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If the heat of 110 J is added to a gaseous system, whose internal energy is 40 J , then the amount of external work done is
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The correct answer is:
70 J
Heat added to the system $\Delta Q=110 \mathrm{~J}$
Internal energy $\Delta u=40 \mathrm{~J}$
External work done is given by
$\Delta W=\Delta Q-\Delta u=110-40=70 \mathrm{~J}$
Internal energy $\Delta u=40 \mathrm{~J}$
External work done is given by
$\Delta W=\Delta Q-\Delta u=110-40=70 \mathrm{~J}$
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