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   "source": [
    "# Chapter 22: Transport Processes in Gas"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex22.1:pg-911"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 22.1 \n",
      "\n",
      "\n",
      " Mean free path = math.exp m,\n",
      " The fraction of molecules have free path longer than 2*lambda =  13.5335283237  percent\n"
     ]
    }
   ],
   "source": [
    "# Given that\n",
    "p = 1.013e5 # Pressure in Pa\n",
    "t = 300 # Temperature in K\n",
    "d = 3.5 # Effective diameter of oxygen molecule in Angstrom \n",
    "r = 2 # Ratio of free path of molecules with the lambda\n",
    "print \"\\n Example 22.1 \\n\"\n",
    "sigma = math.pi*(d*(10**-10))**2\n",
    "n = p/(t*1.38*(10**-23))\n",
    "R = math.exp(-r)\n",
    "print \"\\n Mean free path = math.exp m,\\n The fraction of molecules have free path longer than 2*lambda = \",R*100, \" percent\"\n",
    "# Answer given in the book contain round off error for mean free path."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex22.2:pg-912"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 22.2 \n",
      "\n",
      "\n",
      " Pressure of the gas =  134.236067593  Pa,\n",
      " No of collisions made by a molecule per meter of path = math.exp 38022.8136882\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# Given that\n",
    "lambda1 = (2.63e-5) # Mean free path of the molecules of the gas in m\n",
    "t = 25 # Temperature in degree centigrade\n",
    "r = 2.56e-10 # Radius of the molecules in m\n",
    "print \"\\n Example 22.2 \\n\"\n",
    "sigma = 4*math.pi*r**2\n",
    "n = 0.707/(sigma*lambda1)\n",
    "p = n*(t+273)*(1.38*10**-23)\n",
    "N = 1.0/lambda1\n",
    "print \"\\n Pressure of the gas = \",p,\" Pa,\\n No of collisions made by a molecule per meter of path = math.exp\",N\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex22.3:pg-912"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 22.3 \n",
      "\n",
      "\n",
      " The no of free paths which are longer than, \n",
      " 10 cm =  3679.0 ,\n",
      " 20 cm =  1354.0 ,\n",
      " 50 cm =  68.0 ,\n",
      "\n",
      " The no of free paths which are between,\n",
      " 5 cm and 10 cm =  -2387.0 ,\n",
      " 9.5 cm and 10.5 cm =  -369.0 ,\n",
      " 9.9 cm and 10.1 cm =  -74.0 ,\n",
      "\n",
      " The no of free paths which are exactly 10 cm =  -0.0\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# Given that\n",
    "import math\n",
    "from scipy import integrate \n",
    "lambda1 = 10.0 # Mean free path of the gas in cm\n",
    "N0 = 10000.0 # No of free paths\n",
    "x1 = 10.0 # In cm\n",
    "x2 = 20.0 # In cm\n",
    "x3 = 50.0 # In cm\n",
    "x4 = 5.0 # In cm\n",
    "x5 = 9.5 # In cm\n",
    "x6 = 10.5 # In cm\n",
    "x7 = 9.9 # In cm\n",
    "x8 = 10.1 # In cm\n",
    "print \"\\n Example 22.3 \\n\"\n",
    "# For x>10 cm\n",
    "N1 = N0*(math.exp(-1))\n",
    "# For x>20 cm\n",
    "N2 = N0*(math.exp(-2))\n",
    "# For x>50 cm\n",
    "N3 = N0*(math.exp(-5))\n",
    "def f(x): \n",
    "    y = (-N0/lambda1)*(math.exp((-x)/lambda1)),\n",
    "    return y\n",
    "# For 5>x>10 cm\n",
    "N4,er = integrate.quad( lambda x: (-N0/lambda1)*(math.exp((-x)/lambda1)),x4,x1)\n",
    "# For 9.5>x>10.5 cm\n",
    "N5,e = integrate.quad( lambda x: (-N0/lambda1)*(math.exp((-x)/lambda1)),x5,x6)\n",
    "# For 9.9>x>10.1 cm\n",
    "N6,eor = integrate.quad( lambda x: (-N0/lambda1)*(math.exp((-x)/lambda1)),x7,x8)\n",
    "# For x=10 cm\n",
    "N7,eer = integrate.quad( lambda x: (-N0/lambda1)*(math.exp((-x)/lambda1)),x1,x1)\n",
