If the heat of $110 j$ is added to a gaseous and change in internal energy is $40 J$ , the amount of external work done is:

180 j

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70 j

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110 j

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30 j

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Solution

The correct option is B $70 j$
Given,
$Δ Q = 110 J$
$Δ U = 40 J$
$Δ W = ?$
From 1st law of thermodynamics,
$Δ Q = Δ U + Δ W$
$110 = 40 + Δ W$
$Δ W = 70 J$
The correct option is B.

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