{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 4 : Kinetic theory of gases"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.1 page no : 137\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"t = 273;\t\t\t #temperture of the oxygen molecule in K\n",
"m = 32;\t\t \t#molecular mass of the gas in gm\n",
"r = 8.32*10**7;\t\t\t#molar gas consmath.tant in ergs per mole\n",
"v2 = 33200.;\t\t\t#velocity of the gas in cm/sec\n",
"\n",
"# Calculations\n",
"v1 = ((3*r*t)/m)**(1./2);\t\t\t#rms velocity of the molecule in cm/s\n",
"T = ((v2*v2*m)/(3*r));\t\t\t#temperature of the molecule with sound has velocity in K\n",
"\n",
"# Result\n",
"print 'the rms velocity of the molecule is %.2e cm/s \\\n",
"\\nthe temperature of the molecule is %3.0f K'%(v1,T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the rms velocity of the molecule is 4.61e+04 cm/s \n",
"the temperature of the molecule is 141 K\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.2 page no : 137"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"t1 = 308.;\t\t\t#temperature of the nitrogen molecule in K\n",
"m1 = 28.;\t\t\t#molecular mass of the nitrogen in gm\n",
"m2 = 2.;\t\t\t#molecular mass of the hydrogen molecule in gm\n",
"\n",
"# Calculations\n",
"t2 = (t1*m2/m1);\t\t\t#temperature of the hydrogen molecule in K\n",
"\n",
"# Result\n",
"print 'the temperature of the hydrogen molecule is %3.0fK'%(t2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the temperature of the hydrogen molecule is 22K\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.3 pageno : 138"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"y = 0.00129;\t\t\t#density of the air in gm/cc\n",
"p = 76;\t\t\t#pressure of the nitrogen molecule in cm\n",
"g = 981;\t\t\t#accelaration due to gravity in cm/sec**2\n",
"m = 13.6;\t\t\t#density of the mercury in gm/cc\n",
"\n",
"# Calculations\n",
"v = ((3*p*g*m)/y)**(1./2);\t\t\t#rms velocity of air at ntp in cm/sec\n",
"\n",
"# Result\n",
"print 'the rms velocity of the air is %.2e cm/sec'%(v)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the rms velocity of the air is 4.86e+04 cm/sec\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.4 pageno : 138"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"d = 16*0.000089;\t\t\t#density of the oxygen molecule in gm/cc\n",
"p = 76;\t\t\t#pressure of the air in cm\n",
"g = 981;\t\t\t#gravitaitonal accelaration in cm/sec**2\n",
"m = 13.6;\t\t\t#density of the mercury in gm/cc\n",
"\n",
"# Calculations\n",
"v = ((3*p*g*m)/d)**(1./2);\t\t\t#velocuty of the oxygen molecule in cm/sec\n",
"\n",
"# Result\n",
"print 'velocity of oxygen molecule is %.2e cm/sec'%(v)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"velocity of oxygen molecule is 4.62e+04 cm/sec\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.5 pageno : 138"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"t = 273;\t\t\t#temperature of the hydrogen molecule in K\n",
"n = 6.03*10**23;\t\t\t#1 mole of hydrogen molecules\n",
"r = 8.31*10**7;\t\t\t#universal gas consmath.tant in erg/K/mole\n",
"\n",
"# Calculations\n",
"e = (1.5*r*t)/n;\t\t\t#kinetic energy of the hydrogen molecule in erg\n",
"\n",
"# Result\n",
"print \"the kinetic energy of the hydrogen molecule is %.2e erg\"%e\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the kinetic energy of the hydrogen molecule is 5.64e-14 erg\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.6 page no : 138"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"m = 1.;\t\t\t#mass of the oxygen in gm\n",
"r = 8.31*10**7;\t\t\t#universal gas consmath.tant in erg/K/mole\n",
"t = 320.;\t\t\t#temperature of the oxygen in K\n",
"\t\t\t#for 1gm mole k.e is 1.5rt then for 1 gm oxygen (1/32)(k.e)\n",
"\n",
"# Calculations\n",
"e = (m/32)*(3*r*t/2);\t\t\t#kinetic energy of the oxygen in erg\n",
"\n",
"# Result\n",
"print 'the kinetic energy of the oxygen is %.2e erg'%(e)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the kinetic energy of the oxygen is 1.25e+09 erg\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.7 pageno : 138"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"t = 273;\t\t\t#temperature at ntp in K\n",
