\n", + "# Solution\n", + "print \"Rate constant = 0.693 / t05\"\n", + "K = 0.693 / t05\n", + "print \"Rate constant\", \"{:.4f}\".format(K), \"/ min\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate constant = 0.693 / t05\n", + "Rate constant 0.0173 / min\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page no: 150" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Variables\n", + "t0 = 37.0 # cm^3 of KMnO4\n", + "t5 = 29.8 # cm^3 of KMnO4\n", + "t15 = 19.6 # cm^3 of KMnO4\n", + "t25 = 12.3 # cm^3 of KMnO4\n", + "t45 = 5.00 # cm^3 of KMnO4\n", + "\n", + "# Solution\n", + "K5 = 2.303 / 5 * log10(t0 / t5)\n", + "K15 = 2.303 / 15 * log10(t0 / t15)\n", + "K25 = 2.303 / 25 * log10(t0 / t25)\n", + "K45 = 2.303 / 45 * log10(t0 / t45)\n", + "\n", + "print \"At t = 5 min, K =\", \"{:.3e}\".format(K5), \"/min\"\n", + "print \"At t = 15 min, K =\", \"{:.3e}\".format(K15), \"/min\"\n", + "print \"At t = 25 min, K =\", \"{:.3e}\".format(K25), \"/min\"\n", + "print \"At t = 45 min, K =\", \"{:.3e}\".format(K45), \"/min\"\n", + "print \"As the different values of K are nearly same, so the reaction\",\n", + "print \"is of first-order\"\n", + "K = (K45 + K25 + K5 + K15) / 4\n", + "print \"The average value of K = \", \"{:.3e}\".format(K), \"/min\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At t = 5 min, K = 4.329e-02 /min\n", + "At t = 15 min, K = 4.237e-02 /min\n", + "At t = 25 min, K = 4.406e-02 /min\n", + "At t = 45 min, K = 4.449e-02 /min\n", + "As the different values of K are nearly same, so the reaction is of first-order\n", + "The average value of K = 4.355e-02 /min\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem : 4, Page no: 150" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "t = 60 # min\n", + "x = \"0.5a\"\n", + "K = 5.2 * 10 ** - 3 # / mol L / min\n", + "\n", + "# Solution\n", + "a = 1 / (t * K)\n", + "print \"Initial concentration\", \"{:.3f}\".format(a), \"mol / L\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Initial concentration 3.205 mol / L\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 5, Page no: 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Solution\n", + "print \"99.9 % / 50 % =\",\n", + "times = round((2.303 * log10(100 / (100 - 99.9))) / (2.303 * log10(100 / (100 - 50))))\n", + "print times\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "99.9 % / 50 % = 10.0\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 6, Page no: 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Constants\n", + "R = 1.987 # cal / K / mol\n", + "\n", + "# Variables\n", + "T1 = 273.0 # K\n", + "T2 = 303.0 # K\n", + "K1 = 2.45 * 10 ** -5\n", + "K2 = 162 * 10 ** -5\n", + "\n", + "# Solution\n", + "Ea = log10(K2 / K1) * R * 2.303 / ((T2 - T1) / (T1 * T2))\n", + "print \"The activation energy of the reaction is\", int(Ea), \"cal / mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The activation energy of the reaction is 22968 cal / mol\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 7, Page no: 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "t05 = 30 # minutes\n", + "a = 0.1 # M\n", + "\n", + "# Solution\n", + "print \"For second order reaction,\"\n", + "print \"t0.5 = 1 / Ka\"\n", + "K = 1 / (a * t05)\n", + "print \"The rate constant is\", \"{:.3f}\".format(K), \"mol / lit / min\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For second order reaction,\n", + "t0.5 = 1 / Ka\n", + "The rate constant is 0.333 mol / lit / min\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 8, Page no: 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "\n", + "# Variables\n", + "T = 500 # C\n", + "Pi = 350 # torr\n", + "r1 = 1.07 # torr / s\n", + "r2 = 0.76 # torr / s\n", + "\n", + "# Solution\n", + "print \"1.07 = k(0.95a)^n\"\n", + "print \"0.76 = k(0.80a)^n\"\n", + "n = log(r1 / r2) / log(0.95 / 0.80)\n", + "print \"Hence, order of reaction