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diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter10_2.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter10_2.ipynb new file mode 100644 index 00000000..8834e822 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter10_2.ipynb @@ -0,0 +1,398 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10: Gases and the Kinetic Theory" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex10.1:pg-288" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The pressure in the lungs is Pl= 109289.6 Pa\n" + ] + } + ], + "source": [ + " import math #Example 10_1\n", + " \n", + " #To find out the pressure in Lungs\n", + "h=6 #units in cm Hg\n", + "Pa=76 #Units in cm Hg\n", + "Pl=(h+Pa) #units in cm Hg\n", + "\n", + "Pl=Pl*10**-2 #units in Meters Hg\n", + "g=9.8 #Units in Meters/cm**2\n", + "H=13600 #Constant \n", + "Pl=Pl*H*g #Units in Pa\n", + "print \"The pressure in the lungs is Pl=\",round(Pl,1),\" Pa\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex10.2:pg-288" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Mass per atom is= 1.054e-25 Kg/Atom\n" + ] + } + ], + "source": [ + " import math #Example 10_2\n", + " \n", + " \n", + " #To find the mass of copper atom\n", + "maa=63.5 #Units in Kgs\n", + "n=6.022*10**26 #Units in number of atoms\n", + "Mass=maa/n #units in Kg/atom\n", + "print \"The Mass per atom is=\",round(Mass,28),\"Kg/Atom\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex10.3:pg-289" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The volume associated is 2.46e-29 Meter**3/Atom\n" + ] + } + ], + "source": [ + " import math #Example 10_3\n", + " \n", + " \n", + " #To find te volume associted with mercury atom in liquid mercury\n", + "M=201.0 #Units in Kg/Kmol\n", + "n=6.02*10**26 #units in K mol**-2\n", + "mo=M/n #units in Kg\n", + "n1=13600.0 #units in Kg/Meter**3\n", + "noatoms=n1/mo #units in atoms/Meter**3\n", + "volume_atom=1/noatoms #units in Meter**3/Atom\n", + "print \"The volume associated is \",round(volume_atom,31),\"Meter**3/Atom\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex10.4:pg-289" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volume occupied is V= 22.4 Meter**3/Kmol\n" + ] + } + ], + "source": [ + " import math #Example 10_4\n", + " \n", + " \n", + " #To find the volume that one kilomole of an ideal gas occupies\n", + "p=1.013*10**5 #units in Pa\n", + "t=273.15 #units in K\n", + "n=1 #units in K mol\n", + "R=8314 #units in J/Kmol K\n", + "v=(n*R*t)/p #units in Meter**3/Kmol\n", + "print \"Volume occupied is V=\",round(v,1),\" Meter**3/Kmol\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex10.5:pg-290" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The pressure in the container is P= 249.0 Pa\n" + ] + } + ], + "source": [ + " import math #Example 10_5\n", + " \n", + " \n", + " #To find the gas pressure in the container\n", + "v=5*10**-3 #units in meter**3\n", + "t=300.0 #units in K\n", + "m1=14*10**-6 #Units in Kg\n", + "M=28 #Units in Kg/Kmol \n", + "n=m1/M #units in K mol\n", + "R=8314 #units in J/Kmol K\n", + "p=(n*R*t)/v #units in Meter**3/Kmol\n", + "print \"The pressure in the container is P=\",round(p),\" Pa\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex10.6:pg-291" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The mass of air in flask is= 5.82e-05 Kg\n" + ] + } + ], + "source": [ + " import math #Example 10_6\n", + " \n", + " \n", + " #To determine the mass of the air in flask\n", + "p=1.013*10**5 #Units in Pa\n", + "v=50*10**-6 #Units in meter**3\n", + "M=28.0 #Units in Kg/Mol\n", + "R=8314.0 #units in J/Kmol K\n", + "T=293 #units in K\n", + "m=(p*v*M)/(R*T) #Units in Kg\n", + "print \"The mass of air in flask is=\",round(m,7),\"Kg\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex10.7:pg-292" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The final pressure in the drum is P2= 1.14 atm\n" + ] + } + ], + "source": [ + " import math #Example 10_7\n", + " \n", + " \n", + " #To find out the final pressure in the drum\n", + "p1=1 #Units in atm\n", + "t2=333.0 #units in K\n", + "t1=293.0 #units in K\n", + "p2=p1*(t2/t1) #units in atm\n", + "print \"The final pressure in the drum is P2=\",round(p2,2),\" atm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex10.8:pg-292" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The final volume of gas in terms of original volume is V2= 0.05467 *V1\n" + ] + } + ], + "source": [ + " import math #Example 10_8\n", + " \n", + " \n", + " #To find the final volume of gas\n", + "\n", + "t1=27.0 #Units in Centigrade\n", + "t1=t1+273 #Units in Kelvin\n", + "t2=547.0 #Units in Centigrade\n", + "t2=t2+273 #Units in Kelvin\n", + "t1=27.0 #Units in Centigrade\n", + "t1=t1+273 #Units in Kelvin\n", + "t1=27.0 #Units in Centigrade\n", + "t1=t1+273 #Units in Kelvin\n", + "p2=3700.0 #units in cm Hg\n", + "p1=74.0 #units in cm Hg\n", + "v1_v2=1/((t1/t2)*(p2/p1)) #In terms of V1\n", + "print \"The final volume of gas in terms of original volume is V2=\",round(v1_v2,5),\"*V1\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex10.9:pg-293" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Final pressure is P2= 328.0 K Pa\n" + ] + } + ], + "source": [ + " import math #Example 10_9\n", + " \n", + " \n", + " #To find the pressure after the car has been driven at high speed\n", + "t2=308 #Units in K\n", + "t1=273.0 #Units in K\n", + "p2_p1=(t2)/t1 #In terms of P1\n", + "P1=190.0 #Units in K Pa \n", + "P2=101 #Units in K Pa \n", + "P2=p2_p1*(P1+P2) #Units in K Pa \n", + "print \"The Final pressure is P2=\",round(P2),\" K Pa\" \n", + " #In text book the answer is printed wrong as P2=329 K Pa but the correct answer is 328 K Pa \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex10.10:pg-294" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The nitrogen molecule goes at a speed of V= 517.0 meter/sec\n" + ] + } + ], + "source": [ + " import math #Example 10_10\n", + "\n", + "\n", + " \n", + " #To findout how fast the nitrogen molecule moving in air\n", + "M=28.0 #Units in Kg/Mol\n", + "Na=6.02*10**26 #Units in K mol**-1\n", + "mo=M/Na #Units in Kg\n", + "k=1.38*10**-23 #units in J/K\n", + "T=27+273.0 #Units in K\n", + "v2=(3*k*T)/mo #unit in Meter**2/Sec**2\n", + "v=math.sqrt(v2) #Units in meter/sec\n", + "print \"The nitrogen molecule goes at a speed of V=\",round(v),\" meter/sec\"\n", + " #In text book the answer is printed wrong as v=517 m/sec the correct answer is v=516 meter/ sec\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |