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(A) For an isothermal and reversible expansion of an ideal gas,the heat absorbed $(q)$ is given by the formula:$q = -w = 2.303 \ nRT \ \log \left( \fr

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$955.7$

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(A) For an isothermal and reversible expansion of an ideal gas,the heat absorbed $(q)$ is given by the formula:$q = -w = 2.303 \ nRT \ \log \left( \frac{V_2}{V_1} ight)$Given:$n = 0.75 \ mol$$T = 27 + 273 = 300 \ K$$R = 8.314 \ J \ K^{-1} \ mol^{-1}$$V_1 = 15 \ L, V_2 = 25 \ L$Substituting the values:$q = 2.303 	imes 0.75 	imes 8.314 	imes 300 	imes \log \left( \frac{25}{15} ight)$$q = 2.303 	imes 0.75 	imes 8.314 	imes 300 	imes \log \left( \frac{5}{3} ight)$$q = 2.303 	imes 0.75 	imes 8.314 	imes 300 	imes (\log 5 - \log 3)$$q = 2.303 	imes 0.75 	imes 8.314 	imes 300 	imes (0.70 - 0.48)$$q = 2.303 	imes 0.75 	imes 8.314 	imes 300 	imes 0.22$$q = 955.7 \ J$

Calculate $q$ when $0.75 \ mol$ of an ideal gas expands isothermally and reversibly at $27 \ ^oC$ from a volume of $15 \ L$ to $25 \ L$.
(Given $\log 5 = 0.70, \log 3 = 0.48$)
...... $Joule$

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Calculate $q$ when $0.75 \ mol$ of an ideal gas expands isothermally and reversibly at $27 \ ^oC$ from a volume of $15 \ L$ to $25 \ L$.(Given $\log 5 = 0.70, \log 3 = 0.48$)...... $Joule$