is\", round(n)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1.07 = k(0.95a)^n\n", + "0.76 = k(0.80a)^n\n", + "Hence, order of reaction is 2.0\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 9, Page no: 152" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "print \"Applying steady state approximation to the concentration of NOCl2,\",\n", + "print \"we get\"\n", + "print \"d[NOCl2] / dt = 0 = k1 [NO] [Cl2] - k-1 [NOCl2] - k2 [NO] [NOCl2]\"\n", + "print \"[NOCl2] = k1 [NO] [Cl2] / (k-1 + k2 [NO])\"\n", + "print \"Now, the overall rate of reaction,\"\n", + "print \"d[NOCl2] / dt = k2 [NO] [NOCl2]\"\n", + "print \"From the above equations we get,\"\n", + "print \"(k1 k2 / k-1) * [NO]^2 * [Cl2], if k2[NO] << k-1\"\n", + "print \"k [NO]^2[Cl2], where k = k1 * k2 / k-1\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Applying steady state approximation to the concentration of NOCl2, we get\n", + "d[NOCl2] / dt = 0 = k1 [NO] [Cl2] - k-1 [NOCl2] - k2 [NO] [NOCl2]\n", + "[NOCl2] = k1 [NO] [Cl2] / (k-1 + k2 [NO])\n", + "Now, the overall rate of reaction,\n", + "d[NOCl2] / dt = k2 [NO] [NOCl2]\n", + "From the above equations we get,\n", + "(k1 k2 / k-1) * [NO]^2 * [Cl2], if k2[NO] << k-1\n", + "k [NO]^2[Cl2], where k = k1 * k2 / k-1\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 10, Page no: 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Constant\n", + "R = 1.987 # cal / K / mol\n", + "\n", + "# Variables\n", + "K2_K1 = 4 # factor increase\n", + "T1 = 27 # C\n", + "T2 = 47 # C\n", + "\n", + "# Solution\n", + "T1 += 273.0\n", + "T2 += 273.0\n", + "Ea = log10(4) * 2.303 * R * (T1 * T2 / (T2 - T1))\n", + "print \"The activation energy for the reaction is\", \"{:.2e}\".format(Ea),\n", + "print \"cal /mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The activation energy for the reaction is 1.32e+04 cal /mol\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 11, Page no: 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Variables\n", + "a = 1 # mole\n", + "x = 3 / 4.0 # reaction completed\n", + "\n", + "# Solution\n", + "K = (2.303 / 6) * log10(1 / (1 - x))\n", + "print \"The rate constant is\", \"{:.3f}\".format(K), \"/min\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The rate constant is 0.231 /min\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 12, Page no: 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Solution\n", + "print \"Let the initial concentration be 100, when x = 25\",\n", + "print \" t = 30 minutes\"\n", + "a = 100\n", + "x = 25.0\n", + "t = 30 # minutes\n", + "K = 2.303 / t * log10(a / (a - x))\n", + "t05 = 0.683 / K\n", + "t = 2.303 / K * log10(a / x)\n", + "print \"K =\", \"{:.2e}\".format(K), \"/ min\"\n", + "print \"T0.5 =\", \"{:.2f}\".format(t05), \"min\"\n", + "print \"t =\", \"{:.1f}\".format(t), \"min\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Let the initial concentration be 100, when x = 25 t = 30 minutes\n", + "K = 9.59e-03 / min\n", + "T0.5 = 71.21 min\n", + "t = 144.6 min\n" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_6.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_6.ipynb new file mode 100644 index 00000000..b15856da --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_6.ipynb @@ -0,0 +1,620 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Chapter 6: Electrochemistry" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page no: 180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Variables\n", + "T = 25 # C\n", + "E = 0.296 # V\n", + "Cu = 0.015 # M\n", + "\n", + "# Solution\n", + "Eo = E - 0.0296 * log10(Cu)\n", + "print \"The standard electrode potential is\", \"{:.3f}\".format(Eo), \"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The standard electrode potential is 0.350 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page no: 180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Variables\n", + "T = 25 # C\n", + "Cu = 0.1 # M\n", + "Zn = 0.001 # M\n", + "Eo = 1.1\n", + "\n", + "# Solution\n", + "E = Eo + 0.0296 * log10(Cu / Zn)\n", + "print \"The emf of Daniell cell is\", \"{:.4f}\".format(E), \"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emf of Daniell cell is 1.1592 V\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page no: 180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Constant\n", + "R = 8.314 # J / K\n", + "F = 96500 # C / mol\n", + "\n", + "# Variables\n", + "Cu = 0.15 # M\n", + "Eo = 0.34 # V\n", + "T = 298 # K\n", + "n = 2 # moles\n", + "\n", + "# Solution\n", + "E = Eo + (2.303 * R * T) / (n * F) * log10(Cu)\n", + "print \"The single electrode potential for copper metal is\", \"{:.4f}\".format(E),\n", + "print \"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The single electrode potential for copper metal is 0.3156 V\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 4, Page no: 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variable\n", + "Eo_Cu = 0.3370 # Cu+2 -> Cu\n", + "Eo_Zn = - 0.7630 # Zn -> Zn +2\n", + "\n", + "# Solution\n", + "Eo_cell = Eo_Cu - Eo_Zn\n", + "print \"Daniel cell is, Zn | Zn +2 || Cu+2 | Cu\"\n", + "print \"Eo (cell) is\", Eo_cell, \"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Daniel cell is, Zn | Zn +2 || Cu+2 | Cu\n", + "Eo (cell) is 1.1 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 5, Page no: 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variable\n", + "Eo_Cu = 0.3370 # Cu+2 -> Cu\n", + "Eo_Cd = - 0.4003 # Cd -> Cd +2\n", + "\n", + "# Solution\n", + "Eo_cell = Eo_Cu - Eo_Cd\n", + "print \"Cell is, Cd | Cd +2 || Cu+2 | Cu\"\n", + "print \"Eo (cell) is\", Eo_cell, \"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Cell is, Cd | Cd +2 || Cu+2 | Cu\n", + "Eo (cell) is 0.7373 V\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 6, Page no: 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "F = 96500 # C / mol\n", + "\n", + "# Variables\n", + "n = 2\n", + "T = 25 # C\n", + "Eo_Ag = 0.80 # Ag+ / Ag\n", + "Eo_Ni = - 0.24 # Ni+2 / Ni\n", + "\n", + "# Solution\n", + "Eo_Cell = Eo_Ag - Eo_Ni\n", + "print \"Standard free energy change,\"\n", + "delta_Go = - n * F * Eo_Cell\n", + "print delta_Go, \"J / mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Standard free energy change,\n", + "-200720.0 J / mol\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 7, Page no: 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "print \"-------------------------\"\n", + "print \"Reduction half reaction:\",\n", + "print \"2Fe+3 + 2e- --> 2Fe+2\"\n", + "print \"Oxidation half reaction:\",\n", + "print \"Fe - 2e- --> Fe+2\"\n", + "print \"Overall cell reaction :\",\n", + "print \"2Fe+3 + Fe --> 3Fe+2\"\n", + "\n", + "print \"-------------------------\"\n", + "print \"Reduction half reaction:\",\n", + "print \"Hg+2 + 2e- --> Hg\"\n", + "print \"Oxidation half reaction:\",\n", + "print \"Zn - 2e- --> Zn+2\"\n", + "print \"Overall cell reaction :\",\n", + "print \"Hg+2 + Zn --> Hg + Zn+2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "-------------------------\n", + "Reduction half reaction: 2Fe+3 + 2e- --> 2Fe+2\n", + "Oxidation half reaction: Fe - 2e- --> Fe+2\n", + "Overall cell reaction : 2Fe+3 + Fe --> 3Fe+2\n", + "-------------------------\n", + "Reduction half reaction: Hg+2 + 2e- --> Hg\n", + "Oxidation half reaction: Zn - 2e- --> Zn+2\n", + "Overall cell reaction : Hg+2 + Zn --> Hg + Zn+2\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 8, Page no: 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "F = 96500 # C / mol\n", + "\n", + "# Variables\n", + "E1o = - 2.48 # V\n", + "E2o = 1.61 # V\n", + "\n", + "# Solution\n", + "delta_G1 = - 3 * F * (- 2.48)\n", + "delta_G2 = - 1 * F * 1.61\n", + "print \"delta_G3 = delta_G1 + delta_G2\"\n", + "print \"delta_G3 = - 4 * F * E3o\"\n", + "E3o = (delta_G1 + delta_G2) / (- 4 * F)\n", + "print \"The reduction potential for the half-cell Pt/Ce, Ce+4\",\n", + "print \"is\", \"{:.4f}\".format(E3o), \"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "delta_G3 = delta_G1 + delta_G2\n", + "delta_G3 = - 4 * F * E3o\n", + "The reduction potential for the half-cell Pt/Ce, Ce+4 is -1.4575 V\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 9, Page no: 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "print \"Anodic half reaction :\",\n", + "print \" Cd --> Cd+2 + 2e-\"\n", + "print \"Cathodic half reaction :\",\n", + "print \"2H+ + 2e- --> H2\"\n", + "print \"-\" * 50\n", + "print \"Overall cell reaction :\",\n", + "print \"Cd + 2H+ <--> Cd+2 + H2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Anodic half reaction : Cd --> Cd+2 + 2e-\n", + "Cathodic half reaction : 2H+ + 2e- --> H2\n", + "--------------------------------------------------\n", + "Overall cell reaction : Cd + 2H+ <--> Cd+2 + H2\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 10, Page no: 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Variables\n", + "T = 25 # C\n", + "Cu = 0.1 # M\n", + "Zn = 0.001 # M\n", + "Eo = 1.1 # V\n", + "\n", + "# Solution\n", + "print \"Zn(s) | Zn+2 (0.001M) || Cu+2(0.1M) | Cu(s)\"\n", + "Ecell = Eo + 0.0592 / 2 * log10(Cu / Zn)\n", + "print \"The emf of a Daniell cell is\", \"{:.4f}\".format(Ecell), \"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Zn(s) | Zn+2 (0.001M) || Cu+2(0.1M) | Cu(s)\n", + "The emf of a Daniell cell is 1.1592 V\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 11, Page no: 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Variables\n", + "pH = 7 # O2\n", + "Eo = 1.229 # V\n", + "pO2 = 0.20 # bar\n", + "\n", + "# Solution\n", + "print \"Nearnst's equation at 25C is,\"\n", + "print \"E = Eo - (0.0592 / 2) * log(1 / ([H+]^2 * [pO2]^ (1/2)))\"\n", + "E = Eo - (0.0592 / 2) * log10(1.0 / (((10 ** (- 7)) ** 2) * (pO2 ** (1 / 2.0))))\n", + "print \"The reduction potential for the reduction is\",\n", + "print \"{:.3f}\".format(E), \"V\"\n", + "# descrepency due to calculation error in the book" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Nearnst's equation at 25C is,\n", + "E = Eo - (0.0592 / 2) * log(1 / ([H+]^2 * [pO2]^ (1/2)))\n", + "The reduction potential for the reduction is 0.804 V\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 12, Page no: 183" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "E_KCl = 0.2415 # V\n", + "E_cell = 0.445 # V\n", + "\n", + "\n", + "# Solution\n", + "print \"Emf of the cell is\"\n", + "print \"At 25C,\"\n", + "print \"E = Er - El = Eref - ((RT)/ F) * ln H+\"\n", + "pH = (E_cell - E_KCl) / 0.059\n", + "\n", + "Eo_cell = - 0.8277 # V\n", + "print \"Thus, equilibrium constant for the reaction\"\n", + "print \"\\t 2H2O --> H3O+ + OH- may be calculated as\"\n", + "K = 10 ** (Eo_cell / 0.0591)\n", + "print \"K =\", \"{:.0e}\".format(K)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Emf of the cell is\n", + "At 25C,\n", + "E = Er - El = Eref - ((RT)/ F) * ln H+\n", + "Thus, equilibrium constant for the reaction\n", + "\t 2H2O --> H3O+ + OH- may be calculated as\n", + "K = 1e-14\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 13, Page no: 183" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "EoSn = 0.15 # V\n", + "EoCr = - 0.74 # V\n", + "\n", + "# Solution\n", + "print \"3Sn+4 + 2Cr --> 3Sn+2 + 2Cr+3\"\n", + "Eo_cell = EoSn - EoCr\n", + "n = 6\n", + "K = 10 ** (n * Eo_cell / 0.0591)\n", + "print \"The equillibrium constant for th reaction is\", \"{:.2e}\".format(K)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "3Sn+4 + 2Cr --> 3Sn+2 + 2Cr+3\n", + "The equillibrium constant for th reaction is 2.27e+90\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 14, Page no: 184" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = 25 # C\n", + "Eo = - 0.8277 # V\n", + "\n", + "# Solution\n", + "print \"The reversible reaction,\"\n", + "print \"2H2O <--> H3O+ + OH-\"\n", + "print \"May be divided into two parts.