"\t\t\t#rms velocity of oxygen is 3/2 times its rms velocity at ntp then e1 = (3/2)*e\n",
"\n",
"# Calculations\n",
"t1 = (9.*t/4.);\t\t\t#temperature of the oxygen molecule in K\n",
"\n",
"# Result\n",
"print 'temperature of the oxygen in %3.1f K'%(t1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"temperature of the oxygen in 614.2 K\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.8 page no : 139"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"p = 10;\t\t\t#pressure of the gas in atm\n",
"v = 5000;\t\t\t#volume of the gas in ml\n",
"l = 76;\t\t\t#length of the mercury in barometer in cm\n",
"g = 981;\t\t\t#accelaration due to gravity in cm/sec**2\n",
"d = 13.6;\t\t\t#density of the mercury in gm/cc\n",
"\n",
"# Calculations\n",
"e = 3*p*v*l*g*d;\t\t\t#kinetic energy of the molecule in ergs\n",
"\n",
"# Result\n",
"print 'the kinetic energy of the molecule is %.2e ergs'%(e)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the kinetic energy of the molecule is 1.52e+11 ergs\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.9 page no : 139"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"t = 323;\t\t\t#temperature of the hydrogen molecule in K\n",
"m1 = 1;\t\t\t#mass of the hydrogen molecule in gm\n",
"m2 = 2;\t\t\t#molecular weight of the hydrogen in gm\n",
"r = 8.3*10**7;\t\t\t#universal gas consmath.tant in erg/K/mole\n",
"\n",
"# Calculations\n",
"e = (m1*r*t*3/(m2*2));\t\t\t#kinetic enrgy of the hydrogen molecule in ergs\n",
"\n",
"# Result\n",
"print 'the kinetic energy of the molecule is %.0e ergs'%(e)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the kinetic energy of the molecule is 2e+10 ergs\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.10 page no : 139"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"t1 = 273;\t\t\t#temperature of the hydrogen molecule at n.t.p in K\n",
"\n",
"# Calculations\n",
"#rms value of hydrogen molecule is double to its rms value at n.t.p, so 3rt/m = 4(3rt/m)\n",
"t2 = 4*t1;\t\t\t#temperature of the hydrogen molecule in K\n",
"\n",
"# Result\n",
"print 'the temperature of the hydrogen molecule is %.f K'%(t2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the temperature of the hydrogen molecule is 1092 K\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.11 page no : 139"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"t1 = 273.;\t\t\t#temperature of the hydrogen molecule in K\n",
"t2 = 373;\t\t\t#temperature of the hydrogen molecule in K\n",
"d = 0.0000896;\t\t\t#density of the hydrogen molecule in gm/cc\n",
"p = 76*13.6*981;\t\t\t#pressure of the hydrogen molecule in gm/cm/sec**2\n",
"\n",
"# Calculations\n",
"v0 = (3*p/d)**(0.5);\t\t\t#rms velocity at 0deg.C\n",
"v100 = v0*(t2/t1)**(0.5);\t\t\t#rms velocity at 100deg.C\n",
"\n",
"# Result\n",
"print 'the rms velocity at 0deg.C is %.2e cm/sec \\\n",
"\\nthe rms velocity at 100deg.C is %.3e cm/sec'%(v0,v100)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the rms velocity at 0deg.C is 1.84e+05 cm/sec \n",
"the rms velocity at 100deg.C is 2.154e+05 cm/sec\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.12 page no : 140"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"cp = 6.84;\t\t\t#specific heat at consmath.tant pressure in cal/gm mole/deg.C\n",
"r = 8.31*10**7;\t\t\t#universal gas constant in ergs/gm mole/deg.C\n",
"v = 130000;\t\t\t#velocity of sound in cm/sec\n",
"j = 4.2*10**7;\t\t\t#joules constant in ergs/cal\n",
"\n",
"#CALCULATION\n",
"cv = cp-(r/j);\t\t\t#specific heat at constant volume in gm-mole/deg.C\n",
"y = (cp/cv);\t\t\t#index of co-efficient\n",
"v1 = (3/y)**(0.5)*v;\t\t\t#rms velocity in cm/sec\n",
"\n",
"# Result\n",
"print 'the rms velocity of gas molecule is %.3e cm/sec'%(v1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the rms velocity of gas molecule is 1.898e+05 cm/sec\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.13 page no : 140"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"t = 300;\t\t\t#temperature of the oxygen molecule in K\n",
