If 1 L of an ideal gas at a pressure of 20 | vedclass

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(D) For an isothermal reversible expansion of an ideal gas,the heat absorbed $(q_{rev})$ is equal to the work done by the system $(-w_{rev})$.$q_{rev}

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$100$

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(D) For an isothermal reversible expansion of an ideal gas,the heat absorbed $(q_{rev})$ is equal to the work done by the system $(-w_{rev})$.$q_{rev} = -w_{rev} = 2.303 P_1 V_1 \log \frac{V_2}{V_1}$Given $P_1 = 20 \ atm$,$V_1 = 1 \ L$,and $q_{rev} = 92.12 \ L \ atm$.$92.12 = 2.303 	imes 20 	imes 1 	imes \log \frac{V_2}{1}$$92.12 = 46.06 \log V_2$$\log V_2 = \frac{92.12}{46.06} = 2$$V_2 = 10^2 = 100 \ L$Thus,$X = 100$.

If $1 \ L$ of an ideal gas at a pressure of $20 \ atm$ expands isothermally and reversibly to a final volume of '$X$' $L$ by absorbing $92.12 \ L \ atm$ heat,'$X$' (in $L$) is

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