A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5

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NEET is a competitive examination conducted in India. This question is from the 2017 NEET Chemistry paper.

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This NEET 2017 Chemistry question has a easy difficulty level and is marked as an important question.

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Question

A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be

Accepted Answer

Correct Answer: D — −505 JStep-by-step Solution:The container is well-insulated (adiabatic), so no heat exchange occurs: q = 0.First law of thermodynamics: ΔU = q + w = 0 + w = wWork done by the gas against constant external pressure:w = −Pext × ΔV = −Pext × (Vf − Vi)= −2.5 atm × (4.50 − 2.50) L= −2.5 × 2.0 = −5.0 L·atmConverting to joules (1 L·atm = 101.325 J):w = −5.0 × 101.325 = −506.6 J ≈ −505 JTherefore ΔU = −505 J. The negative sign indicates the internal energy of the gas decreases (gas does work on surroundings).Why other options are incorrect:Option A (+505 J): Incorrect sign; work is done by the gas (expansion), so ΔU must be negative.Option B (1136.25 J): This would result from using an incorrect conversion factor or incorrectly multiplying P × Vf instead of P × ΔV.Option C (−500 J): Close but uses 1 L·atm = 100 J (approximate); the correct value is 101.325 J/L·atm giving −506.6 ≈ −505 J.Key Concept: Adiabatic process: q = 0, therefore ΔU = w. Work in irreversible expansion against constant external pressure: w = −PextΔV. Unit conversion: 1 L·atm = 101.325 J. For gas expansion, w is negative (gas loses energy doing work on surroundings).

A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be

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A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be

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