Thermodynamics Ques 50
50. $110$ joules of heat is added to a gaseous system whose internal energy is $40$ $ J$. Then the amount of external work done is
[1993]
(a) $150 $ $J$
(b) $70 $ $J$
(c) $110 $ $J$
(d) $40 $ $J$
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Answer:
Correct Answer: 50.(b)
Solution:
- (b) $\Delta Q=\Delta U+\Delta W$
$\Rightarrow \Delta W=\Delta Q-\Delta U=110-40=70 $ $J$
BETA
