# Chapter 5 : Nature of Heat¶

## Example 5.1 Page No : 159¶

In :
# Input data
v = 480.  # The velocity of a lead bullet in m/s
Sp = 0.03  # Specific heat of lead cal/g-K

# Calculations
m = 10.  # Let us assume the mass of bullet in gms
V = v * 100  # The velocity of the bullet in cm/s
W = (1. / 2) * m * (V**2)  # The work done in ergs
J = 4.2 * 10**7  # The mechanical equivalent of heat in ergs/calorie
H = W / J  # The amount of heat produced in cals
H1 = H / 2  # Half of the heat energy is used to raise the temperature of the bullet in cals
t = H1 / (m * Sp)  # The rise in the temperature in degree centigrade

# Output
print 'The rise in the temperature is t = %3.2f degree centigrade ' % (t)

The rise in the temperature is t = 457.14 degree centigrade


## Example 5.2 Page No : 162¶

In :
# Input data
t = 1.  # The increase in the temperature of a piece of aluminium in degree centigrade
a = 6. * 10**23  # The number of atoms present in 27 g of aluminium in atoms
Sp = 0.22  # The specific heat of aluminium in cal/g-K
m = 27.  # The amount of aluminium in g
J = 4.2 * 10**7  # The mechanical equivalent of heat in ergs/calorie

# Calculations
H = m * Sp * t  # Heat required to raise the temperature of 27 gms of aluminium by 1 degree centigrade in cals
E = m * Sp * J  # Energy gained by atoms of aluminium in ergs
E1 = E / a  # Increase in energy per atom of aluminium in ergs

# Output
print 'The increase in energy per atom of aluminium is %3.4g ergs ' % (E1)

The increase in energy per atom of aluminium is 4.158e-16 ergs


## Example 5.3 Page No : 168¶

In :
# Input data
h = 50.  # The height from which water falls in metres
m = 100.  # Let us assume the mass of the water in gms
g = 980.  # Gravitational constant in gms/s**2
J = 4.2 * 10**7  # The mechanical equivalent of heat in ergs/calorie

# Calculations
h1 = h * 100  # The height from which water falls in cm
W = m * g * h1  # The work done in ergs
t = W / (J * m)  # The rise in temperature of water in degree centigrade

# Output
print 'The rise in temperature of water is t = %3.3f degree centigrade ' % (t)

The rise in temperature of water is t = 0.117 degree centigrade


## Example 5.4 Page No : 174¶

In :
# Input data
v = 1.  # The volume of oxygen at N.T.P in cm**3
d = 13.6  # The density of mercury in g/cm**3
r = 4.62 * 10**4  # The R.M.S velocity of oxygen molecules at 0 degree centigrade in cm/s
m = 52.8 * 10**-24  # Mass of one molecule of oxygen in g
g = 980.  # Gravitational constant in gms/s**2

# Calculations
P = 76. * g * d  # The pressure in dynes/cm**2
n = ((3 * P) / (m * r**2))  # Number of molecules in 1 cc of oxygen at N.T.P

# Output
print 'The number of molecules in 1 c.c of oxygen at N.T.P is n = %3.4g ' % (n)

The number of molecules in 1 c.c of oxygen at N.T.P is n = 2.696e+19


## Example 5.5 Page No : 177¶

In :
# Input data
t = -100  # The given temperature in degree centigrade

# Calculations
T1 = t + 273  # The given temperature in K
m1 = 1.  # number of hydrogen molecules
m2 = 16.  # number of oxygen molecules
m = m2 / m1  # Number of oxygen molecules to the hydrogen molecules
T2 = (T1 * m) - 273  # The temperature in degree centigrade

# Output
print 'The temperature at which the oxygen molecules have the same root mean square velocity  as that of hydrogen molecules is T2 = %3.0f degree centigrade ' % (T2)

The temperature at which the oxygen molecules have the same root mean square velocity  as that of hydrogen molecules is T2 = 2495 degree centigrade


## Example 5.6 Page No : 180¶

In :
# Input data
t = 27.  # The given temperature in degree centigrade
d = 13.6  # The density of mercury in g/cm**3
g = 980.  # Gravitational constant in gms/s**2
m1 = 16.  # number of oxygen molecules
D = 0.000089  # The density of hydrogen at N.T.P in g/cc
T = 273.  # The temperature at N.T.P in K

# Calculations
P = 76. * g * d  # The pressure in dynes/cm**2
p = m1 * D  # The density of oxygen at N.T.P in g/cc
C = ((3 * P) / (p))**(1. / 2)  # The RMS velocity of oxygen molecule in cm/s
T1 = t + T  # The given temperature in K
# The RMS velocity of the molecules at 27 degree centigrade in cm/s
C1 = C * (T1 / T)**(1. / 2)

