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In thermodynamic process, 200 Joules of heat is given to a gas and 100 Joules of work is also done on it. The change in internal energy of the gas is

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100 J

300 J

419 J

24 J

answer is B.

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Detailed Solution

$\begin{array}{l} ΔQ = ΔU + ΔW; ΔQ = 200 J and ΔW = - 100 J \\ \Rightarrow ΔU = ΔQ - ΔW = 200 - (- 100) = 300 J \end{array}$

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