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(B) The total number of moles of the gas remains constant because the system is closed.Let the volume of the smaller vessel be $V$ and the volume of t

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$6$

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(B) The total number of moles of the gas remains constant because the system is closed.Let the volume of the smaller vessel be $V$ and the volume of the larger vessel be $2V$.Initial moles in the larger vessel: $n_1 = \frac{P_1 V_1}{R T_1} = \frac{8 	imes 2V}{R 	imes 1000} = \frac{16V}{1000R}$.Initial moles in the smaller vessel: $n_2 = \frac{P_2 V_2}{R T_2} = \frac{7 	imes V}{R 	imes 500} = \frac{14V}{1000R}$.Total initial moles: $n_{total} = n_1 + n_2 = \frac{16V + 14V}{1000R} = \frac{30V}{1000R}$.At steady state,the total volume is $V_f = V + 2V = 3V$ and the temperature is $T_f = 600 \ K$.Using the ideal gas law for the final state: $n_{total} = \frac{P_f V_f}{R T_f}$.$\frac{30V}{1000R} = \frac{P_f 	imes 3V}{R 	imes 600}$.Canceling $V$ and $R$ from both sides:$\frac{30}{1000} = \frac{3 P_f}{600}$.$\frac{30}{1000} = \frac{P_f}{200}$.$P_f = \frac{30 	imes 200}{1000} = 6 \ kPa$.

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