    "print \"\\n The no of free paths which are longer than, \\n 10 cm = \",math. ceil(N1) ,\",\\n 20 cm = \",math. ceil(N2) ,\",\\n 50 cm = \",math. ceil(N3) ,\",\\n\\n The no of free paths which are between,\\n 5 cm and 10 cm = \",math.floor(N4) ,\",\\n 9.5 cm and 10.5 cm = \",math.floor(N5) ,\",\\n 9.9 cm and 10.1 cm = \",math.floor(N6) ,\",\\n\\n The no of free paths which are exactly 10 cm = \",N7 \n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex22.4:pg-913"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 22.4 \n",
      "\n",
      "\n",
      " Coefficient of viscosity = math.exp Ns/m**2 2.051171875e-05\n"
     ]
    }
   ],
   "source": [
    "# Given that\n",
    "p = 1.0 # Pressure in atm\n",
    "t = 300.0 # Temperature in K\n",
    "print \"\\n Example 22.4 \\n\"\n",
    "# From previous example, we have\n",
    "m = 5.31e-26 # In kg/molecule\n",
    "v = 445.0 # In m/s\n",
    "sigma = 3.84e-19 # In m**2\n",
    "# Therefore\n",
    "mu = (1.0/3.0)*(m*v/sigma)\n",
    "print \"\\n Coefficient of viscosity = math.exp Ns/m**2\",mu"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex22.5:pg-913"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 22.5 \n",
      "\n",
      "\n",
      " Thermal conductivity  =  0.0  W/mK,\n",
      " If the gas has Maxwellian velocity distribution,\n",
      " Thermal conductivity =  5.98958333333e-05  W/mK\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# Given that\n",
    "p = 1.0 # Pressure in atm\n",
    "t = 300.0 # Temperature in K\n",
    "F = 5.0 # For oxygen gas degree of freedom\n",
    "print \"\\n Example 22.5 \\n\"\n",
    "v = 445.0 # In m/s as given in the book\n",
    "m = 5.31e-26 # Mass of oxygen molecule in kg\n",
    "sigma = 3.84e-19 # As given in the book in m**2\n",
    "k = (1/6)*(v*F*(1.38*10**-23))/sigma\n",
    "# If the gas has Maxwellian velocity distribution,\n",
    "k_ = (1.0/3.0)*(F*(1.38*10**-23)/sigma)*((1.38*10**-23)*t/(math.pi*m))**(1/2)\n",
    "print \"\\n Thermal conductivity  = \",k ,\" W/mK,\\n If the gas has Maxwellian velocity distribution,\\n Thermal conductivity = \",k_ ,\" W/mK\"\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex22.6:pg-914"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 27,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 22.6 \n",
      "\n",
      "\n",
      " Pressure in the cathode ray tube =  0.142844028924  Pa\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "# Given that\n",
    "F = .90 # Fraction of electrons leaving the cathode ray reach the anode without making a collision\n",
    "x = 0.2 # Distance between cathode ray and anode in m\n",
    "d = 3.6e-10 # Diameter of ion in m\n",
    "t = 2000.0 # Temperature of electron in K\n",
    "print \"\\n Example 22.6 \\n\"\n",
    "lambda1 = x/(math.log(1/F))\n",
    "sigma = math.pi*(d**2)\n",
    "n = 4/(sigma*lambda1)\n",
    "p = n*(1.38*10**-23)*(t)\n",
    "print \"\\n Pressure in the cathode ray tube = \",p ,\" Pa\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex22.7:pg-914"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 22.7 \n",
      "\n",
      "\n",
      " No of collisions per sec are made by one molecule with the other molecule = 9962400.07749 \n",
      "The no of molecules strike the flask per sq. cm = 6.11714975845e+20 \n",
      " No of molecules in the flask = 2.44685990338e+22\n"
     ]
    }
   ],
   "source": [
    "# Given that\n",
    "V = 1.0 # Volume of the flask in litre\n",
    "p = 1.0 # Pressure in atm\n",
    "t = 300.0 # Temperature in K\n",
    "r = 1.8e-10 # Radius of oxygen gas molecule in m\n",
    "m = 5.31e-26 # Mass of oxygen molecule in kg\n",
    "print \"\\n Example 22.7 \\n\"\n",
    "n = (p*(1.013e5))/((1.38e-23)*(t)*1000)\n",
    "sigma = 4*math.pi*(r**2)\n",
    "v = ((8*(1.38e-23)*t)/(math.pi*m))**(1/2)\n",
    "z = sigma*n*v*1000\n",
