The maximum work done in expanding 16 g of o | vedclass

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(A) The formula for maximum work done in a reversible isothermal expansion is $w_{max} = -2.303 \ nRT \ \log_{10} \frac{V_2}{V_1}$.Given: Mass of $O_2

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$2.01 	imes 10^3 \ J$

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(A) The formula for maximum work done in a reversible isothermal expansion is $w_{max} = -2.303 \ nRT \ \log_{10} \frac{V_2}{V_1}$.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 $T = 300 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.$V_1 = 5 \ dm^3$,$V_2 = 25 \ dm^3$.$w_{max} = -2.303 	imes 0.5 	imes 8.314 	imes 300 	imes \log_{10} \frac{25}{5}$.$w_{max} = -2.303 	imes 0.5 	imes 8.314 	imes 300 	imes 0.699$.$w_{max} \approx -2008.5 \ J \approx -2.01 	imes 10^3 \ J$.Since the work is done by the system,the magnitude of work done is $2.01 	imes 10^3 \ J$.

The maximum work done in expanding $16 \ g$ of oxygen at $300 \ K$ from an initial volume of $5 \ dm^3$ to a final volume of $25 \ dm^3$ isothermally and reversibly is $(\log_{10} 5 = 0.699)$.

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