If 0.75 mol of an ideal gas expands isotherm | vedclass

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(A) For an isothermal reversible expansion of an ideal gas,the maximum work done is given by the formula: $W_{max} = -2.303 \ nRT \ \log(\frac{V_2}{V_

Vedclass · Accepted Answer

$957$

Vedclass · Answer

(A) For an isothermal reversible expansion of an ideal gas,the maximum work done is given by the formula: $W_{max} = -2.303 \ nRT \ \log(\frac{V_2}{V_1})$.Given values are: $n = 0.75 \ mol$,$T = 27 \ ^{\circ}C = 300 \ K$,$V_1 = 15 \ L$,$V_2 = 25 \ L$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Substituting these values: $W_{max} = -2.303 	imes 0.75 	imes 8.314 	imes 300 	imes \log(\frac{25}{15})$.$W_{max} = -2.303 	imes 0.75 	imes 8.314 	imes 300 	imes \log(1.667)$.$W_{max} = -2.303 	imes 0.75 	imes 8.314 	imes 300 	imes 0.2218 \approx -957.5 \ J$.The magnitude of maximum work obtained is $957.5 \ J$.

If $0.75 \ mol$ of an ideal gas expands isothermally at $27 \ ^{\circ}C$ from $15 \ L$ to $25 \ L$,the maximum work obtained is ...... $J$. (in $.5$)

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