path: root/sample_notebooks/GauravMittal/Ch3.ipynb

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   "source": [
    "# Chapter 3 - Properties of pure substances"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 3.1 Page: 88"
   ]
  },
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     "text": [
      "Example: 3.1 - Page: 88\n",
      "\n",
      "\n",
      "Molar Volume of the gas is 1.29e-02 cubic metres\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 3.1 - Page: 88\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****\n",
    "P = 2*10**5## [Pa]\n",
    "T = 273 + 37## [K]\n",
    "R = 8.314## [J/mol K]\n",
    "#****************\n",
    "# Since the gas behaves ideally:\n",
    "V = R*T/P## [cubic metres]\n",
    "print \"Molar Volume of the gas is %.2e cubic metres\"%(V)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 3.2 Page: 89"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
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   "outputs": [
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     "name": "stdout",
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     "text": [
      "Example: 3.2 - Page: 89\n",
      "\n",
      "\n",
      "The pressure of air after compression is 1200 kPa\n",
      "\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 3.2 - Page: 89\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****\n",
    "V1 = 8## [cubic m]\n",
    "P1 = 300## [kPa]\n",
    "V2 = 2## [cubic m]\n",
    "#**************\n",
    "# Apptying  the ideal gas Eqn. & since the Temperature remains constant:\n",
    "P2 = P1*V1/V2## [kPa]\n",
    "print \"The pressure of air after compression is %d kPa\\n\"%(P2)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 3.3 Page: 89"
   ]
  },
  {
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   "execution_count": 5,
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   "outputs": [
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     "text": [
      "Example: 3.3 - Page: 89\n",
      "\n",
      "\n",
      "The pressure of the remaining air is 0 kPa\n",
      "\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 3.3 - Page: 89\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****\n",
    "V1 = 6## [cubic m]\n",
    "P1 = 500## [kPa]\n",
    "R = 0.287## [kJ/kg K]\n",
    "#*************\n",
    "# Applying the charectarstic equation to the gas initially:\n",
    "# P1*V1 = m1*R*T1.......................................(i)\n",
    "# Applying the charectarstic equation to the gas which was left in the vessel after one-fifth of the gas has been removed:\n",
    "# P2*V2 = m2*R*T2.......................................(ii)\n",
    "# V2 = V1# T2 = T1# m2 = (4/5)*m1# Eqn (ii) becomes:\n",
    "# P2*V1 = (4/5)*m1*R*T1..................................(iii)\n",
    "# Dividing eqn (i) by eqn (iii), we get:\n",
    "P2 = (4/5)*P1## [kPa]\n",
    "print \"The pressure of the remaining air is %d kPa\\n\"%(P2)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 3.4 Page: 90"
   ]
  },
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   "execution_count": 6,
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   "outputs": [
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     "text": [
      "Example: 3.4 - Page: 90\n",
      "\n",
      "\n",
      "Final Temperature of the air when the piston reaches stop is 361.5 K\n",
      "\n",
      "Pressure of air inside the cylinder is 2.51 bar\n",
      "\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 3.4 - Page: 90\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****\n",
    "T1 = 450 + 273## [K]\n",
    "P1 = 3## [bar]\n",
    "#***************\n",
    "# Soluton(a)\n",
    "# From Fig. 3.7, (Page 90)\n",
    "# Since the weight remains the same, therefore, the final pressure is equal to the initial pressure.\n",
    "# Therefore it is a constant pressure process.\n",
    "P2 = P1## [bar]\n",
    "# Volumetric Ratio:\n",
    "V2_by_V1 = 2.5/(2.5 + 2.5)# Applying ideal gas law & P1 = P2\n",
    "T2 = T1*V2_by_V1## [K]\n",
    "print \"Final Temperature of the air when the piston reaches stop is %.1f K\\n\"%(T2)\n",
