Specific Heat Capacity and Mean Free Path – Mean free path: NEET MCQs with Answers

10 free questions on Specific Heat Capacity and Mean Free Path – Mean free path for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Mean free path.

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Question 1

Previous-year question
The value of γ (= Cₚ/Cᵥ) for hydrogen, helium and another ideal diatomic gas X (whose molecules are not rigid but have an additional vibrational mode) are respectively [NEET 2019 (Odisha)]
  1. 7/5, 5/3, 9/7
  2. 5/3, 7/5, 9/7
  3. 5/3, 7/5, 7/5
  4. 7/5, 5/3, 7/5
Show answer
Answer: (A) 7/5, 5/3, 9/7
Explanation: H₂ (rigid diatomic, f = 5): γ = 7/5; He (monoatomic, f = 3): γ = 5/3; X with one vibrational mode (f = 7): γ = 9/7.

Question 2

Previous-year question
Two cylinders A and B of equal capacity are connected to each other via a stop cock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stop cock is suddenly opened. The process is [NEET 2020]
  1. adiabatic
  2. isochoric
  3. isobaric
  4. isothermal
Show answer
Answer: (A) adiabatic
Explanation: The insulation prevents any heat exchange, so the free expansion into the evacuated cylinder is adiabatic (and, for an ideal gas, also happens to leave the temperature unchanged).

Question 3

Previous-year question
In which of the following processes, heat is neither absorbed nor released by a system? [NEET 2019]
  1. isothermal
  2. adiabatic
  3. isobaric
  4. isochoric
Show answer
Answer: (B) adiabatic
Explanation: An adiabatic process is by definition one in which no heat crosses the boundary, so ΔQ = 0.

Question 4

Previous-year question
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then [NEET 2016]
  1. Compressing the gas isothermally will require more work to be done
  2. Compressing the gas through adiabatic process will require more work to be done
  3. Compressing the gas isothermally or adiabatically will require the same amount of work
  4. Which of the case requires more work will depend upon the atomicity of the gas
Show answer
Answer: (B) Compressing the gas through adiabatic process will require more work to be done
Explanation: The adiabat is steeper than the isotherm, so for the same halving of volume the area under the adiabatic curve is greater and more work must be done on the gas.

Question 5

Previous-year question
An ideal gas is compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas? [AIPMT 2015]
  1. Isobaric
  2. Isochoric
  3. Isothermal
  4. Adiabatic
Show answer
Answer: (D) Adiabatic
Explanation: For the same volume reduction the adiabat is steepest, so the pressure rises fastest and the area under the curve — and hence the work done on the gas — is greatest.

Question 6

Previous-year question
A monatomic gas at a pressure P, having a volume V expands isothermally to a volume 2V and then adiabatically to a volume 16V. The final pressure of the gas is (take γ = 5/3) [AIPMT 2014]
  1. 64P
  2. 32P
  3. P/64
  4. 16P
Show answer
Answer: (C) P/64
Explanation: After the isothermal step the pressure is P/2. The adiabatic step from 2V to 16V multiplies it by (1/8)^(5/3) = 1/32, giving P/64.

Question 7

Previous-year question
The mean free path for a gas with molecular diameter d and number density n can be expressed as [NEET 2020]
  1. 1/(√2 n π d²)
  2. 1/(√2 n² π d²)
  3. 1/(√2 n² π² d²)
  4. 1/(√2 n π d)
Show answer
Answer: (A) 1/(√2 n π d²)
Explanation: Taking the collision cross-section as πd² and correcting for relative motion by the factor √2, the mean free path is λ = 1/(√2 πd²n).
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Question 8

Previous-year question
The mean free path of the molecules of a gas whose molecular radius is r is inversely proportional to [AIPMT 2014]
  1. r³
  2. r²
  3. r
  4. √r
Show answer
Answer: (B) r²
Explanation: λ = 1/(√2 πd²n) with d = 2r, so λ ∝ 1/d² ∝ 1/r².

Question 9

Previous-year question
In a gas at 1 atm pressure and 27°C a molecule has a diameter of 5 Å. Find the mean free path of the molecule: [Re-AIPMT 2015]
  1. 3.8 × 10⁻⁸ m
  2. 2 × 10⁻⁸ m
  3. 8.3 × 10⁻⁹ m
  4. 3.2 × 10⁻⁶ m
Show answer
Answer: (A) 3.8 × 10⁻⁸ m
Explanation: λ = kT/(√2 πd²P) = (1.38 × 10⁻²³ × 300)/(1.414 × 3.14 × (5 × 10⁻¹⁰)² × 1.013 × 10⁵) ≈ (4.14 × 10⁻²¹)/(1.12 × 10⁻¹³) ≈ 3.7 × 10⁻⁸ ≈ 3.8 × 10⁻⁸ m.

Question 10

Previous-year question
The mean free path of molecules of a gas (radius ‘r’) is inversely proportional to: [AIPMT 2014]
  1. r³
  2. r²
  3. r
  4. √r
Show answer
Answer: (B) r²
Explanation: λ = 1/(√2 πd²n), where d = 2r is the molecular diameter. Hence λ ∝ 1/d² ∝ 1/r².

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