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In a thermodynamic process, the pressure of a fixed mass of a gas is changed in such a manner that the gas releases $20 \mathrm{~J}$ of heat and $8 \mathrm{~J}$ of work is done on the gas. If the initial internal energy of the gas was $30 \mathrm{~J}$, then the final internal energy will be
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The correct answer is:
$18 \mathrm{~J}$
Given $\Delta \mathrm{Q}=-20 \mathrm{~J}, \mathrm{~W}=-8 \mathrm{~J}$
Using Ist law $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$
$$
\begin{array}{l}
\Rightarrow \Delta Q=-20+8=-12 \mathrm{~J} \\
\mathrm{Uf}=-12+30=18 \mathrm{~J}
\end{array}
$$
Using Ist law $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$
$$
\begin{array}{l}
\Rightarrow \Delta Q=-20+8=-12 \mathrm{~J} \\
\mathrm{Uf}=-12+30=18 \mathrm{~J}
\end{array}
$$
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