110 Joules of heat are added to a gaseous system, whose internal energy is 40 J; then the amount of external work done is :

150 J

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70 J

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110 J

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40 J

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Solution

The correct option is D 70 J
According to the first law of thermodynamics we have,
$Δ U = Q + W$
$40 = 110 + W$
$W = - 70 J$

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