summaryrefslogtreecommitdiff
path: root/Principles_of_Physics_by_F.J.Bueche/Chapter10_2.ipynb
diff options
context:
space:
mode:
Diffstat (limited to 'Principles_of_Physics_by_F.J.Bueche/Chapter10_2.ipynb')
-rw-r--r--Principles_of_Physics_by_F.J.Bueche/Chapter10_2.ipynb398
1 files changed, 398 insertions, 0 deletions
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
+}