Calculate the maximum work done in expanding | vedclass

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(B) The formula for maximum work done in an isothermal reversible expansion is $W_{max} = -2.303 \ nRT \log(\frac{V_2}{V_1})$.Given: Mass of $O_2 = 16

Vedclass · Accepted Answer

$-2010$

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(B) The formula for maximum work done in an isothermal reversible expansion is $W_{max} = -2.303 \ nRT \log(\frac{V_2}{V_1})$.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 $T = 300 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Initial volume $V_1 = 5 \ dm^3$,Final volume $V_2 = 25 \ dm^3$.Substituting the values: $W_{max} = -2.303 	imes 0.5 	imes 8.314 	imes 300 	imes \log(\frac{25}{5})$.$W_{max} = -2.303 	imes 0.5 	imes 8.314 	imes 300 	imes \log(5)$.$W_{max} = -2.303 	imes 0.5 	imes 8.314 	imes 300 	imes 0.699$.$W_{max} \approx -2009.6 \ J \approx -2010 \ J$.

Calculate the maximum work done in expanding $16 \ g$ of oxygen $(O_2)$ at $300 \ K$ and occupying a volume of $5 \ dm^3$ isothermally until the volume becomes $25 \ dm^3$. (in $J$)

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