If the system is a gas in a cylinder with a movable piston, the gas in moving the piston does work. Since force is pressure times area, and area times displacement is volume, work done by the system against a constant pressure P is ∆W = P ∆V where ∆V is the change in volume of the gas.
-- NCERT Class 11 Physics, Ch. 11, p. 231Work Compressing Gas
10 MCQs2 revision cards9-step worked example
Source: NCERT Kinetic TheoryOfficial key: NTA-verifiedLast updated: 10 Oct 2026
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A gas in a cylinder with a movable piston changes its volume by ΔV against a constant pressure P. According to NCERT, what is the work done by the system?
- A.ΔW = V ΔP
- B.ΔW = ΔV / P
- C.ΔW = P ΔV
- D.ΔW = P / ΔV
Tap to see the answer
Answer: C. C is correct. NCERT Class 11 Physics, Chapter 11, page 231 says the work done by the system against a constant pressure P is ΔW = P ΔV, where ΔV is the change in volume of the gas.
A is wrong: A is wrong because the work at constant pressure comes from the change in volume, not from a change in pressure; here the pressure does not change.
B is wrong: B is wrong because dividing the volume change by the pressure gives a quantity with the wrong units for work.
D is wrong: D is wrong because the pressure multiplies the volume change; dividing by ΔV gives the wrong units and the wrong size.
Work Compressing Gas, explained for NEET
The trap is the sign. W = P ΔV is the work done by the gas, so when the gas is compressed ΔV is negative and W is negative: the work is done on the gas, by the surroundings.
For a gas in a cylinder with a movable piston, the work done by the system against a constant pressure P is ΔW = P ΔV, where ΔV is the change in volume of the gas (NCERT Class 11 Physics, Chapter 11, page 231). Use the change in volume, not the final volume.
At constant temperature the pressure changes as the volume changes, so NCERT adds up ΔW = P ΔV over the whole process. For an ideal gas going from V₁ to V₂ this gives W = μRT ln(V₂/V₁) (NCERT Class 11 Physics, Chapter 11, page 234). For V₂ > V₁ the work W is positive, and for V₂ < V₁ it is negative. In an isothermal compression, work is done on the gas by the environment and heat is released (NCERT Class 11 Physics, Chapter 11, page 235).
Why does the heat leave? For an ideal gas the internal energy depends only on temperature, so it does not change in an isothermal process. The heat exchanged equals the work: Q = W (NCERT Class 11 Physics, Chapter 11, page 235). In the Carnot cycle, the heat released by the gas in its isothermal compression is also the work done on the gas by the environment (NCERT Class 11 Physics, Chapter 11, page 238).
At constant pressure the work done by the gas is W = P (V₂ − V₁) = μR (T₂ − T₁) (NCERT Class 11 Physics, Chapter 11, page 235).
Watch-out: write the sign of W as work done by the gas. The phrase "work done on the gas" flips it. Use ΔV = V₂ − V₁, with V₂ the final volume.
How do you solve a Work Compressing Gas question? A worked example
- 1
Given
2.0 mol of an ideal gas at 4.0 × 10² K is compressed isothermally from 3.0 × 10¹ L to 1.0 × 10¹ L. Take R = 8.3 J/(mol K) and ln 3 = 1.10.
- 2
Required
The work done on the gas and the heat released by it.
- 3
Concept
For an isothermal change of an ideal gas, W = μRT ln(V₂/V₁) is the work done by the gas (NCERT Class 11 Physics, Chapter 11, page 234). In an isothermal compression, work is done on the gas and heat is released, and the heat equals the work because the internal energy does not change (NCERT Class 11 Physics, Chapter 11, page 235).
- 4
Formula
W = μRT ln(V₂/V₁), and the work done on the gas is −W.
- 5
Substitution
W = 2.0 × 8.3 × 4.0 × 10² × ln(1.0 × 10¹ / 3.0 × 10¹) = 6.64 × 10³ × ln(1/3).
