NEET Previous Year Questions (2014-2024): Kinetic Theory | Physics Class 11 PDF Download

2023

Q1: The temperature of a gas is –50°C. To what temperature the gas should be heated so that the rms speed is increased by 3 times?     [2023]
(a) 669°C 
(b) 3295°C
(c) 3097 K
(d) 223 K
Ans: (b)  3295°C

Explanation:

The root mean square (RMS) speed v_rms of a gas is given by the equation:

v_rms = √ 3RTm

v_rms ∝ √T

T₁ = 273 - 50 = 223 K

v_rms is increased by 3 times
T₂ = ?

So, final RMS speed = v + 3v = 4v

v4v = √ T₁T₂

√ 116 = √ 223T₂

T₂ = 3568 K

T₂ = 3568 - 273 = 3295°C

2021

Q1: Match column - I and column - II and choose the correct match from the given choices.     [2021]

(a) (A) - (Q), (B) - (P), (C) - (S), (D) - (R)
(b) (A) - (R), (B) - (Q), (C) - (P), (D) - (S)
(c) (A) - (R), (B) - (P), (C) - (S), (D) - (Q)
(d) (A) - (Q), (B) - (R), (C) - (S), (D) - (P)
Ans: (a) (A) - (Q), (B) - (P), (C) - (S), (D) - (R)

Explanation:

1. (A) Root mean square speed of gas molecules:

   Root mean square (RMS) speed of gas molecules is given by the formula:
   v_rms = √ 3RTM

   Hence, it matches with (Q).

2. (B) Pressure exerted by an ideal gas:

   The pressure exerted by an ideal gas can be represented using:
   P = 13 mv²

   Hence, it matches with (P).

3. (C) Average kinetic energy of a molecule:

   The average kinetic energy per molecule is given by:
   E = 32 k_BT

   Hence, it matches with (S).

4. (D) Total internal energy of 1 mole of a diatomic gas:

   The total internal energy for one mole of a diatomic gas is:
   U = 52 RT

   Hence, it matches with (R).

2020

Q1: A cylinder contains hydrogen gas at pressure of 249 kPa and temperature 27°C.
Its density is : (R = 8.3 J mol^-1 K^-1) [2020]
(a) 0.1 kg/m³ 
(b) 0.02 kg/m³ 
(c) 0.5 kg/m³ 
(d) 0.2 kg/m³
Ans: (d) 0.2 kg/m³

Explanation:

Concept:

Ideal Gas Equation:

The ideal gas equation represents the state of a gas and provides a good approximation of its behavior under certain conditions. It is expressed as:

PV = nRT

Where:

• P is the pressure,
• V is the volume,
• n is the number of moles of the gas,
• R is the universal gas constant, and
• T is the temperature.

Using the ideal gas equation, we have:

PV = mM RT     ⟹   ρV = m ⟹ m = ρV

PV = (ρV)RTM

ρ = PMRT

Substituting the values:

ρ = 249 × 10³ × 2 × 10^–38.3 × 300 = 0.2 kg/m³

Q2: The mean free path for a gas, with molecular diameter d and number density n can be expressed as :     [2020]
(a)

(b) 

(c)

(d)

Ans: (d)

Explanation:

The mean free path for gas is given by

Q3: The average thermal energy for a mono-atomic gas is (k_B is Boltzmann constant and T, absolute temperature) [2020]
(a) 1/2k_BT
(b) 3/2k_BT
(c) 5/2k_BT
(d) 7/2k_BT
Ans: (b) 3/2kBT

Explanation:

Thermal Energy

• Thermal energy refers to the energy present within a system that is responsible for its temperature.
• Thermal energy is given by the relation:

ΔE_T = μ_k F d

Where:

• μ_k – Coefficient of friction
• F – Force
• d – Distance

Step 1: Formula for Calculating the Average Thermal Energy

For mono-atomic gas, the total degree of freedom = 3 (translational degrees of freedom), i.e., rotational, vibrational, and translational.

The average thermal energy for a mono-atomic gas (U) is given as:

U = (total degree of freedom) × 12 K_BT

Step 2: Calculating the Average Thermal Energy

By putting the value of the degree of freedom, we get:

U = 32 K_BT

Hence, the average thermal energy for a mono-atomic gas is:

U = 32 K_BT

2019

Q1: Increase in temperature of a gas filled in a container would lead to:    [2019]
(a) Increase in its mass
(b) Increase in its kinetic energy
(c) Decrease in its pressure
(d) Decrease in intermolecular distance
Ans: (b)  Increase in its kinetic energy

Explanation:

(a) Increase in its mass: The mass of the gas remains constant as temperature changes; it is not affected by temperature.

(b) Increase in its kinetic energy: As the temperature of the gas increases, the kinetic energy of the gas molecules increases because the molecules move faster at higher temperatures.

(c) Decrease in its pressure: An increase in temperature would actually lead to an increase in pressure (if the volume is constant), not a decrease.

(d) Decrease in intermolecular distance: The intermolecular distance usually increases when the temperature rises because gas molecules move faster and spread out more.

2018

Q1: At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere ? (Given: Mass of oxygen molecule (m) = 2.76 × 10^–26 kg Boltzmann's constant kB = 1.38 × 10^–23 J K^–1):-    [2018]
(a) 2.508 × 10⁴ K
(b) 8.360 × 10⁴ K
(c) 5.016 × 10⁴ K
(d) 1.254 × 10⁴ K
Ans: (b) 8.360 × 10⁴ K

Explanation:

Mass of oxygen molecule (m) = 2.76 × 10^-26 kg

Boltzmann's constant (k_B) = 1.38 × 10^-23 J·K^-1

Escape velocity (V_escape) = 11200 m/s

Say at temperature T, it attains V_escape.

