NEET Kinetic Theory Previous Year Questions with Solutions

For an ideal gas with spherical molecules of diameter d and number density n, the mean free path lambda is given by:

• • The sqrt(2) factor accounts for the relative motion of colliding molecules.
• • pi * d^2 represents the effective spherical collision cross-section.

Rearranging the formula to isolate the number density n:

• • Number density is inversely proportional to the mean free path.
• • Number density is inversely proportional to the square of molecular diameter.

Calculate the ratio of number densities for Gas A relative to Gas B:

• • lambda_B / lambda_A = 2 (since lambda_A is half of lambda_B).
• • (d_B / d_A)^2 = (1/2)^2 = 1/4 (since d_A is twice d_B).
• • Multiplying gives n_A / n_B = 1/2 ==> n_A = (1/2) * n_B.

Pressure gauges read pressure relative to ambient atmospheric pressure. Therefore, absolute pressure is the sum of gauge pressure and atmospheric pressure:

• • Failure to add 1 atm leads to 0.156 kg (the exact trap in Option 4).

Using the ideal gas equation P₂ V = n₂ R T with V = 0.03 m³ and T = 300 K:

• • Remaining moles in cylinder: n₂ ≈ 14.544 moles.

The number of moles withdrawn from the cylinder is the difference between initial and final moles:

• • Mass withdrawn: Δm = Δn × Molar Mass = 3.656 × 32 g = 116.99 g ≈ 0.117 kg ≈ 0.116 kg.
• • Matches Option (3) with precision.

For an ideal gas, the equation of state is PV = nRT. Expressing temperature T as a function of volume V:

The graph plots T on the vertical axis and V on the horizontal axis. For an isobaric line representing Charles’s law, the slope is:

Because n and R are constants, the slope of the T-V line is directly proportional to pressure P. A steeper line corresponds to a higher pressure:

From the given graph, curve P₁ is the steepest (highest slope), curve P₂ has intermediate slope, and curve P₃ has the lowest slope (gentlest inclination):

Gas law relations must strictly be calculated in Kelvin (absolute temperature):

Root mean square velocity of an ideal gas molecule of molar mass M is given by:

The velocity increases BY 3 times its initial value, which makes the new velocity 4 times the original:

Square the velocity ratio to find the final absolute temperature, then convert back to Celsius:

Convert back to degrees Celsius:

From kinetic theory of gases, the root mean square speed is directly proportional to the square root of absolute temperature:

Derived by considering rate of change of momentum of gas molecules bombarding container walls, where $n$ is number density, $m$ is molecular mass, and $\bar{v}^2$ is mean square speed:

By the equipartition of energy theorem, each translational degree of freedom contributes $\frac{1}{2} k_B T$. For 3 translational degrees of freedom of a single molecule:

A rigid diatomic gas has 5 degrees of freedom (3 translational + 2 rotational). For 1 mole ($n = 1$), the internal energy is:

Starting with the ideal gas law: $$P V = n R T = \left(\frac{m}{M}\right) R T$$ Rearranging for density $$\rho = \frac{m}{V}$$: $$\rho = \frac{P M}{R T}$$

- Absolute Temperature: $$T = 27^\circ\text{C} + 273 = 300\text{ K}$$ - Pressure: $$P = 249\text{ kPa} = 249 \times 10^3\text{ Pa}$$ - Universal Gas Constant: $$R = 8.3\text{ J mol}^{-1}\text{K}^{-1}$$ - Molar mass of hydrogen gas ($$\text{H}_2$$): $$M = 2\text{ g mol}^{-1} = 2 \times 10^{-3}\text{ kg mol}^{-1}$$

$$\rho = \frac{(249 \times 10^3\text{ N m}^{-2}) \times (2 \times 10^{-3}\text{ kg mol}^{-1})}{(8.3\text{ J mol}^{-1}\text{K}^{-1}) \times (300\text{ K})}$$ The factors of $$10^3$$ and $$10^{-3}$$ in the numerator cancel: $$\text{Numerator} = 249 \times 2 = 498\text{ kg m}^{-2}$$ $$\text{Denominator} = 8.3 \times 300 = 2490\text{ J K}^{-1} = 2490\text{ N m}$$ $$\rho = \frac{498}{2490} = \frac{1}{5} = 0.2\text{ kg m}^{-3}$$

The calculated density is exactly $$0.2\text{ kg/m}^3$$, matching Option (2).

The mean free path ($$\lambda$$) is the average distance traveled by a gas molecule between two successive collisions with other gas molecules.

Suppose a molecule of diameter $$d$$ moves with speed $$v$$ while all other molecules are assumed stationary. A collision occurs whenever the center of another molecule lies within a distance $$d$$. In time $$\Delta t$$, it sweeps a collision cylinder of volume: $$V_{\text{coll}} = \pi d^2 (v \Delta t)$$ With number density $$n$$, the number of collisions is $$N_{\text{coll}} = n \pi d^2 v \Delta t$$. The mean free path is the total distance divided by number of collisions: $$\lambda = \frac{v \Delta t}{N_{\text{coll}}} = \frac{1}{n \pi d^2}$$

In reality, all molecules are in random thermal motion. The effective collision frequency depends on the average relative speed $$\langle v_{\text{rel}} \rangle$$. For a Maxwell-Boltzmann velocity distribution: $$\langle v_{\text{rel}} \rangle = \sqrt{2} \langle v \rangle$$ Substituting this relative velocity into the collision rate introduces a factor of $$\sqrt{2}$$ into the denominator: $$\lambda = \frac{1}{\sqrt{2} n \pi d^2}$$

NCERT states verbatim in the Kinetic Theory chapter: "$$\lambda = \frac{1}{\sqrt{2} n \pi d^2}$$". This corresponds uniquely to Option (2).

A monoatomic gas molecule (such as Helium, Neon, or Argon) behaves as a point mass. It can move freely in 3 independent spatial directions (along the x, y, and z Cartesian axes), giving it 3 translational degrees of freedom ($$f = 3$$). It has negligible moment of inertia about its own center, so rotational degrees of freedom are absent.

According to Boltzmann's Law of Equipartition of Energy, in thermal equilibrium at absolute temperature $$T$$, the average energy associated with each independent degree of freedom per molecule is: $$E_{\text{per mode}} = \frac{1}{2} k_B T$$, where $$k_B$$ is Boltzmann's constant.

Multiplying the energy per mode by the total number of degrees of freedom ($$f = 3$$): $$E_{\text{avg}} = 3 \times \left(\frac{1}{2} k_B T\right) = \frac{3}{2} k_B T$$.

Therefore, the average thermal energy of a monoatomic gas atom is $$\frac{3}{2} k_B T$$, matching Option 2.