- 6
Calculation
ln(1/3) = −ln 3 = −1.10, so W = 6.64 × 10³ × (−1.10) = −7.3 × 10³ J. The value ln 3 = 1.10 is a given constant and does not set the number of significant figures; the result carries two, like the given data.
- 7
Final answer
The work done by the gas is −7.3 × 10³ J, so the work done on the gas is +7.3 × 10³ J, and the gas releases 7.3 × 10³ J of heat to the surroundings.
- 8
Common trap
Giving a positive W for the gas, or leaving out the logarithm. The ratio of volumes goes inside ln, and the result for a compression is negative for the work done by the gas.
- 9
Similar NEET-style question
A gas at a constant pressure of 2.0 × 10⁵ Pa is compressed from 5.0 × 10⁻³ m³ to 3.0 × 10⁻³ m³. What is the work done on the gas? (Answer: ΔV = 3.0 − 5.0 = −2.0 L = −2.0 × 10⁻³ m³, so the work done on the gas is −P ΔV = 2.0 × 10⁵ Pa × 2.0 × 10⁻³ m³ = +4.0 × 10² J; the work done by the gas is the opposite sign, −4.0 × 10² J.)
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Can you answer these Work Compressing Gas MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
A gas in a cylinder with a movable piston changes its volume by ΔV against a constant pressure P. According to NCERT, what is the work done by the system?
Show answer and why every option is right or wrong
Answer: C. C is correct. NCERT Class 11 Physics, Chapter 11, page 231 says the work done by the system against a constant pressure P is ΔW = P ΔV, where ΔV is the change in volume of the gas.
Why A is wrong: A is wrong because the work at constant pressure comes from the change in volume, not from a change in pressure; here the pressure does not change.
Why B is wrong: B is wrong because dividing the volume change by the pressure gives a quantity with the wrong units for work.
Why D is wrong: D is wrong because the pressure multiplies the volume change; dividing by ΔV gives the wrong units and the wrong size.
MCQ 2Easy RecallPractice
An ideal gas is compressed isothermally. According to NCERT, what happens?
Show answer and why every option is right or wrong
Answer: A. A is correct. NCERT Class 11 Physics, Chapter 11, page 235 says that in an isothermal compression, work is done on the gas by the environment and heat is released.
Why B is wrong: B is wrong because absorbing heat and doing work is the isothermal expansion; NCERT page 235 gives compression the opposite pair.
Why C is wrong: C is wrong because zero heat exchange describes an adiabatic process; at constant temperature the heat released equals the work done on the gas.
Why D is wrong: D is wrong because for an ideal gas the internal energy depends only on temperature, so it does not change in an isothermal process.
MCQ 3Easy RecallPractice
In an isothermal process of an ideal gas with V₂ < V₁, what is the sign of the work W done by the gas, from W = μRT ln(V₂/V₁)?
Show answer and why every option is right or wrong
Answer: D. D is correct. NCERT Class 11 Physics, Chapter 11, page 235 notes that W is positive for V₂ > V₁ and negative for V₂ < V₁; here V₂ < V₁, so ln(V₂/V₁) is negative and W is negative.
Why A is wrong: A is wrong because a positive W belongs to expansion, V₂ > V₁; in compression the logarithm is negative (trap: sign of the work in compression).
Why B is wrong: B is wrong because the volume changes, so ln(V₂/V₁) is not zero and neither is W.
Why C is wrong: C is wrong because heat released does not make the work done by the gas positive; the heat released equals the work done on the gas.
MCQ 4Direct ApplicationPractice
A gas at a constant pressure of 1.0 × 10⁵ Pa is compressed from 4.0 × 10⁻³ m³ to 3.0 × 10⁻³ m³. What is the work done by the gas?