√ 3k_BTm_O2 = 11200 m/s

On solving,

T = 8.360 × 10⁴ K

2017

Q1: A gas mixture consists of 2 moles of O₂ and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:- [2017]
(a) 15 RT
(b) 9 RT
(c) 11 RT
(d) 4 RT
Ans: (c) 

Explanation:

For O₂ (diatomic gas):

• Oxygen (O₂) is a diatomic gas. For diatomic gases, the internal energy per mole is given by:

  U_O2 = 52 RT per mole

• For 2 moles of O₂:

  U_O2 = 2 × 52 RT = 5RT

For Ar (monatomic gas):

• Argon (Ar) is a monatomic gas. For monatomic gases, the internal energy per mole is given by:

  U_Ar = 32 RT per mole

• For 4 moles of Ar:

  U_Ar = 4 × 32 RT = 6RT

Total Internal Energy of the System:

The total internal energy is the sum of the internal energy of O₂ and Ar:

U_total = U_O2 + U_Ar = 5RT + 6RT = 11RT

Thus, the total internal energy of the system is 11RT.

2016

Q1: The molecules of a given mass of a gas have r.m.s. velocity of 200 ms^-1 at 27°C and 1.0 x 10^-5 Nm^-2 pressure. When the temperature and pressure of the gas are respectively, 127°C and 0.05 x 10⁵Nm², the r.m.s. velocity of velocity of its molecules in ms^-1 is; [2016]
(a) 100 / 3
(b) 100 √2
(c) 400 / √3
(d) 100√2 / 3
Ans: (c)

Explanation:

The root mean square (r.m.s.) velocity (v_rms) of a gas is given by:

v_rms ∝ √T

This means that the r.m.s. velocity is proportional to the square root of the absolute temperature of the gas.

Given that the initial r.m.s. velocity v_rms,i = 200 m/s at an initial temperature T₁ = 27°C = 300 K, we want to find the r.m.s. velocity at the new temperature T₂ = 127°C = 400 K.

Using the relation:

v_rms,2v_rms,i = √ T₂T₁

Substitute the given values:

v_rms,2200 = √ 400300 = √ 43

Thus:

v_rms,2 = 200 × √ 43 = 200 × 2√3 = 400√3 m/s

Q2: A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T.  The mass of each molecule of the gas is m. Which of the following gives the density of the gas ?       [2016]
(a) P/(kT)
(b) Pm/(kT)
(c) P/(kTV)
(d) mkT
Ans: (b)  Pm/(kT)

Explanation:
We know that ideal gas equation is
PV = NkT
Where, P is the pressure
V is the volume
N is the number of molecules
k is the Boltzmann constant
T is the temperature

Now, the mass of the gas sample is given by:

M = N × m

Where m is the mass of one molecule.

The density (ρ) of the gas is:

ρ = MV = N × mV

From the ideal gas law PV = NkT, we can solve for NV (the number density):

NV = PkT

Substituting this into the density equation:

ρ = m × PkT

Thus, the density of the gas is:

ρ = PmkT

2015

Q1: The ratio of the specific heats  in terms of degrees of freedom (n) is given by: [2015]
(a)
(b)
(c)
(d)
Ans:

Explanation:
The ratio of the specific heats γ = C_PC_V is related to the degrees of freedom (n) of the gas molecules.
For a gas:

• C_V(specific heat at constant volume) is given by:

  C_V = n2 R

• C_P (specific heat at constant pressure) is related to C_Vas:

  C_P = C_V + R = n2 R + R = (n + 2)2 R

Now, the ratio:

γ = C_PC_V = (n + 2)2 Rn2 R = n + 2n

Thus, the correct expression for γ is:

γ = 1 + 2n

Q2: Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is  [2015]
(a) 2
(b) 1/2
(c) 2/3
(d) 3/4 
Ans: (d) 3/4

Explanation:
According to an ideal gas equation, the molecular weight of an ideal gas is
M = ρRT / P 

As:

P = ρRTM

where P, T, and ρ are the pressure, temperature, and density of the gas respectively, and R is the universal gas constant.

∴ The molecular weight of A is:

M_A = ρ_ART_AP_A

and that of B is:

M_B = ρ_BRT_BP_B

Hence, their corresponding ratio is:

M_AM_B = ρ_Aρ_B × T_AT_B × P_BP_A1

Here:

• ρ_A/ρ_B = 1.5 = 32
• T_A/T_B = 1
• P_A/P_B = 2

∴ M_AM_B = 32 × 1 × 12 = 34

2014

Q1: The mean free path of molecules of a gas,(radius ‘r’) is inversely proportional to:    [2014] 
(a) r
(b) √r
(c) r³
(d) r²
Ans: (d) r²

Explanation:
The mean free path (λ) of a gas molecule is the average distance a molecule travels between collisions. It is given by the formula:

λ = 1√2 nπd²

Where:

• n is the number density (number of molecules per unit volume).
• d is the diameter of the gas molecule.

Since the diameter d is directly proportional to the radius r of the molecule (d = 2r), the mean free path is inversely proportional to the square of the radius (r²).

Thus, the mean free path is inversely proportional to r².