\"\n", + "print \"2H2O + e- --> 1/2 H2 + OH- (cathode) Eo = -0.8277 V\"\n", + "print \"H2O + 1/2 H2 --> H3O+ + e- (anode) Eo = 0\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reversible reaction,\n", + "2H2O <--> H3O+ + OH-\n", + "May be divided into two parts.\n", + "2H2O + e- --> 1/2 H2 + OH- (cathode) Eo = -0.8277 V\n", + "H2O + 1/2 H2 --> H3O+ + e- (anode) Eo = 0\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 15, Page no: 184" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "E = 0.4 # V\n", + "\n", + "# Solution\n", + "print \"The cell is Pt(H2) | H+, pH2 = 1 atm\"\n", + "print \"The cell reaction is\"\n", + "print \"1/2 H2 --> H+ + e-\"\n", + "pH = E / 0.0591\n", + "print \"pH =\", \"{:.3f}\".format(pH)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The cell is Pt(H2) | H+, pH2 = 1 atm\n", + "The cell reaction is\n", + "1/2 H2 --> H+ + e-\n", + "pH = 6.768\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_7.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_7.ipynb new file mode 100644 index 00000000..229b5da3 --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_7.ipynb @@ -0,0 +1,431 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7: Solid State" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 1, Page no: 209" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "print \"Weiss indices \\t\\t\\t1/2, 2/3, infinity \\t2/3, 2, 1/3\"\n", + "print \"Reciprocal of Weiss indices 2, 3/2, 1/infinity \\t3/2, 1/2, 3\"\n", + "print \"Clear fractions\\t\\t\\t4, 3, 0 \\t3, 1, 6\"\n", + "print \"Miller indices \\t\\t\\t (430) \\t (316)\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Weiss indices \t\t\t1/2, 2/3, infinity \t2/3, 2, 1/3\n", + "Reciprocal of Weiss indices 2, 3/2, 1/infinity \t3/2, 1/2, 3\n", + "Clear fractions\t\t\t4, 3, 0 \t3, 1, 6\n", + "Miller indices \t\t\t (430) \t (316)\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 2, Page no: 209" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "\n", + "# Variable\n", + "a = 450 # pm\n", + "\n", + "# Solution\n", + "d220 = a / sqrt(2 ** 2 + 2 ** 2 + 0)\n", + "print \"Interplanar spacing\", int(d220)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Interplanar spacing 159\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 3, Page no: 209" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "print \"Intercept \\t(a, b ,c)\\t\\t(a, 2b, c) \\t\\t(a, 2b, 2c) \\t\\t(infi, b, -c)\"\n", + "print \"Weiss indices\\t 1, 1, 1 \\t\\t 1, 2, 1 \\t\\t 1, 2, 2 \\t\\t infi, 1, -1\"\n", + "print \"Reciprocals \\t 1, 1, 1 \\t\\t 1, 1/2, 1 \\t\\t 1, 1/2, 1/2\\t\\t 0, 1, -1\"\n", + "print \"Miller indicec\\t (111) \\t\\t (212) \\t\\t (211) \\t\\t (011)\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Intercept \t(a, b ,c)\t\t(a, 2b, c) \t\t(a, 2b, 2c) \t\t(infi, b, -c)\n", + "Weiss indices\t 1, 1, 1 \t\t 1, 2, 1 \t\t 1, 2, 2 \t\t infi, 1, -1\n", + "Reciprocals \t 1, 1, 1 \t\t 1, 1/2, 1 \t\t 1, 1/2, 1/2\t\t 0, 1, -1\n", + "Miller indicec\t (111) \t\t (212) \t\t (211) \t\t (011)\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 4, Page no: 209" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "rNa = 0.98 * 10 ** - 10 # m\n", + "rCl = 1.81 * 10 ** - 10 # m\n", + "\n", + "# Solution\n", + "rr = rNa / rCl\n", + "print \"When the radius ration is\", \"{:.2f}\".format(rr),\n", + "print \", the coordination number is 6.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "When the radius ration is 0.54 , the coordination number is 6.