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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 $∆ U$ of the gas in joules will be

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-750 J

-505 J

+505 J

+1100 J

answer is B.

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Detailed Solution

q = 0; for insulated container
$W = - p (V_{f} - V_{i}) = - 2.5 [4.5 - 2.5]$
= -5 lit-atm = –5(101.33) J
= –505 J (Approx.)
from fi rst law of thermodynamics,
$Δ E = q + W; q = 0$
Therefore, $ΔE = W ∣= - 505 J$ .

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