$$110\,J$$ of heat is added to a gaseous system, whose internal energy is $$40\,J,$$ then the amount of external work done is
A.
$$150\,J$$
B.
$$70\,J$$
C.
$$110\,J$$
D.
$$40\,J$$
Answer :
$$70\,J$$
Solution : Concept
Apply first law of thermodynamics to calculate the required work done.
From first law of thermodynamics $$\Delta Q = \Delta U + \Delta W$$
where, $$\Delta Q =$$ heat given
$$\Delta U =$$ change in intemal energy
$$\Delta W =$$ work done
Here, $$\Delta Q = 110\,J$$
$$\eqalign{
& \Delta U = 40\,J \cr
& \therefore \Delta W = \Delta Q - \Delta U = 110 - 40 = 70\,J \cr} $$
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