# Output
print 'The RMS velocity of the oxygen molecules at 27 degree centigrade is C1 = %3.4g cm/s ' % (C1)

The RMS velocity of the oxygen molecules at 27 degree centigrade is C1 = 4.843e+04 cm/s


## Example 5.7 Page No : 186¶

In :
# Input data
d = 13.6  # The density of mercury in g/cm**3
g = 980.  # Gravitational constant in gms/s**2
m = 3.2  # Mass of oxygen in gms
t = 27.  # The given temperature in degree centigrade
p = 76.  # The pressure in cm of Hg
R = 8.31 * 10**7  # The Universal gas constant in ergs/g mol-K

# Calculations
P = p * g * d  # The given pressure in dynes/cm**2
T = t + 273  # The given temperature in K
V = (T * R) / P  # Volume per g mol of oxygen in cc per g mol
m1 = 32.  # Molecular weight of Oxygen
V1 = V * (m / m1)  # Volume of 3.2 g of oxygen in cc

# Output
print 'The Volume occupied by 3.2 gms of Oxygen is V = %3.0f cc ' % (V1)

The Volume occupied by 3.2 gms of Oxygen is V = 2461 cc


## Example 5.9 Page No : 193¶

In :
# Input data
v = 1.  # The volume of an Ideal gas at N.T.P in m**3
d = 13.6  # The density of mercury in g/cm**3
g = 980.  # Gravitational constant in gms/s**2
p = 76.  # The pressure in cm of Hg
R = 8.31 * 10**7  # The Universal gas constant in ergs/g mol-K
N = 6.023 * 10**23  # The Avogadro number
T = 273.  # The temperature at N.T.P in K

# Calculations
P = p * g * d  # The given pressure in dynes/cm**2
x = (P * N * 10**6) / (R * T)  # Number of molecules in one cubic metre volume

# Output
print 'The number of molecules in one cubic metre of an ideal gas at N.T.P is x = %3.4g ' % (x)

The number of molecules in one cubic metre of an ideal gas at N.T.P is x = 2.689e+25


## Example 5.10 Page No : 196¶

In :
# Input data
v = 1.  # The volume of an ideal gas in litre
d = 13.6  # The density of mercury in g/cm**3
g = 980.  # Gravitational constant in gms/s**2
p = 76.  # The pressure in cm of Hg
R = 8.31 * 10**7  # The Universal gas constant in ergs/g mol-K
N = 6.023 * 10**23  # The Avogadro number
T = 273.  # The temperature at N.T.P in K
t = 136.5  # The given temperature in degree centigrade
p1 = 3.  # The given atmospheric pressure in atm pressure

# Calculations
T1 = T + t  # The given temperature in K
P = p * g * d  # The given pressure in dynes/cm**2
x = (p1 * P * N * 10**3) / (R * T1)  # Number of molecules in one litre volume

# Output
print 'The number of molecules in one litre of an ideal gas volume is x = %3.4g ' % (x)

The number of molecules in one litre of an ideal gas volume is x = 5.378e+22


## Example 5.11 Page No : 203¶

In :
# Input data
v = 1.  # The volume of a gas in cc
d = 13.6  # The density of mercury in g/cm**3
p2 = 10.**-7  # The pressure in cm of Hg
g = 980.  # Gravitational constant in gms/s**2
p1 = 76.  # The pressure in cm of Hg
R = 8.31 * 10**7  # The Universal gas constant in ergs/g mol-K
N = 6.023 * 10**23  # The Avogadro number
T = 273.  # The temperature at N.T.P in K
n1 = 2.7 * 10**19  # The number of molecules per cc of gas at N.T.P
t2 = 0.  # The given temperature in degree centigrade
t3 = 39.  # The given temperature in degree centigrade

# Calculations
P1 = p1 * g * d  # The given pressure in dynes/cm**2
P2 = p2 * g * d  # The given pressure in dynes/cm**2
# The number of molecules per cc of the gas at 0 degree centigrade
n2 = n1 * (P2 / P1)
T2 = t2 + 273  # The given temperature in K
T3 = t3 + 273  # The given temperature in K
# The number of molecules per cc of the gas at 398 degree centigrade
n3 = n2 * (T2 / T3)

# Output
print 'The number of molecules per cc of the gas , \n (1)at 0 degree centigrade and 10^-6 mm pressure of mercury is n2 = %3.4g \n (2)at 39 degree centigrade and 10^-6 mm pressure of mercury is n3 = %3.4g' % (n2, n3)

The number of molecules per cc of the gas ,
(1)at 0 degree centigrade and 10^-6 mm pressure of mercury is n2 = 3.553e+10
(2)at 39 degree centigrade and 10^-6 mm pressure of mercury is n3 = 3.109e+10


