If 100 L of an ideal gas at a pressure of 2 | vedclass

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(C) For an isothermal process of an ideal gas,$\Delta U = 0$. According to the first law of thermodynamics,$\Delta U = q + w = 0$,which implies $w = -

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$10$

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(C) For an isothermal process of an ideal gas,$\Delta U = 0$. According to the first law of thermodynamics,$\Delta U = q + w = 0$,which implies $w = -q$. Given that the system releases $460.6 \ L \ atm$ of heat,$q = -460.6 \ L \ atm$. Therefore,$w = -(-460.6 \ L \ atm) = +460.6 \ L \ atm$. For a reversible isothermal compression,the work done is given by $w = -2.303 nRT \log(\frac{V_2}{V_1})$. Since $PV = nRT$,we can write $w = -2.303 P_1 V_1 \log(\frac{V_2}{V_1})$. Substituting the given values: $460.6 = -2.303 	imes (2 \ atm 	imes 100 \ L) \log(\frac{X}{100})$. $460.6 = -460.6 \log(\frac{X}{100})$. $-1 = \log(\frac{X}{100})$. $\frac{X}{100} = 10^{-1} = 0.1$. $X = 100 	imes 0.1 = 10 \ L$.

If $100 \ L$ of an ideal gas at a pressure of $2 \ atm$ is compressed isothermally and reversibly to a final volume of '$X$' $L$ and releases $460.6 \ L \ atm$ of heat,the final volume '$X$' (in $L$) is

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