If 5 mol of an ideal gas expands from 10 L | vedclass

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(B) For an isothermal reversible expansion,the work done is given by the formula:$w = -2.303 \ nRT \log \left( \frac{V_2}{V_1} \right)$Substituting th

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$28721$

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(B) For an isothermal reversible expansion,the work done is given by the formula:$w = -2.303 \ nRT \log \left( \frac{V_2}{V_1} ight)$Substituting the given values:$n = 5 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$V_1 = 10 \ L$,$V_2 = 100 \ L$$w = -2.303 	imes 5 	imes 8.314 	imes 300 	imes \log \left( \frac{100}{10} ight)$$w = -2.303 	imes 5 	imes 8.314 	imes 300 	imes \log(10)$Since $\log(10) = 1$:$w = -2.303 	imes 5 	imes 8.314 	imes 300 = -28720.713 \ J$Given $w = -x \ J$,we have $-x = -28720.713$,so $x \approx 28721$.

If $5 \ mol$ of an ideal gas expands from $10 \ L$ to a volume of $100 \ L$ at $300 \ K$ under isothermal and reversible conditions,then the work done,$w$,is $-x \ J$. The value of $x$ is $ . . . . . . $. (Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

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