"n = 6.02*10**23;\t\t\t#avagdrao's number\n",
"m = 32/n;\t\t\t#mass of each molecule in oxygen\n",
"k = 1.38*10**(-16);\t\t\t#boltzmann consmath.tant in erg/deg\n",
"\n",
"# Calculations\n",
"v = (8*k*t/(3.14*m))**(0.5);\t\t\t#average velocity of oxygen molecule in cm/sec\n",
"v2 = v*0.022384;\t\t\t#velocity in miles/hrs\n",
"\n",
"# Results\n",
"print 'the avg velocity of oxygen molecule is %.f miles/hour'%(v2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the avg velocity of oxygen molecule is 997 miles/hour\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.14 page no : 140"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"v1 = 2.4;\t\t\t#velocity of first particle in km/sec\n",
"v2 = 2.6;\t\t\t#velocity of second particle in km/sec\n",
"v3 = 3.7;\t\t\t#velocity of third particle in km/sec\n",
"\n",
"# Calculations\n",
"rv = ((v1**2+v2**2+v3**2)/(3))**(0.5);\t\t\t#rms velocity of the particles in km/sec\n",
"mv = (v1+v2+v3)/(3);\t\t\t #mean velocity of the particles in km/sec\n",
"r = rv/mv;\t\t\t #ratio of the rms to mean velocity\n",
"\n",
"# Results\n",
"print 'the ratio of rms to mean velocity is %3.3f'%(r)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the ratio of rms to mean velocity is 1.019\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.15 pageno : 141"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"n = 2.76*10**19;\t\t\t#no.of molecules per cc\n",
"d = 3.36*10**(-8);\t\t\t#diameter of the helium molecule in cm\n",
"\n",
"# Calculations\n",
"mf = 1/((2**(0.5))*3.14*(d**2)*n)\n",
"\n",
"# Result\n",
"print 'the mean free path of the hydrogen molecue is %.2e cm'%(mf)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the mean free path of the hydrogen molecue is 7.23e-06 cm\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.16 page no : 141"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"n = 85*10**(-6);\t\t\t#coefficent of vismath.cosity in dynes/cm**2/velocity gradient\n",
"c = 16*10**4;\t\t\t#velocity in cm/sec\n",
"p = 0.000089;\t\t\t#density in gm/cc\n",
"N = 6.06*10**23/22400;\t\t\t#avagadro number\n",
"a = (2)**(0.5)*(22./7);\t\t\t#constant\n",
"\n",
"# Calculations\n",
"mf = (3*n/(p*c));\t\t\t#mean free path in cm\n",
"cr = c/mf;\t\t\t#collision rate\n",
"d = (1/(a*N*mf))**(0.5);\t\t\t#molecular diameter of hydrogen gas in cm\n",
"\n",
"# Results\n",
"print 'the mean free path is %.2e cm \\\n",
"\\nthe collision rate is %.1e \\\n",
"\\nthe molecular diameter of hydrogen gas is %.1e cm'%(mf,cr,d)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the mean free path is 1.79e-05 cm \n",
"the collision rate is 8.9e+09 \n",
"the molecular diameter of hydrogen gas is 2.2e-08 cm\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.17 page no : 141"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"d = 2*10**(-8);\t\t\t#diameter of the molecule in cm\n",
"k = 1.38*10**(-6);\t\t\t#boltzmann constant in ergs/deg\n",
"t = 273;\t\t\t#temperature at ntp in K\n",
"p = 76*13.6*981;\t\t\t#pressure at ntp in gm/cm/sec**2\n",
"\n",
"# Calculations\n",
"mf = ((k*t)/(2**(0.5)*3.14*(d**2)*p));\t\t\t#mean free path in cm\n",
"\n",
"# Result\n",
"print 'mean free path at ntp is %.1e cm'%(mf)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"mean free path at ntp is 2.1e+05 cm\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.18 page no : 141"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"t = 288;\t\t\t#temperature in K\n",
"k = 1.38*10**(-16);\t\t\t#boltzmann constant in erg/deg\n",
"N = 6.02*10**23;\t\t\t#avagadro number\n",
"m = 32/N;\t\t\t#mass of each oxygen molecule in gm\n",
"v = 196*10**-6;\t\t\t#viscosity in poise\n",
"\n",
"# Calculations\n",
"av = ((8*k*t/(3.14*m))**0.5);\t\t\t#average velocity in cm/sec\n",
"d = (m*av/(3*3.14*2**(0.5)*v))**0.5;\t\t\t#diameter of the molecule in cm\n",
"\n",
"# Results\n",
