{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 3: Statistical Mechanics"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.1, Page 132"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import *\n",
"\n",
"#Variable Declaration\n",
"m = 5.32e-26; # Mass of one oxygen molecule, kg\n",
"k_B = 1.38e-23; # Boltzmann constant, J/K\n",
"T = 200; # Temperature of the system, K\n",
"v = 100; # Speed of the oxygen molecules, m/s\n",
"dv = 1; # Increase in speed of the oxygen molecules, m/s\n",
"\n",
"#Calculations\n",
"P = 4*pi*(m/(2*pi*k_B*T))**(3./2)*exp((-m*v**2)/(2*k_B*T))*v**2*dv;\n",
"\n",
"#Result\n",
"print \"The probability that the speed of oxygen molecule is %4.2e\"%P\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The probability that the speed of oxygen molecule is 6.13e-04\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.2, Page 132"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Declaration\n",
"A = 32; # Gram atomic mass of oxygen, g/mol\n",
"N_A = 6.023e+026; # Avogadro's number, per kmol\n",
"m = A/N_A; #mass of the molecule, kg\n",
"k_B = 1.38e-23; # Boltzmann constant, J/K \n",
"T = 273; # Temperature of the gas, K\n",
"\n",
"#Calculations&Results\n",
"v_av = 1.59*sqrt(k_B*T/m); # Average speed of oxygen molecule, m/s\n",
"print \"The average speed of oxygen molecule is = %3d m/s\"%v_av\n",
"v_rms = 1.73*sqrt(k_B*T/m); # The mean square speed of oxygen molecule, m/s \n",
"print \"The root mean square speed of oxygen gas molecule is = %3d m/s\"%(ceil(v_rms))\n",
"v_mp = 1.41*sqrt(k_B*T/m); # The most probable speed of oxygen molecule, m/s \n",
"print \"The most probable speed of oxygen molecule is = %3d m/s\"%(ceil(v_mp)) #incorrect answer in the textbook\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The average speed of oxygen molecule is = 423 m/s\n",
"The root mean square speed of oxygen gas molecule is = 461 m/s\n",
"The most probable speed of oxygen molecule is = 376 m/s\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.3, Page 133"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Declaration\n",
"m_H = 2; # Gram molecular mass of hydrogen, g\n",
"m_O = 32.; # Gram molecular mass of oxygen, g\n",
"k_B = 1.38e-23; # Boltzmann constant, J/K \n",
"v_avO = 1.; # For simplicity average speed of oxygen gas molecule is assumed to be unity, m/s\n",
"v_avH = 2*v_avO; # The average speed of hydrrogen gas molecule, m/s\n",
"T_O = 300.; # Temperature of oxygen gas, K\n",
"\n",
"#Calculations\n",
"# As v_avO/v_av_H = sqrt(T_O/T_H)*sqrt(m_H/m_O), solving for T_H\n",
"T_H = (v_avH/v_avO*sqrt(m_H/m_O)*sqrt(T_O))**2; # Temperature at which the average speed of hydrogen gas molecules is the same as that of oxygen gas molecules, K\n",
"\n",
"#Result\n",
"print \"Temperature at which the average speed of hydrogen gas molecules is the same as that of oxygen gas molecules at 300 K = %2d\"%T_H\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Temperature at which the average speed of hydrogen gas molecules is the same as that of oxygen gas molecules at 300 K = 75\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.4, Page 133"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Declaration\n",
"v_mp = 1; # Most probable speed of gas molecules, m/s\n",
"dv = 1.01*v_mp-0.99*v_mp; # Change in most probable speed, m/s\n",
"v = v_mp; # Speed of the gas molecules, m/s\n",
"\n",
"#Calculations\n",
"Frac = 4/sqrt(pi)*1/v_mp**3*exp(-v**2/v_mp**2)*v**2*dv; \n",
"\n",
"#Result\n",
"print \"The fraction of oxygen gas molecules within one percent of most probable speed = %5.3f\"%Frac\n",
"#rounding-off error"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The fraction of oxygen gas molecules within one percent of most probable speed = 0.017\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.5, Page 134"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"#Variable Declaration\n",
