Khopdi Kharab: Kinetic Theory

Diameter of an oxygen molecule, d= 3Å
Radius,r= 1.5 Å = 1.5 × 10^–8cm
Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm³
Molecular volume of oxygen gas,
Where,N is Avogadro’s number = 6.023 × 10²³molecules/mole

Ratio of the molecular volume to the actual volume of oxygen
= 3.8 × 10^–4
The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as:
PV= nRT
Where,
Ris the universal gas constant = 8.314 J mol^–1K^–1
n= Number of moles = 1
T= Standard temperature = 273 K
P= Standard pressure = 1 atm = 1.013 × 10⁵Nm^–2

= 0.0224 m³
= 22.4 litres
Hence,the molar volume of a gas at STP is 22.4 litres.
(a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratio is equal. μR (μ is the number of moles and R is the universal gas constant) is a constant quality. It is not dependent on the pressure of the gas.
(b) The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature T₁ is closer to the dotted plot than the curve of the gas at temperature T₂. A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore, T₁ > T₂ is true for the given plot.
(c) The value of the ratio PV/T, where the two curves meet, is μR. This is because the ideal gas equation is given as:
PV = μRT

Where,
P is the pressure
T is the temperature
V is the volume
μ is the number of moles
R is the universal constant
Molecular mass of oxygen = 32.0 g
Mass of oxygen = 1 × 10^–3 kg = 1 g
R = 8.314 J mole^–1 K^–1
∴
= 0.26 J K^–1
Therefore, the value of the ratio PV/T, where the curves meet on the y-axis, is
0.26 J K^–1.
(d) If we obtain similar plots for 1.00 × 10^–3 kg of hydrogen, then we will not get the same value of PV/T at the point where the curves meet the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u).
We have:

R = 8.314 J mole^–1 K^–1
Molecular mass (M) of H₂ = 2.02 u

m = Mass of H₂
∴

= 6.3 × 10^–2 g = 6.3 × 10^–5 kg
Hence, 6.3 × 10^–5 kg of H₂ will yield the same value of PV/T.
Volumeof oxygen, V₁= 30 litres = 30 × 10^–3m³
Gauge pressure, P₁= 15 atm = 15 × 1.013 × 10⁵Pa
Temperature,T₁= 27°C = 300 K
Universal gas constant, R= 8.314 J mole^–1K^–1
Let the initial number of moles of oxygen gas in the cylinder be n₁.
The gas equation is given as:
P₁V₁= n₁RT₁

= 18.276
But,
Where,
m₁= Initial mass of oxygen
M= Molecular mass of oxygen = 32 g
∴m₁= n₁M= 18.276 × 32 = 584.84 g
Aftersome oxygen is withdrawn from the cylinder, the pressure and temperature reduces.
Volume,V₂= 30 litres = 30 × 10^–3m³
Gauge pressure, P₂= 11 atm = 11 × 1.013 × 10⁵Pa
Temperature,T₂= 17°C = 290 K
Letn₂be the number of moles of oxygen left in the cylinder.
The gas equation is given as:
P₂V₂= n₂RT₂

= 13.86
But,
Where,
m₂is the mass of oxygen remaining in the cylinder
∴m₂= n₂M= 13.86 × 32 = 453.1 g
The mass of oxygen taken out of the cylinder is given by the relation:
Initial mass of oxygen in the cylinder –Final mass of oxygen in the cylinder
=m₁– m₂
= 584.84 g – 453.1 g
= 131.74 g
= 0.131 kg
Therefore, 0.131 kg of oxygen is taken out of the cylinder.
Volume of the air bubble, V₁ = 1.0 cm³ = 1.0 × 10^–6 m³
Bubble rises to height, d = 40 m
Temperature at a depth of 40 m, T₁ = 12°C = 285 K
Temperature at the surface of the lake, T₂ = 35°C = 308 K
The pressure on the surface of the lake:
P₂ = 1 atm = 1 ×1.013 × 10⁵ Pa
The pressure at the depth of 40 m:
P₁ = 1 atm + dρg
Where,
ρ is the density of water = 10³ kg/m³
g is the acceleration due to gravity = 9.8 m/s²
∴P₁ = 1.013 × 10⁵ + 40 × 10³ × 9.8 = 493300 Pa
We have:
Where, V₂ is the volume of the air bubble when it reaches the surface