Show answer and why every option is right or wrong
Answer: B. B is correct. ΔW = P ΔV (NCERT Class 11 Physics, Chapter 11, page 231) with ΔV = 3.0 × 10⁻³ − 4.0 × 10⁻³ = −1.0 × 10⁻³ m³ gives ΔW = 1.0 × 10⁵ × (−1.0 × 10⁻³) = −1.0 × 10² J.
Why A is wrong: A is wrong because it drops the sign; a compression has a negative ΔV, so the work done by the gas is negative.
Why C is wrong: C is wrong because −1.0 × 10⁵ J takes ΔV as 1.0, a number in litres, as if it were in cubic metres; the change is 1.0 × 10⁻³ m³.
Why D is wrong: D is wrong because 3.0 × 10² J uses the final volume instead of the change in volume (trap: final volume used for ΔV).
MCQ 5Direct ApplicationPractice
In an isothermal compression of an ideal gas, 6.0 × 10² J of work is done on the gas. How much heat does the gas release?
Show answer and why every option is right or wrong
Answer: D. D is correct. For an ideal gas the internal energy does not change in an isothermal process, so the heat released equals the work done on the gas (NCERT Class 11 Physics, Chapter 11, page 235).
Why A is wrong: A is wrong because zero heat exchange is the adiabatic case; at constant temperature the work done on the gas leaves as heat.
Why B is wrong: B is wrong because it assumes half of the work is stored as internal energy, but for an ideal gas at constant temperature none is.
Why C is wrong: C is wrong because the heat released cannot exceed the work done on the gas; Q = W exactly, with no extra heat.
MCQ 6Direct ApplicationPractice
One mole of an ideal gas is heated at constant pressure from 3.0 × 10² K to 3.4 × 10² K. Take R = 8.3 J/(mol K). What is the work done by the gas?
Show answer and why every option is right or wrong
Answer: A. A is correct. At constant pressure W = P (V₂ − V₁) = μR (T₂ − T₁) (NCERT Class 11 Physics, Chapter 11, page 235), so W = 1.0 × 8.3 × 40 = 332 J, which is 3.3 × 10² J.
Why B is wrong: B is wrong because 2.8 × 10³ J is R times the final temperature alone, 8.3 × 3.4 × 10²; the work uses the temperature difference.
Why C is wrong: C is wrong because 2.5 × 10³ J is R times the initial temperature alone; the work uses the temperature difference.
Why D is wrong: D is wrong because heating at constant pressure makes the gas expand, so the work done by the gas is positive, not negative.
MCQ 7CalculationPractice
One mole of an ideal gas at 3.0 × 10² K is compressed isothermally, its volume falling from 2.0 × 10¹ L to 1.0 × 10¹ L. Take R = 8.3 J/(mol K) and ln 2 = 0.69. What is the work done by the gas?
Show answer and why every option is right or wrong
Answer: B. B is correct. W = μRT ln(V₂/V₁) (NCERT Class 11 Physics, Chapter 11, page 234) = 1.0 × 8.3 × 3.0 × 10² × ln(1/2) = 2.49 × 10³ × (−0.69) = −1.7 × 10³ J. The negative sign means 1.7 × 10³ J is done on the gas (page 235).
Why A is wrong: A is wrong because it has the sign of an expansion; V₂ < V₁ makes ln(V₂/V₁) negative, so the work done by the gas is negative.
Why C is wrong: C is wrong because −2.5 × 10³ J is μRT with the logarithm left out; the factor ln(V₂/V₁) must be applied.
Why D is wrong: D is wrong because −7.5 × 10² J uses the base-10 logarithm, 0.30, instead of the natural logarithm, 0.69.
MCQ 8CalculationPractice
A gas at a constant pressure of 1.5 × 10⁵ Pa is compressed from 6.0 × 10⁻³ m³ to 2.0 × 10⁻³ m³. What is the work done on the gas by the surroundings?
