Kinetic Interpretation Temperature

Answer: A. The kinetic interpretation of temperature gives ⟨KE⟩ = (3/2)kT per molecule, where k is the Boltzmann constant (NCERT Class 11 Physics, Chapter 12, page 250).

Why B is wrong: B gives (1/2)kT, which is the energy per degree of freedom, not the total translational KE (which has 3 translational degrees of freedom, giving 3 × (1/2)kT = (3/2)kT).

Why C is wrong: C gives (3/2)RT, which is the translational KE per mole, not per molecule. The stem asks for the energy of a single molecule, requiring Boltzmann constant k, not the gas constant R.

Why D is wrong: D gives (5/2)kT, which corresponds to 5 degrees of freedom (a rigid diatomic molecule's total energy including rotational). The question asks for translational KE only, which involves 3 translational degrees of freedom.

Answer: C. k = R/Nₐ, where Nₐ is Avogadro's number. This follows from PV = nRT = NkT, giving R = Nₐk (NCERT Class 11 Physics, Chapter 12, page 250).

Why A is wrong: A gives k = RNₐ, which inverts the relationship. Since R is the per-mole constant and k is the per-molecule constant, k must be smaller than R, not larger.

Why B is wrong: B gives k = Nₐ/R, which has incorrect dimensions. k has units J/K, R has units J/(mol·K), so k = R/Nₐ is the dimensionally consistent relation.

Why D is wrong: D gives k = R²/Nₐ, which has units J²·mol/(mol·K²·mol) — dimensionally incorrect for a quantity measured in J/K.

Answer: B. ⟨KE⟩ = (3/2)kT depends only on temperature, not on molecular mass. At thermal equilibrium, all ideal gas species share the same average translational KE (NCERT Class 11 Physics, Chapter 12, page 250).

Why A is wrong: A incorrectly assumes heavier molecules carry more KE. While heavier molecules move slower (lower v_rms), their average translational KE is (3/2)kT — mass-independent.

Why C is wrong: C incorrectly assumes lighter molecules carry more KE. Lighter molecules do move faster, but the product (1/2)mv² averages to the same (3/2)kT for all species at the same T.

Why D is wrong: D is wrong because monatomic gas molecules do possess translational KE. In fact, translational motion is the only kinetic energy mode for monatomic gases, and it equals (3/2)kT per molecule.

Answer: C. ⟨KE⟩ = (3/2)kT = (3/2)(1.38 × 10⁻²³)(300) = (3/2)(4.14 × 10⁻²¹) = 6.21 × 10⁻²¹ J.

Why A is wrong: A gives (1/2)kT = 2.07 × 10⁻²¹ J, the energy per single degree of freedom, not the total translational KE (which sums over 3 translational degrees).

Why B is wrong: B gives (1)(kT) = 4.14 × 10⁻²¹ J, which omits the (3/2) factor. The translational KE has 3 degrees of freedom, each contributing (1/2)kT.

Why D is wrong: D gives 6.21 × 10⁻²³ J, which is off by a factor of 100 — an arithmetic error in handling the powers of 10 during multiplication.

Answer: D. At thermal equilibrium, average translational KE = (3/2)kT for every molecule regardless of species. The ratio is 1:1.

Why A is wrong: A incorrectly uses the molar mass ratio (4:40 = 1:10), confusing KE with speed. Lighter molecules have higher v_rms but the same average translational KE at thermal equilibrium.

Why B is wrong: B inverts the molar mass ratio, assuming lighter means more energetic. While helium atoms move faster, (1/2)mv² averages to the same (3/2)kT.

Why C is wrong: C uses the square-root of the mass ratio, which would be relevant for comparing v_rms values (v_rms ∝ 1/√M), not kinetic energies. The KE ratio at the same T is always 1:1.

Answer: B. Total translational KE = N × (3/2)kT = nNₐ × (3/2)kT = n × (3/2)RT. For n = 2 moles: 2 × (3/2)RT = 3RT.

Why A is wrong: A gives the translational KE of a single molecule, not the total for 2 moles. Must multiply by the total number of molecules (2Nₐ).

Why C is wrong: C gives (3/2)RT, which is the total translational KE for 1 mole. The question specifies 2 moles, so the answer must be doubled.

Why D is wrong: D gives 3kT. This confuses k (per-molecule) with R (per-mole) and also mishandles the factor of 2 moles. The correct per-molecule total for 2 moles uses R, not k.

Answer: A. Average translational KE = (3/2)kT. It depends only on temperature, not on molecular species or pressure. At the same T, both gases have equal average translational KE per molecule.

Why B is wrong: B incorrectly links lighter mass to higher KE. H₂ molecules move faster, but their average translational KE is (3/2)kT — identical to O₂ at the same temperature.

Why C is wrong: C incorrectly links heavier molecular mass to higher KE. The mass difference affects v_rms (O₂ moves slower), not the average translational KE, which is (3/2)kT for both.

Why D is wrong: D is wrong because (3/2)kT contains no pressure term. Pressure affects macroscopic properties like PV, but the average translational KE per molecule at a fixed T is pressure-independent.

Answer: A. (3/2)kT = 1.0 eV = 1.6 × 10⁻¹⁹ J. So T = (2 × 1.6 × 10⁻¹⁹) / (3 × 1.38 × 10⁻²³) = 3.2 × 10⁻¹⁹ / 4.14 × 10⁻²³ = 7.73 × 10³ K.

Why B is wrong: B uses (1)kT = 1.6 × 10⁻¹⁹ instead of (3/2)kT, giving T = 1.6 × 10⁻¹⁹ / 1.38 × 10⁻²³ ≈ 1.16 × 10⁴ K. The correct formula has the 3/2 factor.

Why C is wrong: C doubles the answer from option B, compounding the error of omitting the 3/2 factor with a spurious factor of 2.

Why D is wrong: D is half the correct value: it writes 3kT instead of (3/2)kT, so T = 1.6 × 10⁻¹⁹ / (3 × 1.38 × 10⁻²³) ≈ 3.87 × 10³ K. Using (1/2)kT would instead give 2.32 × 10⁴ K (option C).