"n = 5.; # Number of distinguishable particles which are to be distributed among cells\n",
"n1 = [5, 4, 3, 3, 2]; # Possible occupancy of particles in first cell\n",
"n2 = [0 ,1, 2, 1, 2]; # Possible occupancy of particles in second cell\n",
"n3 = [0 ,0, 0, 1, 1]; # Possible occupancy of particles in third cell\n",
"BIG_W = 0.;\n",
"\n",
"\n",
"print(\"_____________________________________\");\n",
"print(\"n1 n2 n3 5/(n1!n2!n3!)\");\n",
"print(\"_____________________________________\");\n",
"for i in range(5):\n",
" W = math.factorial(n)/(math.factorial(n1[i])*math.factorial(n2[i])*math.factorial(n3[i]));\n",
" if BIG_W < W:\n",
" BIG_W = W;\n",
" ms = [n1[i], n2[i] ,n3[i]];\n",
"\n",
" print \"%d %d %d %d\"%(n1[i], n2[i], n3[i], W);\n",
"\n",
"print \"_____________________________________\";\n",
"print \"The macrostates of most probable distribution with thermodynamic probability %d are:\"%(BIG_W);\n",
"print \"(%d, %d, %d), (%d, %d, %d) and (%d, %d, %d)\"%(ms[0], ms[1], ms[2], ms[1], ms[2], ms[0],ms[2], ms[0], ms[1]);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"_____________________________________\n",
"n1 n2 n3 5/(n1!n2!n3!)\n",
"_____________________________________\n",
"5 0 0 1\n",
"4 1 0 5\n",
"3 2 0 10\n",
"3 1 1 20\n",
"2 2 1 30\n",
"_____________________________________\n",
"The macrostates of most probable distribution with thermodynamic probability 30 are:\n",
"(2, 2, 1), (2, 1, 2) and (1, 2, 2)\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.6, Page 135"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Declaration\n",
"g1 = 4; # Intrinsic probability of first cell\n",
"g2 = 2; # Intrinsic probability of second cell\n",
"k = 2; # Number of cells \n",
"n = 8; # Number of distinguishable particles\n",
"n1 = 8; # Number of cells in first compartment\n",
"n2 = n - n1; # Number of cells in second compartment\n",
"\n",
"#Calculations\n",
"W = factorial(n)*1/factorial(n1)*1/factorial(n2)*(g1)**n1*(g2)**n2;\n",
"\n",
"#Result\n",
"print \"The thermodynamic probability of the macrostate (8,0) = %5d\"%W\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The thermodynamic probability of the macrostate (8,0) = 65536\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.7, Page 135"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"#Variable Declaration\n",
"def st(val):\n",
" str1 = \"\"\n",
" if val == 3 :\n",
" str1 = 'aaa';\n",
" elif val == 2 :\n",
" str1 = 'aa'; \n",
" elif val == 1 :\n",
" str1 = 'a';\n",
" elif val == 0:\n",
" str1 = '0'; \n",
" return str1\n",
"\n",
"g = 3; # Number of cells in first compartment\n",
"n = 3; # Number of bosons\n",
"p = 3;\n",
"r = 1; # Index for number of rows\n",
"print(\"All possible meaningful arrangements of three particles in three cells are:\")\n",
"print(\"__________________________\");\n",
"print(\"Cell 1 Cell 2 Cell 3\");\n",
"print(\"__________________________\");\n",
"\n",
"for i in range(0,g+1):\n",
" for j in range(0,n+1):\n",
" for k in range(0,p+1):\n",
" if (i+j+k == 3):\n",
" print \"%4s %4s %4s\"%(st(i), st(j), st(k)); \n",
" print \"__________________________\";\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"All possible meaningful arrangements of three particles in three cells are:\n",
"__________________________\n",
"Cell 1 Cell 2 Cell 3\n",
"__________________________\n",
" 0 0 aaa\n",
" 0 a aa\n",
" 0 aa a\n",
" 0 aaa 0\n",
"__________________________\n",
" a 0 aa\n",
" a a a\n",
" a aa 0\n",
"__________________________\n",
" aa 0 a\n",
" aa a 0\n",
"__________________________\n",
" aaa 0 0\n",
"__________________________\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.8, Page 136"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Declaration\n",
"g1 = 3; # Number of cells in first compartment\n",
"g2 = 4; # Number of cells in second compartment\n",
"k = 2; # Number of compartments\n",
"n1 = 5; # Number of bosons\n",
"n2 = 0; # Number of with no bosons\n",
"\n",
"#Calculations\n",
