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(B) For a thermally isolated system,the total internal energy remains constant because $\Delta Q = 0$ and $\Delta W = 0$. Since the initial temperatur

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$0.6$

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(B) For a thermally isolated system,the total internal energy remains constant because $\Delta Q = 0$ and $\Delta W = 0$. Since the initial temperatures of both boxes are the same $(T)$,the final equilibrium temperature $T_f$ will also be $T$. Using the ideal gas law $PV = nRT$,we can find the number of moles in each box: $n_1 = \frac{P_1 V_1}{RT} = \frac{1 	imes V}{RT} = \frac{V}{RT}$ $n_2 = \frac{P_2 V_2}{RT} = \frac{0.5 	imes 4V}{RT} = \frac{2V}{RT}$ When the valve is opened,the total number of moles $n_{total} = n_1 + n_2$ occupies the total volume $V_{total} = V_1 + V_2 = V + 4V = 5V$ at temperature $T$. $n_{total} = \frac{V}{RT} + \frac{2V}{RT} = \frac{3V}{RT}$ Using the ideal gas law for the final state: $P_f V_{total} = n_{total} R T$ $P_f (5V) = \left(\frac{3V}{RT}ight) RT$ $P_f (5V) = 3V$ $P_f = \frac{3}{5} \, atm = 0.6 \, atm$.

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