"print 'diameter of the molecule is %.1e cm'%(d)\n",
"print \"Note : answer is slightly different because of rounding error\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"diameter of the molecule is 3.0e-08 cm\n",
"Note : answer is slightly different because of rounding error\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.19 page no : 142"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"mf = 15;\t\t\t#mean free path in cm\n",
"t = 300;\t\t\t#temperature of oxygen molecule in K\n",
"d = 3*10**(-8);\t\t\t#diameter of the molecule in cm\n",
"N = 6.02*10**23;\t\t\t#avagadro number\n",
"r = 8.32*10**7;\t\t\t#universal gas constant in ergs/mole/deg\n",
"a = (2**(0.5))*(22./7);\n",
"\n",
"#CALCULATIONS\n",
"p = (r*t)/(N*a*(d**2)*mf);\t\t\t#pressure of the oxygen molecule in dynes/sq.cm\n",
"\n",
"# Result\n",
"print 'the pressure of the oxygen molecule is %3.3f dynes/sq.cm'%(p)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the pressure of the oxygen molecule is 0.691 dynes/sq.cm\n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.20 pageno : 142"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"k = 5.64*10**-14;\t\t\t#kinetic energy of the hydrogen molecule ergs\n",
"t = 273;\t\t\t#temperature of the oxygen molecule in K\n",
"r = 8.32*10**7;\t\t\t#universal gas constant in ergs \n",
"\n",
"# Calculations\n",
"N = (3./2)*(r*t/k);\t\t\t#avagadro number\n",
"\n",
"# Result\n",
"print 'the avagadro number is %.2e'%(N)\n",
"print \"Note : answer is slightly different because of rounding error\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the avagadro number is 6.04e+23\n",
"Note : answer is slightly different because of rounding error\n"
]
}
],
"prompt_number": 30
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.21 pageno : 143"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"q = 5000;\t\t\t#total number of molecules\n",
"e = 2.7183;\t\t\t#constant value\n",
"t1 = 0.5;\t\t\t#distance travled to the mean free path\n",
"t2 = 1;\t\t\t#distance travelled to the mean free path\n",
"\n",
"#CALCULATONS\n",
"p1 = q*(e**-t1);\t\t\t#n0.of molecules having no collision in traversing a dismath.tance t1\n",
"p2 = q*(e**-t2);\t\t\t#n0.of molecules having no collision in traversing a dismath.tance t2\n",
"\n",
"#OUPUT\n",
"print 'the no. of molecules having no collision in traversing distance equal to 0.5 times the mean free path is %.f \\\n",
"\\nthe no. of molecules having no collision in traversing a distance equal to 1 time the mean free path is %.f'%(p1,p2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the no. of molecules having no collision in traversing distance equal to 0.5 times the mean free path is 3033 \n",
"the no. of molecules having no collision in traversing a distance equal to 1 time the mean free path is 1839\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.22 page no : 143"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"t = 38380;\t\t\t#temperature of the molecule in K\n",
"k = 1.38*10**-16;\t\t\t#boltzman consmath.tant of one electron in ergs/K\n",
"e = 1.6*10**-12;\t\t\t#charge of one electron volts\n",
"\n",
"#CALCULATIOS\n",
"mk = 1.5*k*t/e;\t\t\t#mean kinetic energy per atom in ev\n",
"\n",
"# Result\n",
"print 'the mean kinetic energy of the molecule is %3.2f ev'%(mk) \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the mean kinetic energy of the molecule is 4.97 ev\n"
]
}
],
"prompt_number": 32
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.23 pageno : 143"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"v = 1.7*10**-4;\t\t\t#vismath.cosity of the air molecule in cgs\n",
"d = 0.00129;\t\t\t#density of the molecule in gm/ml\n",
"p = 76*13.6*981;\t\t\t#pressure of the molecule in gm/cm/sec**2\n",
"\n",
"# Calculations\n",
"r = (3*p/d)**(0.5);\t\t\t#rms velocity of the molecule in cm/sec\n",
"mf = (3*v/(d*r));\t\t\t#mean free path in cm\n",
"cf = r/mf;\t\t\t#collision frequency\n",
"\n",
"# Result\n",
"print 'the mean free path is %.1e cm \\\n",
"\\nthe collision frequency is %.e'%(mf,cf)\n",
"print \"Note : answer is slightly different because of rounding error\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the mean free path is 8.1e-06 cm \n",
"the collision frequency is 6e+09\n",
"Note : answer is slightly different because of rounding error\n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.24 page no : 143"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# The pressure of the gas\n",
"\n",
"# Variables\n",
"t2 = 296.4;\t\t\t#temperature of the first plate in K\n",
"t1 = 304.7;\t\t\t#temperature of the second plate in K\n",
"f = 1.6*10**-2;\t\t\t#force repelled cold is dynes/sq.cm\n",
"\n",
"# Calculations\n",
"p = (4*f*t2/(t1-t2));\t\t\t#pressure of the gas in dynes/sq.cm\n",
"\n",
"# Result\n",
"print 'the pressure of the gas is %3.3f dynes/sq.cm'%(p)\n",
"print \"Note : mistake in answer in book. Please calculate manually.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the pressure of the gas is 2.285 dynes/sq.cm\n",
"Note : mistake in answer in book. Please calculate manually.\n"
]
}
],
"prompt_number": 40
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.25 page no : 144"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"mf = 28.5*10**-6;\t\t\t#mean free path in cm\n",
"d = 0.000178;\t\t\t#density of helium in gm/ml\n",
"m = 6*10**-24;\t\t\t#mass of the helium atom in gm\n",
"a = (2**(0.5))*3.14;\t\t\t#constant\n",
"\n",
"# Calculations\n",
"d = (m/(a*d*mf))**(0.5);\t\t\t#diameter of the size in cm\n",
"\n",
"# Result\n",
"print 'the size of the helium atom is %.2e cm'%(d)\n",
"print \"Note : answer is slightly different because of rounding error\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the size of the helium atom is 1.63e-08 cm\n",
"Note : answer is slightly different because of rounding error\n"
]
}
],
"prompt_number": 42
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.26 page no : 144"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"a1 = 0*10**-4;\t\t\t#first horizontal print lacement in cm\n",
"a2 = 5.6*10**-4;\t\t\t#second horizontal print lacement in cm\n",
"a3 = -4.7*10**-4;\t\t\t#third horzontal print lacement in cm\n",
"a4 = -10.8*10**-4;\t\t\t#fourth horizontal print lacement in cm\n",
"a5 = 6.6*10**-4;\t\t\t#fifth horizontal print lacement print lacement in cm\n",
"a6 = -9.8*10**-4;\t\t\t#sixth horizontal print lacement in cm\n",
"a7 = -11.2*10**-4;\t\t\t#7th horizontal print lacement in cm\n",
"a8 = -4.0*10**-4;\t\t\t#8th horizontal print lacement in cm\n",
"a9 = 15.0*10**-4;\t\t\t#9thhorizontal print lacement in cm\n",
"a10 = 19.1*10**-4;\t\t\t#10th horizontal print lacement in cm\n",
"a11 = 16.0*10**-4;\t\t\t#11ht horizontal print lacement in cm\n",
"T = 293;\t\t\t#temperature of the particle in K\n",
"v = 0.01;\t\t\t#viscosity in cgs\n",
"r = 1.15*10**-5;\t\t\t#radius of the particle in cm\n",
"R = 8.32*10**7;\t\t\t#universal gas constant in kj/kg mole\n",
"t = 30;\t\t\t#time for observation of each in sec\n",
"\n",
"# Calculations\n",
"x = (a1**2+a2**2+a3**2+a4**2+a5**2+a6**2+a7**2+a8**2+a9**2+a10**2+a11**2)/11\n",
"n = R*T*t/(x*3*3.14*v*r);\t\t\t#no.of molecules in the observation \n",
"\n",
"# Result\n",
"print 'the value of n is %.1e'%(n)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the value of n is 5.7e+23\n"
]
}
],
"prompt_number": 43
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.27 page no : 144"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"m = 6.*10**-24;\t\t\t#mass of the helium atom in gm\n",
"k = 1.38*10**-16;\t\t\t#boltzmann consmath.tant in erg\n",
"t1 = 100.;\t\t\t#temperature in K\n",
"t2 = 900.;\t\t\t#temperature in K\n",
"\n",
"# Calculations\n",
"r = (t1/t2)**(3./2)*(2.7183**(m*(1./(2*k))*10**8*(1-(1./9))));\t\t\t#fractional change in the no.of helium atoms\n",
"\n",
"#OUPUT\n",
"print 'the fractional change in the no.of helium atoms %.3f'%(r)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the fractional change in the no.of helium atoms 0.256\n"
]
}
],
"prompt_number": 45
}
],
"metadata": {}
}
]
}