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

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

NTA AbhyasNTA Abhyas 2022

Solution:

$\Delta Q=\Delta U+\Delta W$
Given, $\Delta Q=200J$ and $\Delta W=-100J$
$\Rightarrow \Delta U=\Delta Q-\Delta W$
$\Rightarrow \Delta U=200-\left(- 100\right)=300J$