Gas Laws, Kinetic Theory of Gases and Molecular Motion – Equation of state of a perfect gas: NEET MCQs with Answers

15 free questions on Gas Laws, Kinetic Theory of Gases and Molecular Motion – Equation of state of a perfect gas 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: Equation of state of a perfect gas, work done on compressing a gas, Kinetic theory of gases - assumptions, the concept of pressure. Kinetic interpretation of temperature: RMS speed of gas molecules, Avogadro's number.

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Question 1

Previous-year question
Three vessels of equal capacity contain gases at the same temperature and pressure: helium (monoatomic), fluorine (diatomic) and sulphur hexafluoride (polyatomic). The correct statement is [Re-NEET 2022]
  1. all vessels contain unequal numbers of respective molecules
  2. the root mean square speed of molecules is the same in all three cases
  3. the root mean square speed of helium is the largest
  4. the root mean square speed of sulphur hexafluoride is the largest
Show answer
Answer: (C) the root mean square speed of helium is the largest
Explanation: Equal V, P and T mean equal numbers of molecules (Avogadro). v_rms ∝ 1/√M, and helium has the least molar mass, so its rms speed is the largest.

Question 2

Previous-year question
The temperature at which the rms speed of atoms in neon gas equals the rms speed of hydrogen molecules at 15 °C is (atomic mass of neon = 20.2 u, molecular mass of H₂ = 2 u) [NEET 2022 (Overseas)]
  1. 2.9 K
  2. 0.15 × 10³ K
  3. 0.29 × 10³ K
  4. 2.9 × 10³ K
Show answer
Answer: (D) 2.9 × 10³ K
Explanation: Equal v_rms ⇒ T/M is equal: T_Ne = 288 × 20.2/2 ≈ 2909 K ≈ 2.9 × 10³ K.

Question 3

Previous-year question
An ideal gas at 0 °C and atmospheric pressure P has volume V. The percentage increase in temperature needed to expand it to 3V at constant pressure is [NEET 2024 (Sawai Madhopur)]
  1. 200 %
  2. 300 %
  3. 50 %
  4. 100 %
Show answer
Answer: (A) 200 %
Explanation: V ∝ T at constant P, so T must go from 273 K to 819 K, an increase of 546 K, i.e. (546/273) × 100 = 200 %.

Question 4

Previous-year question
A container of volume 200 cm³ contains 0.2 mole of hydrogen and 0.3 mole of argon. The pressure of the system at 200 K (R = 8.3 J K⁻¹ mol⁻¹) is [NEET 2023 (Manipur)]
  1. 6.15 × 10⁵ Pa
  2. 6.15 × 10⁴ Pa
  3. 4.15 × 10⁵ Pa
  4. 4.15 × 10⁶ Pa
Show answer
Answer: (D) 4.15 × 10⁶ Pa
Explanation: P = nRT/V = 0.5 × 8.3 × 200/(200 × 10⁻⁶) = 830/(2 × 10⁻⁴) = 4.15 × 10⁶ Pa.

Question 5

Previous-year question
A cylinder contains hydrogen gas at a pressure of 249 kPa and a temperature of 27 °C. Its density is (R = 8.3 J mol⁻¹ K⁻¹) [NEET 2020]
  1. 0.2 kg m⁻³
  2. 0.1 kg m⁻³
  3. 0.02 kg m⁻³
  4. 0.5 kg m⁻³
Show answer
Answer: (A) 0.2 kg m⁻³
Explanation: ρ = PM/RT = (2.49 × 10⁵ × 2 × 10⁻³)/(8.3 × 300) = 498/2490 = 0.2 kg m⁻³.

Question 6

Previous-year question
Raising the temperature of a gas confined in a container would lead to [NEET 2019]
  1. an increase in its mass
  2. an increase in its kinetic energy
  3. a decrease in its pressure
  4. a decrease in the intermolecular distance
Show answer
Answer: (B) an increase in its kinetic energy
Explanation: In a closed rigid container the mass, volume and mean separation are fixed. Heating raises the mean molecular kinetic energy (and hence the pressure).

Question 7

Previous-year question
At what temperature will the rms speed of oxygen molecules be just sufficient for them to escape from the earth's atmosphere? (mass of an oxygen molecule = 2.76 × 10⁻²⁶ kg, k_B = 1.38 × 10⁻²³ J K⁻¹) [NEET 2018]
  1. 2.508 × 10⁴ K
  2. 8.360 × 10⁴ K
  3. 1.254 × 10⁴ K
  4. 5.016 × 10⁴ K
Show answer
Answer: (B) 8.360 × 10⁴ K
Explanation: Put (3/2)k_BT = (1/2)mv_e² with v_e = 11.2 km s⁻¹. Then T = mv_e²/3k_B = (2.76 × 10⁻²⁶ × 1.2544 × 10⁸)/(3 × 1.38 × 10⁻²³) ≈ 8.36 × 10⁴ K.
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Question 8

Previous-year question
Molecules of a given mass of gas have an rms speed of 200 m s⁻¹ at 27 °C and 1.0 × 10⁵ N m⁻². When the temperature and pressure are 127 °C and 0.05 × 10⁵ N m⁻², the rms speed becomes [NEET 2016]
  1. 100√2 m s⁻¹
  2. 400/√3 m s⁻¹
  3. 100√2/3 m s⁻¹
  4. 100/3 m s⁻¹
Show answer
Answer: (B) 400/√3 m s⁻¹
Explanation: v_rms = √(3RT/M) depends only on temperature, so v₂ = 200 × √(400/300) = 200 × 2/√3 = 400/√3 m s⁻¹.

