2 mol of an ideal gas are expanded isotherma | vedclass

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(D) Given: $n = 2 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$V_1 = 20 \ L$,$V_2 = 40 \ L$. Formula for work done in an isothermal rever

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$-3457.97$

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(D) Given: $n = 2 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$V_1 = 20 \ L$,$V_2 = 40 \ L$. Formula for work done in an isothermal reversible expansion: $W = -2.303 \ n \ R \ T \ \log_{10}\left(\frac{V_2}{V_1}ight)$ Step $1$: Substitute the values: $W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes \log_{10}\left(\frac{40}{20}ight)$ Step $2$: Simplify the logarithmic term: $\log_{10}(2) \approx 0.3010$ Step $3$: Calculate: $W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes 0.3010$ $W \approx -3457.97 \ J$ Thus,the work done is $-3457.97 \ J$.

$2 \ mol$ of an ideal gas are expanded isothermally and reversibly from $20 \ L$ to $40 \ L$ at $300 \ K$. Calculate work done. $(R=8.314 \ J \ K^{-1} \ mol^{-1})$ (in $J$)

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