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11.0 L of an ideal gas at constant external pressure of $5 \mathrm{~atm}$ is compressed isothermally to a final volume of one liter. The heat absorbed and work done respectively, during this compression (in $\mathrm{L}$ atm) are
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The correct answer is:
$-50,50$
Work done $=\mathrm{w}=-\mathrm{p}_{\mathrm{ext}} \Delta \mathrm{V}=-5(1-11)$ $=+50 \mathrm{~L} \mathrm{~atm}$
Now, at constant pressure and temperature:-
$\begin{aligned}
& \Delta \mathrm{U}=\mathrm{q}+\mathrm{w}=\mathrm{q}+\mathrm{p} \Delta \mathrm{V}=0 \\
& \Rightarrow \mathrm{q}=-\mathrm{w}=-50 \mathrm{~L} \text { atm. }
\end{aligned}$
Now, at constant pressure and temperature:-
$\begin{aligned}
& \Delta \mathrm{U}=\mathrm{q}+\mathrm{w}=\mathrm{q}+\mathrm{p} \Delta \mathrm{V}=0 \\
& \Rightarrow \mathrm{q}=-\mathrm{w}=-50 \mathrm{~L} \text { atm. }
\end{aligned}$
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