NEETAceToday's NEET Daily

Perfect-gas equation and work in gas compression: 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.

Practise Kinetic Theory of Gases freeAll Kinetic Theory of Gases questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. An ideal gas is compressed quasi-statically at a constant pressure of 4.0 × 10⁵ N m⁻² from an initial volume of 8.0 × 10⁻³ m³ to a final volume of 3.0 × 10⁻³ m³. The work done on the gas during this compression is:

  1. -2000 J
  2. 1200 J
  3. 2000 J
  4. 3200 J
Show answer and explanation

Answer: (C) 2000 J

Work done by the gas is W_by = P ΔV = 4.0 × 10⁵ × (3.0 × 10⁻³ - 8.0 × 10⁻³) = -2000 J. Therefore, the work done on the gas during the compression is W_on = -W_by = +2000 J.

medium

2. An ideal gas undergoes a process in which its pressure is directly proportional to its volume, P = a V, where a is a constant. If the gas is compressed from an initial volume V₁ to a final volume V₂, the work done ON the gas is:

  1. a (V₁² - V₂²)
  2. (a / 2) (V₂² - V₁²)
  3. a (V₁ - V₂)
  4. (a / 2) (V₁² - V₂²)
Show answer and explanation

Answer: (D) (a / 2) (V₁² - V₂²)

Work done by the gas is W_by = ∫ P dV from V₁ to V₂ = ∫ (a V) dV = (a / 2) (V₂² - V₁²). Since the gas is compressed, V₂ < V₁, so W_by is negative. The work done ON the gas is W_on = -W_by = (a / 2) (V₁² - V₂²).

hard

3. One mole of an ideal monoatomic gas undergoes a compression according to the process P = c/V², where c is a constant. The initial state is (P₀, V₀) and the final volume is V₀/2. The work done by the gas during this process is:

  1. -P₀V₀
  2. -2 P₀V₀
  3. -0.5 P₀V₀
  4. -1.5 P₀V₀
Show answer and explanation

Answer: (A) -P₀V₀

Work done is given by W = ∫ P dV = ∫ (c/V²) dV from V₀ to V₀/2. This integrates to -c [1/V] from V₀ to V₀/2 = -c(2/V₀ - 1/V₀) = -c/V₀. Since c = P₀V₀², W = -P₀V₀² / V₀ = -P₀V₀.

medium

4. An ideal gas is subjected to a cyclic process consisting of three steps: an isobaric expansion from (P₀, V₀) to (P₀, 3V₀), an isochoric cooling from (P₀, 3V₀) to (P₀/2, 3V₀), and a straight-line path on the P-V diagram returning directly from (P₀/2, 3V₀) to (P₀, V₀). The net work done by the gas in this cycle is:

  1. 3P₀V₀ / 2
  2. P₀V₀
  3. P₀V₀ / 2
  4. 2P₀V₀
Show answer and explanation

Answer: (C) P₀V₀ / 2

On the P-V diagram, the cycle forms a right-angled triangle with vertices at (P₀, V₀), (P₀, 3V₀), and (P₀/2, 3V₀). The base is (3V₀ - V₀) = 2V₀ and the height is (P₀ - P₀/2) = P₀/2. The area enclosed by the clockwise cycle is (1/2) × base × height = (1/2) × 2V₀ × (P₀/2) = P₀V₀ / 2.

easy

5. At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure. This is:

  1. Charles's law
  2. Avogadro's law
  3. Dalton's law
  4. Boyle's law
Show answer and explanation

Answer: (D) Boyle's law

Boyle's law: PV = constant at constant temperature. Charles's law says V ∝ T at constant pressure.

NCERT Class 11

hard

6. Two moles of an ideal gas are taken through a process in which temperature varies with volume as T = α V, where α is a positive constant. If the volume of the gas increases from V₀ to 2V₀, the work done by the gas is (where R is the universal gas constant):

  1. 2 α R V₀
  2. α R V₀
  3. 4 α R V₀
  4. 0.5 α R V₀
Show answer and explanation

Answer: (A) 2 α R V₀

From the ideal gas equation, P = nRT/V. Since T = αV, we have P = nR(αV)/V = nαR = constant. Thus, the process is isobaric with constant pressure P = 2αR (for n = 2 moles). The work done is W = P ΔV = (2αR)(2V₀ - V₀) = 2αRV₀.