    "# Solution (b)\n",
    "# When the piston rests ot the stops, the pressure exerted by the weight, air & the atmosphere will be different. But there will beno further decrease in volume.\n",
    "# This is a constant volume process.\n",
    "T3 = 273 + 30## [K]\n",
    "# Applying ideal gas law & V2 = V3\n",
    "P3 = T3*P2/T2## [bar]\n",
    "print \"Pressure of air inside the cylinder is %.2f bar\\n\"%(P3)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 3.5 Page: 95"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
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   "outputs": [
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     "output_type": "stream",
     "text": [
      "Example: 3.5 - Page: 95\n",
      "\n",
      "\n",
      "The temperature at which ammonia exists in the cylinder is 321.5 K\n",
      "\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 3.5 - Page: 95\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****\n",
    "m = 1.373## [kg]\n",
    "P = 1.95*10**(6)## [Pa]\n",
    "V = 0.1## [cubic m]\n",
    "a = 422.546*10**(-3)## [cubic m/square mol]\n",
    "b = 37*10**(-6)## [cubic m/mol]\n",
    "M = 17*10**(-3)## [kg/mol]\n",
    "R = 8.314## [J/mol K]\n",
    "#****************\n",
    "n = m/M## [moles]\n",
    "Vm = V/n## [molar volume, cubic m]\n",
    "# Applying Van der Waals equation of state:\n",
    "T = (P + (a/Vm**2))*((Vm - b)/R)## [K]\n",
    "print \"The temperature at which ammonia exists in the cylinder is %.1f K\\n\"%(T)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 3.10 Page: 103"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
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   "outputs": [
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     "text": [
      "Example: 3.10 - Page: 103\n",
      "\n",
      "\n",
      "Value of derivative is 23.98 bar/OC\n",
      "\n",
      "The pressure generated by heating at constant Volume is 240.84 Pa\n",
      "\n",
      "The change in Volume is -0.038 cubic cm/g\n",
      "\n"
     ]
    }
   ],
   "source": [
    "from math import exp\n",
    "print \"Example: 3.10 - Page: 103\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****\n",
    "beeta = 1.487*10**(-3)## [1/OC]\n",
    "alpha = 62*10**(-6)## [1/bar]\n",
    "V1 = 1.287## [cubic cm /g]\n",
    "#************\n",
    "# Solution (a)\n",
    "# The value of derivative (dP/dT) at constant V:\n",
    "# dV/V = beeta*dT - alpha*dP\n",
    "# dV = 0\n",
    "# dP/dT = beeta/alpha\n",
    "# Value = dP/dT\n",
    "Value = beeta/alpha## [bar/OC]\n",
    "print \"Value of derivative is %.2f bar/OC\\n\"%(Value)\n",
    "# Solution (b)\n",
    "P1 = 1## [bar]\n",
    "T1 = 20## [OC]\n",
    "T2 = 30## [OC]\n",
    "# Applying the same equation:\n",
    "P2 = P1 +(beeta/alpha)*(T2 - T1)## [bar]\n",
    "print \"The pressure generated by heating at constant Volume is %.2f Pa\\n\"%(P2)\n",
    "# Solution (c)\n",
    "T2 = 0## [OC]\n",
    "T1 = 20## [OC]\n",
    "P2 = 10## [bar]\n",
    "P1 = 1## [bar]\n",
    "# The change in Volume can be obtained as:\n",
    "V2 = V1*exp((beeta*(T2 - T1)) - alpha*(P2 - P1))## [cubic cm/g]\n",
    "deltaV = V2 - V1## [cubic cm/g]\n",
    "print \"The change in Volume is %.3f cubic cm/g\\n\"%(deltaV)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 3.11 Page: 107"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example: 3.11 - Page: 107\n",
      "\n",
      "\n",
      "Acentric factor is 0.6447\n"
     ]
    }
   ],
   "source": [
    "from math import log10\n",
    "print \"Example: 3.11 - Page: 107\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****\n",
    "Tc = 513.9## [K]\n",
    "Pc = 61.48*10**5## [Pa]\n",
    "#************\n",
    "Tr = 0.7\n",
    "T = Tr*Tc - 273.15## [OC]\n",
    "P_sat = 10**(8.112 - (1592.864/(T + 226.184)))## [mm Hg]\n",
    "P_sat = P_sat*101325/760## [Pa]\n",
    "Pr_sat = P_sat/Pc## [Pa]\n",
    "omega = -1 - log10(Pr_sat)## [Acentric factor]\n",
    "print \"Acentric factor is %.4f\"%(omega)#"
   ]
  }
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