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 Δ U of the gas in joules will be :

1. \(-500\) J

2. \(-505\) J

3. \(+505\) J

4. 1136.25 J

Doubtly Asked 30 Oct 2026 NEET UG 2017 Practice test neet Save WhatsApp Report

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Answer: 2. \(-505\) J

Key idea: well insulated means adiabatic, so q = 0 and \(\Delta U = w\).

\(w=-p_{ext}\Delta V=-2.5\ \mathrm{atm}\times(4.50-2.50)\ \mathrm{L}=-5\ \mathrm{L\,atm}\).

Taking 1 L atm \(\approx\) 101 J: \(w\approx-5\times101=-505\) J. So \(\Delta U\approx-505\) J (the gas cools while doing work).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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