    "N = (1.0/4.0)*(n*0.1*v)\n",
    "print \"\\n No of collisions per sec are made by one molecule with the other molecule =\", z,\"\\nThe no of molecules strike the flask per sq. cm =\",N,\"\\n No of molecules in the flask =\",n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex22.8:pg-915"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 22.8 \n",
      "\n",
      "\n",
      " Time = 1.00003111262 s\n"
     ]
    }
   ],
   "source": [
    "# Given that\n",
    "lambda1 = 2.0 # Mean free path in cm\n",
    "T = 300.0 # Temperature in K\n",
    "r = 0.5 # As half of the molecules did not make any collision\n",
    "print \"\\n Example 22.8 \\n\"\n",
    "x = lambda1*(math.log(1/r))\n",
    "v = 445.58 # For oxygen at 300K in m/s\n",
    "t = x/(v*100)\n",
    "print \"\\n Time =\", math.exp(t), \"s\"\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex22.9:pg-915"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 22.9 \n",
      "\n",
      "\n",
      " Pressure = 1.03636998072 N/m**2\n"
     ]
    }
   ],
   "source": [
    "\n",
    "# Given that\n",
    "f = 0.9 # Fraction of electrons leaving the cathode ray and reaching the anode without making any collision\n",
    "x = 20.0 # Distance between cathode ray tube and anode in cm\n",
    "sigma = 4.07e-19 # Collision cross section of molecules in m**2\n",
    "T = 2000 # Temperature in K\n",
    "print \"\\n Example 22.9 \\n\"\n",
    "lambda1 = (x*0.01)/(math.log(1.0/f))\n",
    "n = 1/(sigma*lambda1)\n",
    "p = n*(1.38e-23)*T\n",
    "print \"\\n Pressure =\", math.exp(p), \"N/m**2\"\n",
    "# The answer given in the book contains round off error.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex22.10:pg-916"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 22.10 \n",
      "\n",
      "\n",
      " Initial concentration gradient of reactive molecules = 0.0  molecules/m**4, \n",
      " The no of reactive molecules per sec cross a cross section at the mid point of the tube from left to right = 0.9 molecules/m**2,\n",
      " The no of reactive molecules per sec cross a cross section at the mid point of the tube from right to left = 4.08598425576e-12  molecule/m**2,\n",
      " Initial net rate of diffusion =  0.0112863158384 g/m**2-s\n"
     ]
    }
   ],
   "source": [
    "# Given that\n",
    "l = 2.0 # Length of tube in m\n",
    "a = 1e-4 # Cross section of the tube in m**2\n",
    "p = 1.0 # Pressure in atm\n",
    "t = 0 # Temperature in degree centigrade\n",
    "r = 0.5 # Fraction of the carbon atoms which are radioactive C14\n",
    "sigma = 4e-19 # Collision cross section area in m**2\n",
    "print \"\\n Example 22.10 \\n\"\n",
    "n = (p*1.01325e+5)/((1.38e-23)*(t+273))\n",
    "C_g = -n/l\n",
    "m = (46/6.023)*10**-26 # In kg/molecule\n",
    "v = (2.55*(1.38e-23)*(t+273)/m)**(1/2.0)\n",
    "lambda1 = (1.0/(sigma*n))\n",
    "gama = (1.0/4)*(v*n) - (1/6.0)*(v*lambda1*(C_g))\n",
    "gama_ = (1/4.0)*(v*n) + (1.0/6.0)*(v*lambda1*(C_g))\n",
    "x = (1.0/4)*(v*n)\n",
    "y = (1.0/6)*(v*lambda1*(C_g))\n",
    "d = (1.0/6)*(v*lambda1*(-1*C_g))*2*(m)\n",
    "a=x+y\n",
    "b=x-y\n",
    "print \"\\n Initial concentration gradient of reactive molecules =\",math.exp (C_g),\" molecules/m**4, \\n The no of reactive molecules per sec cross a cross section at the mid point of the tube from left to right =\",f , \"molecules/m**2,\\n The no of reactive molecules per sec cross a cross section at the mid point of the tube from right to left =\",e ,\" molecule/m**2,\\n Initial net rate of diffusion = \",d*1000 ,\"g/m**2-s\"\n",
    "# The answer for lambda given in the book conatains calculation error\n",
    "# The answers contains calculation error\n"
   ]
  }
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