medium

7. An ideal gas undergoes a process represented by P = a - bV on a P-V indicator diagram, where a and b are positive constants. The gas is compressed from an initial volume V₁ = a/(2b) to a final volume V₂ = a/(4b). The work done by the gas during this compression is:

  1. -3a² / (32b)
  2. -5a² / (32b)
  3. 3a² / (16b)
  4. -7a² / (32b)
Show answer and explanation

Answer: (B) -5a² / (32b)

Work done by the gas is W = ∫ P dV = ∫ (a - bV) dV from V₁ to V₂. Evaluating: [aV - bV²/2] from a/(2b) to a/(4b). At V₂, W₂ = a(a/4b) - b(a²/16b²)/2 = a²/4b - a²/32b = 7a²/(32b). At V₁, W₁ = a(a/2b) - b(a²/4b²)/2 = a²/2b - a²/8b = 3a²/(8b) = 12a²/(32b). Thus, W = W₂ - W₁ = 7a²/(32b) - 12a²/(32b) = -5a²/(32b).

medium

8. A given mass of an ideal gas is compressed from volume V to V/2 via three different quasi-static processes: Process 1 is isobaric, Process 2 is isothermal, and Process 3 is adiabatic. If W₁, W₂, and W₃ represent the magnitude of work done on the gas in the respective processes, then:

  1. W₁ > W₂ > W₃
  2. W₃ > W₁ > W₂
  3. W₂ > W₁ > W₃
  4. W₃ > W₂ > W₁
Show answer and explanation

Answer: (D) W₃ > W₂ > W₁

During compression from V to V/2, the pressure rises steepest in adiabatic compression (P ∝ V^(-γ)), moderately in isothermal compression (P ∝ V^(-1)), and remains constant in isobaric compression. Since the area under the P-V curve represents the work done, the area is largest for adiabatic, followed by isothermal, and smallest for isobaric. Hence, W₃ > W₂ > W₁.

medium

9. One mole of an ideal gas is compressed at a constant temperature T from an initial volume 3V to a final volume V. The work done on the gas during this isothermal compression is:

  1. R T ln 3
  2. -R T ln 3
  3. 3 R T ln 3
  4. (1/3) R T ln 3
Show answer and explanation

Answer: (A) R T ln 3

Work done by the gas during isothermal compression is W_by = n R T ln(V_f / V_i) = 1 × R T ln(V / 3V) = -R T ln 3. The work done ON the gas is the negative of work done by the gas: W_on = -W_by = R T ln 3.

medium

10. An ideal gas undergoes a compression such that its pressure varies with volume according to the relation P = αV, where α is a positive constant. If the volume decreases from V₀ to V₀/2, the work done on the gas during this compression is:

  1. (3/4) αV₀²
  2. (1/8) αV₀²
  3. (3/8) αV₀²
  4. (1/2) αV₀²
Show answer and explanation

Answer: (C) (3/8) αV₀²

The work done on the gas is W_on = -∫ P dV from V₀ to V₀/2. Thus, W_on = -α [V²/2] from V₀ to V₀/2 = -α/2 [(V₀/2)² - V₀²] = -α/2 [-3V₀²/4] = (3/8) αV₀².

easy

11. A rigid container of volume V contains n moles of an ideal gas at temperature T. If half of the gas molecules escape through a tiny leak while the temperature of the container is raised to 3T, the final pressure inside the container becomes:

  1. 3 times the initial pressure
  2. 0.75 times the initial pressure
  3. 1.5 times the initial pressure
  4. 0.5 times the initial pressure
Show answer and explanation

Answer: (C) 1.5 times the initial pressure

Initially, P₁ = nRT/V. After the leak, the remaining moles are n₂ = n/2, and the temperature is T₂ = 3T. The new pressure is P₂ = (n/2)R(3T)/V = 1.5 (nRT/V) = 1.5 P₁.

medium

12. In terms of the number of molecules N and Boltzmann constant kB, the ideal gas equation is:

  1. PV = NkBT
  2. PV = kBT/N
  3. PV = NRT
  4. P = NkBT V
Show answer and explanation

Answer: (A) PV = NkBT

PV = nRT = (N/NA)RT = NkBT, since kB = R/NA ≈ 1.38 × 10⁻²³ J K⁻¹.

NCERT Class 11

Keep going with Kinetic Theory of Gases. All 289 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Kinetic Theory of Gases