= 5.263 × 10^–6 m³ or 5.263 cm³
Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm³.
Volume of the room, V= 25.0 m³
Temperature of the room, T= 27°C = 300 K
Pressure in the room, P= 1 atm = 1 × 1.013 × 10⁵Pa
The ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T) can be written as:
PV= k_BNT
Where,
K_Bis Boltzmann constant = 1.38 × 10^–23m² kg s^–2K^–1
Nis the number of air molecules in the room
∴
= 6.11 × 10²⁶molecules
Therefore, the total number of air molecules in the given room is 6.11 × 10²⁶.
(i) At room temperature, T= 27°C = 300 K
Average thermal energy
Wherek is Boltzmann constant = 1.38 × 10^–23m² kg s^–2K^–1
∴
= 6.21 × 10^–21J
Hence,the average thermal energy of a helium atom at room temperature (27°C) is 6.21 × 10^–21J.
(ii) On the surface of the sun, T= 6000 K
Average thermal energy

= 1.241 × 10^–19J
Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10^–19J.
(iii) At temperature, T= 10⁷K
Average thermal energy

= 2.07 × 10^–16J
Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10^–16J.
Yes.All contain the same number of the respective molecules.
No. The root mean square speed of neon is the largest.
Since the three vessels have the same capacity, they have the same volume.
Hence, each gas has the same pressure, volume, and temperature.
According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro’s number, N= 6.023 × 10²³.
The root mean square speed (v_rms) of a gas of mass m, and temperature T, is given by the relation:

Where,k is Boltzmann constant
Forthe given gases, kand T are constants.
Hencev_rmsdepends only on the mass of the atoms, i.e.,

Therefore, the root mean square speed of the molecules in the three cases is not the same. Among neon, chlorine, and uranium hexafluoride, the mass of neon is the smallest. Hence, neon has the largest root mean square speed among the given gases.
Temperatureof the helium atom, T_He= –20°C= 253 K
Atomic mass of argon, M_Ar= 39.9 u
Atomic mass of helium, M_He= 4.0 u
Let,(v_rms)_Arbe the rms speed of argon.
Let(v_rms)_Hebe the rms speed of helium.
The rms speed of argon is given by:
…(i)
Where,
Ris the universal gas constant
T_Aris temperature of argon gas
The rms speed of helium is given by:
…(ii)
It is given that:
(v_rms)_Ar=(v_rms)_He

= 2523.675 = 2.52 × 10³K
Therefore, the temperature of the argon atom is 2.52 × 10³K.
Mean free path = 1.11 × 10^–7m
Collision frequency = 4.58 × 10⁹s^–1
Successive collision time ≈500 × (Collision time)
Pressure inside the cylinder containing nitrogen, P= 2.0 atm = 2.026 × 10⁵Pa
Temperature inside the cylinder, T= 17°C =290 K
Radius of a nitrogen molecule, r= 1.0 Å = 1 × 10¹⁰m
Diameter,d = 2 × 1 × 10¹⁰= 2 × 10¹⁰m
Molecular mass of nitrogen, M= 28.0 g = 28 × 10^–3kg
The root mean square speed of nitrogen is given by the relation:

Where,
Ris the universal gas constant = 8.314 J mole^–1K^–1
= 508.26 m/s
The mean free path (l) is given by the relation:

Where,
kis the Boltzmann constant = 1.38 × 10^–23kg m²s^–2K^–1

= 1.11 × 10^–7m
Collision frequency
= 4.58 × 10⁹s^–1
Collision time is given as:

