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Specific heat capacities, mean free path and Avogadro number: NEET MCQs with answers

49 questions on this topic of Kinetic Theory of Gases (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A cylinder contains 1 mole of a rigid diatomic gas at temperature T. Heat is supplied slowly to the gas such that it expands at constant pressure. The ratio of the increase in internal energy (ΔU) to the work done by the gas (W) during this process is:

  1. 2 / 5
  2. 5 / 2
  3. 7 / 2
  4. 2 / 7
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Answer: (B) 5 / 2

For a rigid diatomic gas, Cv = (5/2)R and Cp = (7/2)R. At constant pressure, ΔU = n·Cv·ΔT = (5/2)R·ΔT and W = P·ΔV = n·R·ΔT = R·ΔT. Thus, the ratio ΔU / W = [(5/2)R·ΔT] / [R·ΔT] = 5/2.

medium

2. A container of fixed volume contains an ideal gas at 27 °C. If the gas is heated to 927 °C, the average collision frequency of a molecule with other molecules changes by a factor of:

  1. 2
  2. 1
  3. 4
  4. 16
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Answer: (A) 2

The collision frequency is given by ν = v_avg / λ. In a fixed volume, number density n is constant, so λ is constant. Since v_avg is proportional to √T, ν is proportional to √T. T increases from 300 K to 1200 K (a factor of 4), so the collision frequency increases by a factor of √4 = 2.

hard

3. Two moles of an ideal monatomic gas (γ = 5/3) are mixed with three moles of an ideal diatomic gas (rigid, γ = 7/5). The effective adiabatic exponent γ_mix of the mixture is:

  1. 1.40
  2. 1.53
  3. 1.48
  4. 1.60
Show answer and explanation

Answer: (C) 1.48

For the monatomic gas, Cv1 = 1.5R and Cp1 = 2.5R (n₁ = 2). For the diatomic gas, Cv2 = 2.5R and Cp2 = 3.5R (n₂ = 3). Total Cv = [2(1.5R) + 3(2.5R)] / 5 = (3R + 7.5R) / 5 = 10.5R / 5 = 2.1R. Total Cp = Cv + R = 3.1R. Therefore, γ_mix = Cp / Cv = 3.1 / 2.1 = 31 / 21 ≈ 1.476 ≈ 1.48.

medium

4. An ideal gas is enclosed in a rigid cylinder at temperature T. If the collision frequency of a molecule with other molecules is ν, how does ν depend on temperature T?

  1. ν ∝ T
  2. ν ∝ √T
  3. ν ∝ T²
  4. ν is independent of T
Show answer and explanation

Answer: (B) ν ∝ √T

Collision frequency ν = v_avg / λ. In a rigid container of fixed volume, the mean free path λ is constant because number density n is constant. Since average speed v_avg ∝ √T, it follows that ν ∝ √T.

easy

5. At a given absolute temperature T and pressure P, the mean free path of gas molecules is λ. If both the absolute temperature and pressure of the gas are doubled, the new mean free path will be:

  1. λ / 2
  2. 4λ
  3. 2λ
  4. λ
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Answer: (D) λ

The mean free path of gas molecules is given by λ = kT / (√2 π d² P). Since both T and P are doubled, the ratio T/P remains unchanged. Therefore, the mean free path remains equal to λ.

hard

6. An ideal gas expands along the path P = αV, where α is a positive constant. If the gas is monatomic, what is the molar heat capacity of the gas for this process?

  1. 3 R
  2. R
  3. 2 R
  4. (5/2) R
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Answer: (C) 2 R

The process equation is P V⁻¹ = constant, which is a polytropic process P V^x = constant with x = -1. The molar heat capacity is C = Cv + R / (1 - x). For a monatomic gas, Cv = (3/2)R. Substituting x = -1 gives C = (3/2)R + R / (1 - (-1)) = (3/2)R + R/2 = 2R.

easy

7. If NA is the Avogadro number, kB is the Boltzmann constant, and m is the mass of a single molecule of an ideal gas having molar mass M, which of the following relations is correct?

  1. NA = m × M
  2. kB = M × NA
  3. M = m × NA
  4. kB = m / NA
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Answer: (C) M = m × NA

By definition, the molar mass M is the mass of one mole of substance, which contains Avogadro's number NA of molecules. Therefore, M = m × NA.

easy

8. Consider the following statements regarding the mean free path of gas molecules: Statement I: The mean free path is inversely proportional to the square of the molecular diameter. Statement II: In a closed vessel of fixed volume, the mean free path increases when the gas is heated. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because λ = 1 / (√2 π n d²), so λ ∝ 1/d². Statement II is incorrect because in a closed container of fixed volume, the number density n = N/V is constant, so heating the gas does not change the mean free path.

easy

9. For an ideal gas, the difference between molar specific heats Cp - Cv equals:

  1. Zero
  2. R
  3. 2R
  4. R/2
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Answer: (B) R

Mayer's relation Cp - Cv = R follows from the first law: at constant pressure, extra heat is needed to do work PΔV = RΔT per mole.

NCERT Class 11

medium

10. An ideal gas has an adiabatic index γ = 1.4. The molar heat capacity of this gas in a quasi-static process given by P V² = constant is:

  1. R / 2
  2. 3R / 2
  3. R
  4. 2R
Show answer and explanation

Answer: (B) 3R / 2

For a polytropic process P Vⁿ = constant, C = Cv + R / (1 - n). For γ = 1.4 = 7/5, Cv = R / (γ - 1) = 5R/2. Here n = 2, so C = 5R/2 + R / (1 - 2) = 5R/2 - R = 3R/2.

medium

11. Consider the following two statements regarding Avogadro's number and molar properties of gases: Statement I: Under standard temperature and pressure (STP: 0 °C, 1 atm), one mole of any ideal gas occupies approximately 22.4 litres. Statement II: The value of Avogadro's number depends directly on the ambient temperature and pressure at which the gas is maintained. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are true.
  2. Both Statement I and Statement II are false.
  3. Statement I is true but Statement II is false.
  4. Statement I is false but Statement II is true.
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Answer: (C) Statement I is true but Statement II is false.

Statement I is true because Avogadro's law implies that one mole of any ideal gas occupies 22.4 L at standard temperature and pressure (273.15 K, 1 atm). Statement II is false because Avogadro's number is a universal fundamental constant (approximately 6.022 × 10²³ mol⁻¹) and does not depend on thermodynamic variables such as temperature or pressure.

hard

12. A gas mixture consists of equal masses of Helium (molar mass = 4 g/mol) and Oxygen (molar mass = 32 g/mol). Treating both as ideal gases (rigid molecules), the molar heat capacity at constant volume of the mixture is:

  1. (17/9)R
  2. (37/18)R
  3. (19/12)R
  4. (29/18)R
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Answer: (D) (29/18)R

Let mass of each be m. Moles of He: n1 = m/4; moles of O2: n2 = m/32. For monatomic He, Cv1 = 3R/2; for diatomic O2, Cv2 = 5R/2. Cv_mix = (n1·Cv1 + n2·Cv2) / (n1 + n2) = [(m/4)(3R/2) + (m/32)(5R/2)] / [m/4 + m/32]. Multiplying numerator and denominator by 32/m gives [(8)(3R/2) + (1)(5R/2)] / (8 + 1) = [12R + 2.5R] / 9 = 14.5R / 9 = 29R / 18.

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