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 5, Page no: 210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "rLi = 68 # pm\n", + "rF = 136. # pm\n", + "\n", + "# Solution\n", + "rr = rLi / rF\n", + "print \"Radius ratio =\", rr\n", + "print \"The structure of LiF is scc and C.N of Li+ = 6\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radius ratio = 0.5\n", + "The structure of LiF is scc and C.N of Li+ = 6\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 6, Page no: 210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sin\n", + "\n", + "\n", + "# Variables\n", + "lamda = 2 * 10 ** - 10 # m\n", + "theta = 30 # degrees\n", + "\n", + "# Solution\n", + "print \"For first-order reflection\"\n", + "d = lamda / (2 * sin(theta))\n", + "dist = 2 * d\n", + "print \"Hence, distance between planes is\", \"{:.0e}\".format(abs(dist)), \"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For first-order reflection\n", + "Hence, distance between planes is 2e-10 m\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7, Page no: 210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "\n", + "# Variables\n", + "r = 174.6 # pm\n", + "\n", + "# Solution\n", + "a = r * sqrt(8)\n", + "print \"For 200 plane: h = 2, k = 0, l = 0\"\n", + "d200 = a / sqrt(2 ** 2)\n", + "print \"d200 =\", \"{:.1f}\".format(d200), \"pm\"\n", + "print \"For 200 plane: h = 2, k = 2, l = 0\"\n", + "d220 = a / sqrt(2 ** 2 + 2 ** 2)\n", + "print \"d220 =\", d220, \"pm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For 200 plane: h = 2, k = 0, l = 0\n", + "d200 = 246.9 pm\n", + "For 200 plane: h = 2, k = 2, l = 0\n", + "d220 = 174.6 pm\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 8, Page no: 210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "N = 6.023 * 10 ** 23\n", + "\n", + "# Variables\n", + "wt = 55.6\n", + "p = 0.29 # nm\n", + "n = 2\n", + "\n", + "# Solution\n", + "print \"For BCC pattern,\"\n", + "print \"number of atoms per unit cell = 2\"\n", + "d = n * (wt * 10 ** -3) / (N * (p * 10 ** - 9) ** 3)\n", + "print \"Density of the metal is\", \"{:.2e}\".format(d), \"kg / m^3\"\n", + "print \"Number of nearest enighbour for BCC = 8\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For BCC pattern,\n", + "number of atoms per unit cell = 2\n", + "Density of the metal is 7.57e+03 kg / m^3\n", + "Number of nearest enighbour for BCC = 8\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 9, Page no: 210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "print \"Intercept \\t2a/2, 2b, c/3\"\n", + "print \"Weiss indices\\t1, 2, 1/3\"\n", + "print \"Reciprocals \\t1, 1/2, 3\"\n", + "print \"Miller indices\\t (216)\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Intercept \t2a/2, 2b, c/3\n", + "Weiss indices\t1, 2, 1/3\n", + "Reciprocals \t1, 1/2, 3\n", + "Miller indices\t (216)\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 10, Page no: 211" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "N = 6.023 * 10 ** 23\n", + "\n", + "# Variables\n", + "D = 0.53 # g / cm ^ 3\n", + "MM = 6.94 # g / mol\n", + "n = 2\n", + "\n", + "# Solution\n", + "print \"For BCC pattern,\"\n", + "print \"number of atoms per unit cell = 2\"\n", + "V = D * N / (n * MM)\n", + "V = 1 / V\n", + "print \"Volume of a unit cell of lithium metal is\", \"{:.2e}\".format(V), \"cc\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For BCC pattern,\n", + "number of atoms per unit cell = 2\n", + "Volume of a unit cell of lithium metal is 4.35e-23 cc\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 11, Page no: 211" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "\n", + "print \"AB remain in BCC structure if the edge length is a then body diagonal\",\n", + "print \"is root(3)a\"\n", + "print \"root(3)a = 2(r+ + r-)\"\n", + "A = (sqrt(3) * 0.4123 - 2 * 0.81) / 2\n", + "print \"A+ =\", \"{:.2f}\".format(A), \"nm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "AB remain in BCC structure if the edge length is a then body diagonal is root(3)a\n", + "root(3)a = 2(r+ + r-)\n", + "A+ = -0.45 nm\n" + ] + } + ], + "prompt_number": 25 + } + ], + "metadata": {} + } + ] +}
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