Calculate the work done during the reversible | vedclass

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(D) The work done in a reversible isothermal expansion is given by the formula: $W = -2.303 \ nRT \ \log \left( \frac{V_2}{V_1} \right)$.Given values:

Vedclass · Accepted Answer

$-3434$

Vedclass · Answer

(D) The work done in a reversible isothermal expansion is given by the formula: $W = -2.303 \ nRT \ \log \left( \frac{V_2}{V_1} ight)$.Given values: $n = 2 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 298 \ K$,$V_2 = 20 \ L$,$V_1 = 10 \ L$.Substituting these values: $W = -2.303 	imes 2 	imes 8.314 	imes 298 	imes \log \left( \frac{20}{10} ight)$.$W = -2.303 	imes 2 	imes 8.314 	imes 298 	imes \log(2)$.$W = -2.303 	imes 2 	imes 8.314 	imes 298 	imes 0.3010 \approx -3434.9 \ J$.

Calculate the work done during the reversible isothermal expansion of $2 \ mol$ of an ideal gas at $298 \ K$ from $10 \ L$ to $20 \ L$ in $J$. (in $.9$)

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