Show answer and why every option is right or wrong
Answer: C. C is correct. The work done by the gas is ΔW = P ΔV = 1.5 × 10⁵ × (2.0 × 10⁻³ − 6.0 × 10⁻³) = −6.0 × 10² J (NCERT Class 11 Physics, Chapter 11, page 231). The work done on the gas is the opposite sign, +6.0 × 10² J.
Why A is wrong: A is wrong because −6.0 × 10² J is the work done by the gas; the question asks for the work done on it, which has the opposite sign.
Why B is wrong: B is wrong because 3.0 × 10² J is the pressure times the final volume, 1.5 × 10⁵ × 2.0 × 10⁻³, not times the change.
Why D is wrong: D is wrong because 9.0 × 10² J is the pressure times the initial volume, 1.5 × 10⁵ × 6.0 × 10⁻³, not times the change.
MCQ 9Concept TrapPractice
An ideal gas is compressed isothermally. What happens to its internal energy?
Show answer and why every option is right or wrong
Answer: C. C is correct: for an ideal gas the internal energy depends only on temperature, so it does not change in an isothermal process, and the heat exchanged equals the work, Q = W. Based on NCERT Class 11 Physics, Chapter 11, page 235.
Why A is wrong: A is wrong because the temperature is constant, and for an ideal gas the internal energy depends only on temperature.
Why B is wrong: B is wrong because the temperature does not fall, so the internal energy does not decrease.
Why D is wrong: D is wrong because the internal energy keeps its value; constant does not mean zero.
MCQ 10CalculationPractice
A gas at a constant pressure of 2.0 × 10⁵ Pa expands from 1.0 × 10⁻³ m³ to 3.0 × 10⁻³ m³. What is the work done by the gas?
Show answer and why every option is right or wrong
Answer: A. A is correct: W = P ΔV = 2.0 × 10⁵ × (3.0 − 1.0) × 10⁻³ = +400 J, positive because the gas expands. Based on NCERT Class 11 Physics, Chapter 11, page 231.
Why B is wrong: B is wrong because the sign is negative only for a compression, where ΔV is negative.
Why C is wrong: C is wrong because 600 J uses the final volume 3.0 × 10⁻³ m³ instead of the change in volume.
Why D is wrong: D is wrong because 200 J uses the initial volume 1.0 × 10⁻³ m³ instead of the change in volume.
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Work Compressing Gas: quick recall before you leave
2 cards
What to remember before solving Work Compressing Gas questions
5 NCERT lines
Suppose an ideal gas goes isothermally (at temperature T ) from its initial state (P1, V1) to the final state (P2, V 2). At any intermediate stage with pressure P and volume change from V to V + ∆V (∆V small) ∆W = P ∆ V
-- NCERT Class 11 Physics, Ch. 11, p. 234Note from Eq. (11.12) that for V2 > V1, W > 0; and for V2 < V1, W < 0. That is, in an isothermal expansion, the gas absorbs heat and does work while in an isothermal compression, work is done on the gas by the environment and heat is released.
-- NCERT Class 11 Physics, Ch. 11, p. 235In an isobaric process, P is fixed. Work done by the gas is W = P (V2 – V1) = µ R (T2 – T1)
-- NCERT Class 11 Physics, Ch. 11, p. 235Isothermal compression of the gas from (P3, V3, T2) to (P4, V4, T2). Heat released (Q2) by the gas to the reservoir at temperature T2 is given by Eq. (11.12). This is also the work done (W3 → 4) on the gas by the environment.
-- NCERT Class 11 Physics, Ch. 11, p. 238Work Compressing Gas: NEET previous year questions (PYQs) with answers
8 questions in Kinetic Theory
No question in our NEET 2020–2025 set targets this topic directly.
More in Kinetic Theory: 3 exam traps and mistakes · 8 formulas · 5 question patterns from its other lessons.
Sources
NCERT refs: Class 11 Physics Chapter 11, p.231 | Class 11 Physics Chapter 11, p.234 | Class 11 Physics Chapter 11, p.235 | Class 11 Physics Chapter 11, p.238
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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