"W_50 = factorial(g1+n1-1)*factorial(g2+n2-1)/(factorial(n1)*factorial(g1-1)*factorial(n2)*factorial(g2-1));\n",
"\n",
"#Result\n",
"print \"The probability for the macrostate (5,0) is = %2d\"%W_50\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The probability for the macrostate (5,0) is = 21\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.11, Page 138"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Declaration\n",
"r = 1.86e-10; # Radius of Na, angstrom\n",
"m = 9.1e-31; # Mass of electron,in kg\n",
"h = 6.62e-34; # Planck's constant, J-s\n",
"N = 2; # Number of free electrons in a unit cell of Na\n",
"\n",
"#Calculations\n",
"a = 4*r/sqrt(3); # Volume of Na, m\n",
"V = a**3; # Volume of the unit cell of Na, meter cube\n",
"E = h**2/(2*m)*(3*N/(8*pi*V))**(2./3);\n",
"\n",
"#Result\n",
"print \"The fermi energy of the Na at absolute zero is = %4.2e J\"%E\n",
"#rounding-off error"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The fermi energy of the Na at absolute zero is = 5.02e-19 J\n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.12, Page 13"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Declaration\n",
"m = 9.1e-31; # mass of electron, kg\n",
"h = 6.62e-34; # Planck's constant, J-s\n",
"V = 108/10.5*1e-06; # Volume of 1 gm mole of silver, metre-cube\n",
"N = 6.023e+023; # Avogadro's number\n",
"\n",
"#Calculations\n",
"E_F = h**2/(2*m)*(3*N/(8*pi*V))**(2./3); # Fermi energy at absolute zero, J\n",
"\n",
"#Result\n",
"print \"The fermi energy of the silver at absolute zero = %4.2e J\"%E_F\n",
"#rounding-off error"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The fermi energy of the silver at absolute zero = 8.80e-19 J\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.13, Page 13"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Declaration\n",
"pbe = 24.2e22 #electrons/cm^3\n",
"pcs = 0.91e22 #electrons/cm^3\n",
"efbe = 14.44 #ev\n",
"\n",
"#Calculations\n",
"Efcs = efbe*((pcs/pbe)**(2./3))\n",
"\n",
"#Result\n",
"print \"Fermi energy of free electrons in cesium = %.3f eV\"%Efcs\n",
"#rounding-off error"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Fermi energy of free electrons in cesium = 1.621 eV\n"
]
}
],
"prompt_number": 46
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.14, Page 140"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Declaration\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"m = 9.1e-31; # Mass of the elecron, kg \n",
"h = 6.63e-34; # Planck's constant, Js\n",
"EF = 4.72*e; # Fermi energy of free electrons in Li, J\n",
"\n",
"#Calculations\n",
"rho = 8*pi/3*(2*m*EF/h**2)**(3./2); # Electron density at absolute zero, electrons/metre-cube\n",
"\n",
"#Result\n",
"print \"The electron density in lithium at absolute zero = %4.2e electrons/metre-cube\"%rho\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The electron density in lithium at absolute zero = 4.63e+28 electrons/metre-cube\n"
]
}
],
"prompt_number": 47
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.15, Page 140"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Declaration\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"k_B = 1.38e-023; # Boltzmann constant, J/K\n",
"f_E = 0.01; # Probability that a state with energy 0.5 eV above the Fermi energy is occupied by an electron, eV \n",
"delta_E = 0.5; # Energy difference (E-Ef)of fermi energy, eV\n",
"\n",
"#Calculations\n",
"# Since f_E = 1/(exp((E-Ef)/(k_B*T))+1), solvinf for T \n",
"T = delta_E/(log((1-f_E)/f_E)*k_B/e); # Temperature at which the level above the fermi level is occupied by the electron, K\n",
"\n",
"#Result\n",
"print \"The temperature at which the level above the fermi level is occupied by the electron = %4d K\"%ceil(T)\n",
"#rounding-off error"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature at which the level above the fermi level is occupied by the electron = 1262 K\n"
]
}
],
"prompt_number": 49
}
],
"metadata": {}
}
]
}