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 ........ $J$
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(b) $\Delta Q = \Delta U + \Delta W$; $\Delta Q = 200J$ and $\Delta W = - 100J$
==> $\Delta U = \Delta Q - \Delta W = 200 - ( - 100) = 300J$
==> $\Delta U = \Delta Q - \Delta W = 200 - ( - 100) = 300J$
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