6 moles of an ideal gas expand isothermally a | vedclass

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(D) For an isothermal reversible expansion,the work done is given by the formula: $W = -2.303 	imes nRT 	imes \log(\frac{V_2}{V_1})$.Given values: $

Vedclass · Accepted Answer

$34.465$

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(D) For an isothermal reversible expansion,the work done is given by the formula: $W = -2.303 	imes nRT 	imes \log(\frac{V_2}{V_1})$.Given values: $n = 6 \ mol$,$T = 27 \ ^oC = 300 \ K$,$V_1 = 1 \ L$,$V_2 = 10 \ L$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Substituting the values: $W = -2.303 	imes 6 	imes 8.314 	imes 300 	imes \log(\frac{10}{1})$.$W = -2.303 	imes 6 	imes 8.314 	imes 300 	imes 1$.$W = -34465.3 \ J = -34.465 \ kJ$.The magnitude of maximum work done is $34.465 \ kJ$.

$6$ moles of an ideal gas expand isothermally and reversibly from a volume of $1 \ L$ to a volume of $10 \ L$ at $27 \ ^oC$. What is the maximum work done? (in $kJ$)

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