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Under isothermal condition a gas expands from $0.2 \mathrm{dm}^3$ to $0.8 \mathrm{dm}^3$ against a constant pressure of 2 bar at $300 \mathrm{~K}$. Find the work done by the gas.
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The correct answer is:
$-120 \mathrm{~J}$
$\begin{aligned} \mathrm{W} & =-\mathrm{PdV} \\ & =-2(0.8-0.2) \mathrm{bar}-\mathrm{dm}^2 \\ & =-2(0.6) \times 10^2 \mathrm{~J} \\ & =-120 \mathrm{~J}\end{aligned}$
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