Equation of State Perfect Gas

Answer: A. R is the universal gas constant with SI value 8.314 J mol⁻¹ K⁻¹, as defined in NCERT Class 11 Physics Chapter 12, page 246.

Why B is wrong: B confuses R (molar gas constant) with k (Boltzmann constant). k = 1.38 × 10⁻²³ J K⁻¹ appears in the molecular form PV = NkT, not the molar form.

Why C is wrong: C has the correct name but drops the per-mole unit. R has dimensions of energy per mole per kelvin, so the unit must include mol⁻¹.

Why D is wrong: D mislabels R as the Boltzmann constant. Boltzmann's constant k and gas constant R are related by R = Nₐk but are distinct quantities.

Answer: B. R = Nₐk, where Nₐ is Avogadro's number. This connects the molar form (nRT) to the molecular form (NkT), per NCERT Class 11 Physics Chapter 12, page 246.

Why A is wrong: A inverts the relationship. R = k/Nₐ would make R smaller than k, but R ≈ 8.314 J mol⁻¹ K⁻¹ is far larger than k ≈ 1.38 × 10⁻²³ J K⁻¹.

Why C is wrong: C gives a dimensionally inconsistent expression. Nₐ/k has units of mol⁻¹/(J K⁻¹) = K mol⁻¹ J⁻¹, which does not match R's units of J mol⁻¹ K⁻¹.

Why D is wrong: D rearranges incorrectly. From R = Nₐk, solving for k gives k = R/Nₐ, not k = NₐR.

Answer: B. At low pressure and high temperature, intermolecular forces are negligible and molecular volume is small compared to container volume, so the ideal gas approximation holds. NCERT Class 11 Physics Chapter 12, page 246.

Why A is wrong: A describes conditions where real-gas deviations are largest — molecules are close together (high P) and intermolecular attractions are significant (low T).

Why C is wrong: C has high temperature (good) but high pressure increases the ratio of molecular volume to container volume, causing deviation from ideality.

Why D is wrong: D is partially correct (low pressure) but low temperature increases intermolecular attraction, pushing the gas away from ideal behavior.

Answer: D. T = PV/(nR) = (1.0 × 10⁵ × 0.050)/(2 × 8.314) = 5000/16.628 ≈ 301 K. Direct substitution into PV = nRT.

Why A is wrong: A uses n = 1 instead of n = 2, giving T = 5000/8.314 ≈ 602 K. Always check the number of moles before substituting.

Why B is wrong: B is a rounding trap — the exact calculation gives ≈ 300.7 K, which rounds to 301 K, not 300 K. In NEET, pick the closest option.

Why C is wrong: C likely results from a unit or arithmetic error (dividing by an extra factor). Verify each substitution step carefully.

Answer: A. A is correct. At constant volume, P₁/T₁ = P₂/T₂. T₁ = 27 + 273 = 300 K and P₂ = 2P₁, so T₂ = 2 × 300 = 600 K. The options here are asked in Celsius, so convert back: 600 − 273 = 327 °C.

Why B is wrong: B is wrong because 600 is the answer in KELVIN, not in Celsius. Doubling the absolute temperature gives 600 K; reporting that same number with a °C label skips the conversion back, and 600 °C would be 873 K — nearly three times the starting temperature.

Why C is wrong: C doubles the Celsius value (2 × 27 = 54 °C) instead of converting to kelvin first. This is the classic temperature-conversion trap: PV = nRT requires absolute temperature.

Why D is wrong: D equals T₁ (the initial temperature in kelvin), not T₂. The gas was heated, so the final temperature must be higher than 300 K.

Answer: C. At constant pressure, V ∝ T (Charles's law from PV = nRT). Tripling T triples V. V₂ = 3V.

Why A is wrong: A gives V/3, which would apply if volume were inversely proportional to temperature. At constant P, the relationship is direct, not inverse.

Why B is wrong: B gives 9V, which would apply if V ∝ T². Volume is linearly proportional to T at constant P, not quadratically.

Why D is wrong: D gives V√3, which would apply if V ∝ √T. That scaling applies to RMS speed, not volume. At constant P, V scales linearly with T.

Answer: C. Using P₁V₁/T₁ = P₂V₂/T₂ with V₂ = V₁/2 and T₂ = 200 K: P₂ = P₁ × (V₁/V₂) × (T₂/T₁) = 2.0 × 10⁵ × 2 × (200/400) = 2.0 × 10⁵ Pa.

Why A is wrong: A applies only the temperature change (halving T halves P) but ignores the volume change. Both variables change simultaneously.

Why B is wrong: B applies only the volume change (halving V doubles P) but ignores the temperature change. The cooling exactly cancels the compression effect.

Why D is wrong: D squares the volume factor (2² = 4) and multiplies by the temperature ratio incorrectly. The combined gas law uses linear ratios, not squared ones.

Answer: D. Step 1 (constant P): V₂ = V₁ × T₂/T₁ = 0.030 × (400/300) = 0.040 m³. Step 2 (constant T = 400 K): P₃ = P₂ × V₂/V₃ = 1.0 × 10⁵ × (0.040/0.030) = 1.33 × 10⁵ Pa.

Why A is wrong: A assumes the gas returns to its initial state. While the volume returns to 0.030 m³, the temperature is now 400 K (not 300 K), so the pressure must be higher than the initial value.

Why B is wrong: B inverts the volume ratio in the isothermal step, computing P₂ × (V₃/V₂) = 1.0 × 10⁵ × (0.030/0.040) = 0.75 × 10⁵ Pa. The correct form is P₃ = P₂ × V₂/V₃.

Why C is wrong: C likely results from using T in °C instead of K somewhere in the two-step calculation. Always convert: 27 °C = 300 K, 127 °C = 400 K.