## Example 5.12 Page No : 208¶

In :
# Input data
T = 300.  # The given temperature in K
R = 8.3 * 10**7  # The Universal gas constant in ergs/g mol-K

# Calculations
# The total random kinetic energy per gram -molecule of oxygen in joules
E = ((3. / 2) * (R * T)) / 10**7

# Output
print 'The total random kinetic energy of one gm-molecule of oxygen at 300 K is K.E = %3.0f joules' % (E)

The total random kinetic energy of one gm-molecule of oxygen at 300 K is K.E = 3735 joules


## Example 5.13 Page No : 213¶

In :
# Input data
T = 300.  # The given temperature in K
k = 1.38 * 10**-16  # Boltzmann constant in erg/molecule-deg

# Calculations
E = (3. / 2) * k * T  # The average Kinetic energy of a molecule in ergs

# Output
print 'The Average Kinetic energy of a molecule of a gas at 300 K is K.E = %3.4g ergs ' % (E)

The Average Kinetic energy of a molecule of a gas at 300 K is K.E = 6.21e-14 ergs


## Example 5.14 Page No : 220¶

In :
# Input data
R = 8.32  # Universal gas constant in joules/mole-K
t = 727.  # The given temperature in degree centigrade
N = 6.06 * 10**23  # The Avogadro number

# Calculations
T = 273. + t  # The given temperature in K
k = R / N  # Boltzmann constant in joules/mol-K
E = (3. / 2) * k * T  # Mean translational kinetic energy per molecule in joules

# Output
print 'The mean translational kinetic energy per molecule is K.E = %3.4g joule ' % (E)

The mean translational kinetic energy per molecule is K.E = 2.059e-20 joule


## Example 5.15 Page No : 224¶

In :
# Input data
T = 300.  # The given temperature in K
M = 28.  # Molecular weight of nitrogen in g
R = 8.3 * 10**7  # The Universal gas constant in ergs/g mol-K

# Calculations
E = (3. / 2) * R * T  # The total random kinetic energy of nitrogen in ergs
# The total random kinetic energy of one gram of nitrogen at 300 K in joule
E1 = E / (M * 10**7)

# Output
print 'The total random kinetic energy of one gram of nitrogen at 300 K is K.E = %3.1f joule ' % (E1)

The total random kinetic energy of one gram of nitrogen at 300 K is K.E = 133.4 joule


## Example 5.16 Page No : 228¶

In :
# Input data
T = 200.  # The given temperature in K
m = 2.  # Given mass of Helium in g
M = 4.  # Molecular weight of helium in g
R = 8.3 * 10**7  # The Universal gas constant in ergs/g mol-K

# Calculations
# The energy for 2 g of helium in joules
E = (m * (3. / 2) * (R * T) / (M)) / 10**7

# Output
print 'The total random kinetic energy of 2 g of helium at 200 K is K.E = %3.0f joules' % (E)

The total random kinetic energy of 2 g of helium at 200 K is K.E = 1245 joules


## Example 5.17 Page No : 233¶

In :
# Input data
T = 300.  # The given temperature in K
R = 8.3 * 10**7  # The Universal gas constant in ergs/g mol-K
M = 221.  # The molecular weight of mercury

# Calculations
# The root mean square velocity of a molecule of mercury vapour at 300 K
# in cm/s
C = ((3 * R * T) / (M))**(1. / 2)

# Output
print 'The root mean square velocity of a molecule of mercury vapour at 300 K is C = %3.4g cm/s ' % (C)

The root mean square velocity of a molecule of mercury vapour at 300 K is C = 1.839e+04 cm/s


## Example 5.18 Page No : 239¶

In :
# Input data
T = 300.  # The given temperature in K
M = 32.  # Molecular weight of oxygen
R = 8.3 * 10**7  # The Universal gas constant in ergs/g mol-K

# Calculations
# Total random kinetic energy of 1 g molecule of oxygen in ergs
E = (3. / 2) * R * T
# The required speed of one gram molecule of oxygen in cm/s
v = ((E) * (2 / M))**(1. / 2)

# Output
print 'The required speed of one gram molecule of oxygen is v = %3.2g cm/s ' % (v)

The required speed of one gram molecule of oxygen is v = 4.8e+04 cm/s


## Example 5.19 Page No : 242¶

In :
# Input data
v = 8.  # The speed of the earths first satellite in km/s
R = 8.3 * 10**7  # The Universal gas constant in ergs/g mol-K
M = 2.  # Molecular weight of hydrogen

# Calculations
V = v * 10**5  # The speed of the earths first satellite in cm/s
T = (M * V**2) / (3 * R)  # The temperature at which it becomes equal in K

