If 2.5 moles of an ideal gas at a certain tem | vedclass

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(C) The formula for reversible isothermal work is: $W = -2.303 nRT \log \left(\frac{V_2}{V_1}\right)$Given: $n = 2.5 \ mol$, $V_1 = 2 \ dm^3$, $V_2 =

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$345$

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(C) The formula for reversible isothermal work is: $W = -2.303 nRT \log \left(\frac{V_2}{V_1}ight)$Given: $n = 2.5 \ mol$, $V_1 = 2 \ dm^3$, $V_2 = 20 \ dm^3$, $W = -16.5 \ kJ = -16500 \ J$, $R = 8.314 \ J \ K^{-1} \ mol^{-1}$Substituting the values: $-16500 = -2.303 	imes 2.5 	imes 8.314 	imes T 	imes \log \left(\frac{20}{2}ight)$$-16500 = -2.303 	imes 2.5 	imes 8.314 	imes T 	imes \log(10)$Since $\log(10) = 1$, we have: $-16500 = -2.303 	imes 2.5 	imes 8.314 	imes T$$-16500 = -47.87 	imes T$$T = \frac{16500}{47.87} \approx 344.68 \ K$Rounding to the nearest value, $T \approx 345 \ K$.

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