Kinetic Theory Question Answers: NCERT Class 11 Physics

Diameter of an oxygen molecule, d= 3Å

Radius, r = d /2 = 3/2 = 1.5 Å = 1.5 × 10^-8 cm

Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm³

Molecular volume of oxygen gas, V  =  4/3 π r³ . N

Where, N is Avogadro's number = 6.023 × 10²³ molecules/mole

∴  V  =  4/3 x 3.14 x (1.5 × 10^-8 )³  x 6.023 × 10²³

= 8.51 cm³

Ratio of the molecular volume to the actual volume of oxygen = 8.51 / 22400 =  3.8 × 10^-4

Length of 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

Let h 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

∴ h  =  - 24 +- underroot [(24)²  + 4 x 1 x 1140]  / 2 x 1

= 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:

R₁ / R₂ = Underroot M₂/M₁

Where, M₁ is the molecular mass of hydrogen = 2.020 g 

M₂ is the molecular mass of the unknown gas

∴ M₂  =M₁ (R₁ / R₂)²

=2.02 (28.7 / 7.2)²

= 32.09 g

32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.

The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as:

PV= nRT

Where, R is 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 × 10⁵ Nm^-2

∴  V  =  nRT / P

= 1 x 8.314 x 273  /  1.013 x 10⁵

= 0.0224 m³ = 22.4 litres

Hence, the molar volume of a gas at STP is 22.4 litres.

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 + dpg

Where, p 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:  P₁V₁ / T₁  = P₂V₂ / T₂

Where, V₂ is the volume of the air bubble when it reaches the surface

 V₂  =  P₁V₁T₂  /  T₁P₂

= 493300 x (1.0 x 10^-6) 308  /  285 x 1.013 x 10⁵

= 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 × 10⁵ Pa

The ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T) can be written as:

PV = k_BNT

Where, K_B is Boltzmann constant = 1.38 × 10^-23 m² kg s^-2K^-1

N is the number of air molecules in the room

∴ N  =  PV / K_B T

= 1.013 x 10⁵ x 25 /  1.38 x 10^-23 x 300

= 6.11 × 10²⁶ molecules

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 = 3/2 kT

Where k is Boltzmann constant = 1.38 × 10^-23m² kg s^-2K^-1

∴ 3/2 kT = 3/2 x 1.38 x 10^-38 x 300

= 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  =  3/2 kT

= 3/2 x 1.38 x 10^-38 x 6000

= 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 = 3/2 kT

= 3/2 x 1.38 x 10^-38 x 10⁷

= 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:

v_rms  =  underroot  3kT / m

Where, k is Boltzmann constant

For the given gases, k and T are constants.

Hence v_rmsdepends only on the mass of the atoms, i.e.,

v_rms  ∝  underroot 1/m

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.