# Output
print 'The temperature at which the r.m.s velocity of a hydrogen molecule  will be equal to the speed of earths first satellite is T = %3.4g K' % (T)

The temperature at which the r.m.s velocity of a hydrogen molecule  will be equal to the speed of earths first satellite is T = 5141 K


## Example 5.20 Page No : 248¶

In :
# Input data
t1 = 0.  # The given temperature in degree centigrade

# Calculations
T1 = t1 + 273  # The given temperature in K
# The temperature at which the r.m.s velocity of a gas be half its value
# at 0 degree centigrade in K
T2 = (1. / 2)**2 * T1
T21 = T2 - 273  # The required temperature in degree centigrade

# Output
print 'The required temperature is  T2 = %3.2f K  (or) %3.2f degree centigrade ' % (T2, T21)

The required temperature is  T2 = 68.25 K  (or) -204.75 degree centigrade


## Example 5.21 Page No : 252¶

In :
# Input data
n = 1.66 * 10**-4  # The viscosity of the gas in dynes/cm**2
C = 4.5 * 10**4  # The R.M.S velocity of the molecules in cm/s
d = 1.25 * 10**-3  # The density of the gas in g/cc
N = 6.023 * 10**23  # The Avogadro number
V = 22400.  # The volume of a gas at N.T.P in cc
pi = 3.142  # The mathematical constant of pi

# Calculations
L = (3 * n) / (d * C)  # The mean free path of the molecules of the gas in cm
F = (C / L)  # The frequency collision in per sec
n = N / V  # Number of molecules per cc
# Molecular diameter of the gas molecules in cm
D = 1. / ((1.414 * pi * n * L)**(1. / 2))

# Output
print '(1)The mean free path of the molecules of the gas is %3.0g cm \n (2)The frequency of collision is N = %3.0g /sec \n (3)Molecular diameter of the gas molecules is d = %3.0g cm ' % (L, F, D)

(1)The mean free path of the molecules of the gas is 9e-06 cm
(2)The frequency of collision is N = 5e+09 /sec
(3)Molecular diameter of the gas molecules is d = 3e-08 cm


## Example 5.22 Page No : 255¶

In :
# Input data
n = 2.25 * 10**-4  # The viscosity of the gas in dynes/cm**2
C = 4.5 * 10**4  # The RMS velocity of the molecules in cm/s
d = 10.**-3  # The density of the gas in g/cc

# Calculations
L = (3 * n) / (d * C)  # The mean free path of the molecules in cm

# Output
print 'The mean free path of the molecules is %3g cm ' % (L)

The mean free path of the molecules is 1.5e-05 cm


## Example 5.23 Page No : 261¶

In :
# Input data
d = 2. * 10**-8  # The molecular diameter in cm
n = 3. * 10**19  # The number of molecules per cc
pi = 3.14  # Mathematical constant of pi

# Calculations
L = 1. / ((pi * (d)**2 * n))  # The mean free path of a gas molecule in cm

# Output
print 'The mean free path of a gas molecule is %3.0g cm ' % (L)

The mean free path of a gas molecule is 3e-05 cm


## Example 5.24 Page No : 265¶

In :
# Input data
p = 760.  # The given pressure in mm of Hg
T = 273.  # The temperature of the chamber in K
V = 22400.  # The volume of the gas at N.T.P in cc
p1 = 10.**-6  # The pressure in the chamber in mm of mercury pressure
N = 6.023 * 10**23  # The Avogadro number
d = 2. * 10**-8  # Molecular diameter in cm
pi = 3.14  # Mathematical constant of pi

# Calculations
# The number of molecules per cm**3 in the chamber in molecules/cm**3
n = (N * p1) / (V * p)
# The mean free path of the gas molecules in the chamber in cm
L = 1. / (pi * (d)**2 * n)

# Output
print 'The mean free path of gas molecules in a chamber is %3.4g cm ' % (L)

The mean free path of gas molecules in a chamber is 2.25e+04 cm


## Example 5.25 Page No : 270¶

In :
# Input data
Tc = 132.  # The given temperature in K
Pc = 37.2  # The given pressure in atms
R = 82.07  # Universal gas constant in cm**3 atoms K**-1

# Calculations
# Vander Waals constant in atoms cm**6
a = (27. / 64) * ((R)**2 * (Tc)**2) / Pc
b = ((R * Tc) / (8 * Pc))  # Vander Waals constant in cm**3

# Output
print 'The Van der Waals constants are , \n (1) a = %3.4g atoms cm^6 \n (2) b = %3.2f cm^3 ' % (a, b)

The Van der Waals constants are ,
(1) a = 1.331e+06 atoms cm^6
(2) b = 36.40 cm^3