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(D) The work done in an isothermal reversible expansion is given by the formula: $W = \mu RT \ln\left(\frac{V_2}{V_1}\right)$.Here,$\mu$ is the number

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$2.3 	imes 900\, R\, \log_{10} 2$

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(D) The work done in an isothermal reversible expansion is given by the formula: $W = \mu RT \ln\left(\frac{V_2}{V_1}ight)$.Here,$\mu$ is the number of moles,$R$ is the universal gas constant,$T$ is the absolute temperature,$V_1$ is the initial volume,and $V_2$ is the final volume.Given: Mass of oxygen $(m)$ = $96\, g$,Molar mass of $O_2$ $(M)$ = $32\, g/mol$.Number of moles $\mu = \frac{m}{M} = \frac{96}{32} = 3\, mol$.Temperature $T = 27^{\circ}C = 27 + 273 = 300\, K$.Initial volume $V_1 = 70\, L$,Final volume $V_2 = 140\, L$.Converting natural logarithm to base $10$: $\ln(x) = 2.303 \log_{10}(x)$.Substituting the values: $W = 3 	imes R 	imes 300 	imes 2.3 \log_{10}\left(\frac{140}{70}ight)$.$W = 900 	imes 2.3 	imes R 	imes \log_{10}(2)$.Thus,the work done is $2.3 	imes 900\, R\, \log_{10} 2$.

In an isothermal reversible expansion,if the volume of $96\, g$ of oxygen at $27^{\circ}C$ is increased from $70\, L$ to $140\, L$,then the work done by the gas will be

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