= 3.93 × 10^–13s
Time taken between successive collisions:

= 2.18 × 10^–10s
∴
Hence, the time taken between successive collisions is 500 times the time taken for a collision.
Lengthof the narrow bore, L= 1 m = 100 cm
Length of the mercury thread, l= 76 cm
Length of the air column between mercury and the closed end, l_a= 15 cm
Since the boreis held vertically in air with the open end at the bottom, the mercury length that occupies the air space is: 100 – (76 + 15) = 9 cm
Hence, the total length of the air column = 15 + 9 = 24 cm
Leth cm of mercury flow out as a result of atmospheric pressure.
∴Length of the air column in the bore= 24 + hcm
And, length of the mercury column = 76 – hcm
Initial pressure, P₁= 76 cm of mercury
Initial volume, V₁= 15 cm³
Final pressure, P₂= 76 – (76 – h) = h cm of mercury
Final volume, V₂= (24 + h) cm³
Temperature remains constant throughout the process.
∴P₁V₁= P₂V₂
76 × 15 = h (24 + h)
h²+ 24h –1140 = 0

= 23.8 cm or –47.8 cm
Height cannot be negative. Hence, 23.8 cm of mercurywill flow out from the boreand 52.2 cm of mercury will remain in it. The length of the air column will be 24 + 23.8 = 47.8 cm.
Rate of diffusion of hydrogen, R₁= 28.7 cm³s^–1
Rate of diffusion of another gas, R₂= 7.2 cm³s^–1
According to Graham’s Law of diffusion, we have:

Where,
M₁is the molecular mass of hydrogen = 2.020 g
M₂is the molecular mass of the unknown gas

= 32.09 g
32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.
According to the law of atmospheres, we have:
n₂= n₁exp [-mg(h₂–h₁)/kBT] …(i)
Where,
n₁is thenumber density at heighth₁, and n₂is the number density at height h₂
mg is the weight of the particle suspended in the gas column
Density of the medium = ρ'
Density of the suspended particle = ρ
Mass of one suspended particle = m'
Mass of the medium displaced = m
Volume of a suspended particle = V
According to Archimedes’principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as:
Weight of the medium displaced – Weight of the suspended particle
=mg– m'g

Gas constant, R =k_BN
…(iii)
Substituting equation (ii) in place of mg in equation (i) and then using equation (iii), we get:
n₂= n₁exp [-mg(h₂–h₁)/k_BT]
= n₁exp [-(h₂–h₁)]
= n₁exp [-(h₂–h₁)]
Substance

Radius (Å)

Carbon (diamond)

Gold

Nitrogen (liquid)

Lithium

Fluorine (liquid)

1.29

1.59

1.77

1.73

1.88

Atomic mass of a substance = M
Density of the substance = ρ
Avogadro’s number = N= 6.023 × 10²³
Volume of each atom
Volume of Nnumber of molecules N …(i)
Volume of one mole of a substance = … (ii)
N=

For carbon:
M= 12.01 × 10^–3kg,
ρ =2.22 × 10³kg m^–3
= 1.29 Å
Hence,the radius of a carbon atom is 1.29 Å.
For gold:
M= 197.00 × 10^–3kg
ρ =19.32 × 10³kg m^–3
= 1.59 Å
Hence,the radius of a gold atom is 1.59 Å.
For liquid nitrogen:
M= 14.01 × 10^–3kg
ρ =1.00 × 10³kg m^–3
= 1.77 Å
Hence,the radius of a liquid nitrogen atom is 1.77 Å.
For lithium:
M= 6.94 × 10^–3kg
ρ =0.53 × 10³kg m^–3
= 1.73 Å
Hence,the radius of a lithium atom is 1.73 Å.
Forliquid fluorine:
M= 19.00 × 10^–3kg
ρ =1.14 × 10³kg m^–3
= 1.88 Å
Hence,the radius of a liquid fluorine atom is 1.88 Å.