Kinetic Theory of Gases — JEE Main Questions

1. In the kinetic theory of gases, pressure is proportional to the mean square speed, while mean free path depends on the average speed. This difference arises because:

1. pressure involves momentum transfer per collision, which depends on the square of velocity, while collision frequency depends on relative speed, which averages to √2 times the average speed.
2. pressure involves the average of the square of velocity because force is proportional to velocity squared, while mean free path uses average speed because it is defined as the average distance between collisions.
3. pressure uses root-mean-square speed because it is derived from the ideal gas law, while mean free path uses average speed because it is derived from the Maxwell-Boltzmann distribution.
4. pressure uses mean square speed because the number of collisions per second with the wall is proportional to velocity, while mean free path uses average speed because the collision cross-section is independent of speed.

Answer: pressure involves momentum transfer per collision, which depends on the square of velocity, while collision frequency depends on relative speed, which averages to √2 times the average speed.

Pressure derivation: momentum transfer per collision = 2mv_x, collision rate = v_x/(2L), so force ∝ mv_x², giving pressure ∝ ⟨v²⟩. Mean free path: collision frequency depends on relative speed; averaging over Maxwell-Boltzmann distribution gives ⟨v_rel⟩ = √2 v_avg, so λ = 1/(√2 nπd²). Thus, pressure uses mean square speed, mean free path uses average speed via relative speed.

2. Which of the following statements about degrees of freedom is correct?

1. A degree of freedom is a dependent coordinate needed to specify the molecule's state.
2. A diatomic gas at room temperature has 7 degrees of freedom.
3. A monatomic gas has 3 translational degrees of freedom.
4. Rotational degrees of freedom for a diatomic molecule are 3.

Answer: A monatomic gas has 3 translational degrees of freedom.

A monatomic gas has only translational motion, so it has 3 degrees of freedom (x, y, z). At room temperature, a diatomic gas has 5 degrees of freedom (3 translational + 2 rotational). A degree of freedom is an independent coordinate, not dependent. For a diatomic molecule, rotation about the bond axis is negligible, so only 2 rotational degrees exist.

3. In the derivation of mean free path, why is a factor of √2 introduced?

1. Because the average speed of molecules is √2 times the relative speed
2. Because the number density is multiplied by √2
3. Because the average relative speed of molecules is √2 times the average speed
4. Because the molecular diameter is divided by √2

Answer: Because the average relative speed of molecules is √2 times the average speed

The mean free path derivation uses the average relative speed between molecules. For a Maxwellian gas, the average relative speed v_rel = √2 v_avg, where v_avg is the average speed of a single molecule. This factor arises because all molecules move, not just one. Substituting v_rel into the collision cylinder argument gives λ = 1/(√2 π d² n).

4. A monatomic gas molecule has how many degrees of freedom?

Answer: 3.0

A monatomic gas molecule (like He or Ne) is a single atom. It has only three translational degrees of freedom (motion along x, y, z axes). Rotational degrees are negligible because the moment of inertia about any axis through the atom is nearly zero. Vibration is absent. Hence f = 3.

5. A linear polyatomic molecule like CO₂ has 3 translational, 2 rotational, and 2 vibrational degrees of freedom at high temperature. What is the total number of degrees of freedom?

Answer: 7.0

At high temperature, all degrees of freedom are active. For a linear polyatomic molecule, translational = 3, rotational = 2 (rotation about bond axis negligible), and vibrational = 2 (one vibrational mode contributes 2 quadratic terms). Total f = 3 + 2 + 2 = 7.

6. In the Maxwell speed distribution for an ideal gas, which of the following correctly orders the three characteristic speeds?

1. v_avg < v_mp < v_rms
2. v_rms < v_avg < v_mp
3. v_mp < v_avg < v_rms
4. v_mp < v_rms < v_avg

Answer: v_mp < v_avg < v_rms

The Maxwell distribution is skewed to the right. The most probable speed v_mp is the peak, the average speed v_avg is slightly higher due to the tail, and the rms speed v_rms is highest because it weights higher speeds more. Hence v_mp < v_avg < v_rms.

7. An ideal gas is compressed isothermally to half its volume. What happens to its mean free path λ and collision frequency ν?

1. λ halves, ν doubles
2. λ doubles, ν halves
3. λ unchanged, ν doubles
4. λ halves, ν unchanged

Answer: λ halves, ν doubles

For isothermal compression, temperature T constant. Number density n doubles (since volume halves). Mean free path λ = 1/(√2 π d² n) so λ halves. Collision frequency ν = v_avg/λ; v_avg depends only on T (constant), so ν doubles because λ halves.

8. The mean free path of a gas at temperature T and pressure P is λ. If the temperature is doubled at constant pressure, what is the new mean free path?

1. λ/2
2. λ
3. 4λ
4. 2λ

Answer: 2λ

The mean free path is λ = k_B T / (√2 π d² P). At constant pressure P, λ ∝ T. So if T is doubled, λ becomes 2λ. This is because at constant pressure, heating the gas reduces number density n, increasing the average distance between collisions.

9. According to the law of equipartition of energy, what is the average energy per molecule associated with each quadratic degree of freedom?

1. (1/2) k_B T
2. k_B T
3. (3/2) k_B T
4. R T

Answer: (1/2) k_B T

The law of equipartition states that each quadratic term in the energy expression contributes an average energy of (1/2)k_B T per molecule. For example, translational kinetic energy has three quadratic terms, giving total (3/2)k_B T.

10. An ideal gas undergoes isochoric heating so that its temperature doubles. What happens to its mean free path λ?

1. λ doubles
2. λ remains unchanged
3. λ becomes half
4. λ becomes four times

Answer: λ remains unchanged

For an ideal gas, λ = 1/(√2 π d² n). In isochoric heating, volume is constant, so number density n = N/V is constant. Hence λ remains unchanged. Alternatively, λ ∝ T/P, and at constant volume P ∝ T, so T/P constant, λ constant.

11. At the same temperature, which statement about the average translational kinetic energy of ideal gases is correct?

1. 1 g of H2 has the same KE as 1 g of O2.
2. Both (c) and (d) are correct.
3. 1 molecule of H2 has the same KE as 1 molecule of O2.
4. 1 mole of H2 has the same KE as 1 mole of O2.

Answer: Both (c) and (d) are correct.

Average translational KE per molecule = (3/2)k_B T, independent of mass, so (c) is correct. Per mole, KE = (3/2)RT, also independent of gas, so (d) is correct. Hence both (c) and (d) are correct, making (a) the right choice.

12. At room temperature, the number of degrees of freedom of an oxygen molecule is:

Answer: 5.0

Oxygen (O₂) is a diatomic molecule. At room temperature, it has 3 translational and 2 rotational degrees of freedom (rotation about two axes perpendicular to the bond). The vibrational mode is frozen. So total f = 5.