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Calculate the amount of work done during isothermal expansion of a gas from a volume of $4 \mathrm{dm}^{3}$ to $6 \mathrm{dm}^{3}$ against a constant external pressure of 3 atmosphere?
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The correct answer is:
$-607 \cdot 8 \mathrm{~J}$
$\begin{aligned} \mathrm{V}_{1}=4 \mathrm{dm}^{3}, & \mathrm{~V}_{2}=6 \mathrm{dm}^{3}, \quad \mathrm{P}_{\mathrm{ex}}=3 \mathrm{~atm} \\ \text { Now, } \mathrm{W}=&-\mathrm{P}_{\mathrm{ex}}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)=-3(6-4) \\=&-6 \mathrm{~L} . \mathrm{atm}=-6 \times 101.3 \mathrm{~J} \\ \mathrm{~W}=&-607.8 \mathrm{~J} \end{aligned}$
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