Question 9

Previous-year question
Two vessels separately contain ideal gases A and B at the same temperature. The pressure of A is twice that of B, and the density of A is 1.5 times that of B. The ratio of the molecular weights of A and B is [AIPMT 2014]
  1. 3/4
  2. 2
  3. 1/2
  4. 2/3
Show answer
Answer: (A) 3/4
Explanation: From ρ = PM/RT we get M = ρRT/P. Hence M_A/M_B = (ρ_A/ρ_B)(P_B/P_A) = 1.5 × (1/2) = 3/4.

Question 10

Previous-year question
At what temperature will the rms speed of oxygen molecules become just sufficient for escape from the Earth’s atmosphere? (Given: mass of oxygen molecule m = 2.76 × 10⁻²⁶ kg; Boltzmann’s constant k_B = 1.38 × 10⁻²³ J K⁻¹) [NEET 2018]
  1. 5.016 × 10⁴ K
  2. 8.360 × 10⁴ K
  3. 2.508 × 10⁴ K
  4. 1.254 × 10⁴ K
Show answer
Answer: (B) 8.360 × 10⁴ K
Explanation: Set v_rms = v_escape = 11.2 km/s = 1.12 × 10⁴ m/s. From (3/2)k_BT = ½mv²: T = mv²/(3k_B) = (2.76 × 10⁻²⁶ × (1.12 × 10⁴)²)/(3 × 1.38 × 10⁻²³) ≈ 8.36 × 10⁴ K.

Question 11

Previous-year question
The molecules of a given mass of a gas have rms velocity of 200 m/s at 27°C and 1.0 × 10⁵ N/m² pressure. When the temperature and pressure of the gas are 127°C and 0.05 × 10⁵ N/m² respectively, then the rms velocity of its molecules in m/s is: [AIPMT 2016]
  1. 100√2
  2. 400/√3
  3. 100√2/3
  4. 100/3
Show answer
Answer: (B) 400/√3
Explanation: v_rms = √(3RT/M) depends only on temperature — the pressure change is irrelevant. v₂ = 200 × √(400/300) = 200 × (2/√3) = 400/√3 m/s.

Question 12

Previous-year question
In Maxwell’s velocity distribution curve, the area under the graph: [Re-AIPMT 2015]
  1. Increases when temperature is increased
  2. Decreases when temperature is increased
  3. Remains same at all temperatures
  4. Depends on the pressure of the gas
Show answer
Answer: (C) Remains same at all temperatures
Explanation: The area under the (N–v) distribution curve represents the total number of molecules, which is fixed for a given sample. Raising the temperature flattens and broadens the curve and shifts its peak to higher speeds, but the enclosed area stays the same.

Question 13

Previous-year question
Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is: [Re-AIPMT 2015]
  1. 1/2
  2. 2/3
  3. 3/4
  4. 2
Show answer
Answer: (C) 3/4
Explanation: ρ = PM/(RT). So ρ_A/ρ_B = (P_A M_A)/(P_B M_B) ⇒ 1.5 = 2 × (M_A/M_B) ⇒ M_A/M_B = 0.75 = 3/4.

Question 14

Previous-year question
At 10°C the value of the density of a fixed mass of an ideal gas divided by its pressure is x. At 110°C this ratio is: [AIPMT 2008]
  1. (10/110) x
  2. (283/383) x
  3. x
  4. (383/283) x
Show answer
Answer: (B) (283/383) x
Explanation: ρ/P = M/(RT) ∝ 1/T. Ratio at 383 K = x × (283/383).

Question 15

The Faraday constant is F = N_A e and the molar gas constant is R = N_A k_B, where N_A is Avogadro's number. Given F = 9.65 × 10⁴ C mol⁻¹ and R = 8.31 J mol⁻¹ K⁻¹, the value of e/k_B is closest to
  1. 1.2 × 10⁷ K V⁻¹
  2. 8.6 × 10⁻⁵ K V⁻¹
  3. 1.2 × 10⁴ K V⁻¹
  4. 6.0 × 10²³ K V⁻¹
Show answer
Answer: (C) 1.2 × 10⁴ K V⁻¹
Explanation: Dividing the two relations cancels N_A, so e/k_B = F/R. This gives 9.65 × 10⁴ / 8.31 ≈ 1.16 × 10⁴. The unit C K/J is K/V because 1 J/C = 1 V.

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