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(A) The formula for maximum work done in an isothermal reversible expansion is $W_{max} = -nRT \ln(\frac{V_2}{V_1})$.Given: Mass of $O_2 = 16 \, g$,Mo

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$-2.01 	imes 10^3 \, J$

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(A) The formula for maximum work done in an isothermal reversible expansion is $W_{max} = -nRT \ln(\frac{V_2}{V_1})$.Given: Mass of $O_2 = 16 \, g$,Molar mass of $O_2 = 32 \, g/mol$.Number of moles $n = \frac{16}{32} = 0.5 \, mol$.Temperature $T = 300 \, K$,$R = 8.314 \, J \cdot K^{-1} \cdot mol^{-1}$.$V_1 = 5 \, dm^3$,$V_2 = 25 \, dm^3$.$W_{max} = -0.5 	imes 8.314 	imes 300 	imes \ln(\frac{25}{5})$.$W_{max} = -0.5 	imes 8.314 	imes 300 	imes \ln(5)$.$W_{max} = -1247.1 	imes 1.609 \approx -2006.6 \, J \approx -2.01 	imes 10^3 \, J$.

The maximum work done in expanding $16 \, g$ of $O_2$ gas isothermally at $300 \, K$ from an initial volume of $5 \, dm^3$ to a final volume of $25 \, dm^3$ is:

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