The work done when 2 mol of an ideal gas exp | vedclass

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(D) Given: $n = 2 \ mol$,$V_1 = 1 \ L$,$V_2 = 10 \ L$,$T = 300 \ K$,$R = 0.0083 \ kJ \ K^{-1} \ mol^{-1}$.For a reversible isothermal expansion,the wo

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$11.5$

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(D) Given: $n = 2 \ mol$,$V_1 = 1 \ L$,$V_2 = 10 \ L$,$T = 300 \ K$,$R = 0.0083 \ kJ \ K^{-1} \ mol^{-1}$.For a reversible isothermal expansion,the work done is given by the formula:$W = -2.303 \ nRT \log \frac{V_2}{V_1}$.Substituting the values:$W = -2.303 	imes 2 	imes 0.0083 	imes 300 	imes \log \frac{10}{1}$.$W = -2.303 	imes 2 	imes 0.0083 	imes 300 	imes 1$.$W = -11.47 \ kJ \approx -11.5 \ kJ$.The magnitude of work done by the system is $11.5 \ kJ$.

The work done when $2 \ mol$ of an ideal gas expands reversibly and isothermally from a volume of $1 \ L$ to $10 \ L$ at $300 \ K$ is $(R = 0.0083 \ kJ \ K^{-1} \ mol^{-1})$ (in $kJ$)

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