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(B) The total number of moles of $H_2$ gas remains constant when the valve is opened.$n_{total} = n_A + n_B$Using the ideal gas equation $PV = nRT$,we

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(B) The total number of moles of $H_2$ gas remains constant when the valve is opened.$n_{total} = n_A + n_B$Using the ideal gas equation $PV = nRT$,we have $n = \frac{PV}{RT}$.$n_A = \frac{4 \ atm 	imes 2 \ L}{R 	imes 400 \ K} = \frac{8}{400R} = \frac{0.02}{R}$$n_B = \frac{8 \ atm 	imes 4 \ L}{R 	imes 800 \ K} = \frac{32}{800R} = \frac{0.04}{R}$$n_{total} = \frac{0.02}{R} + \frac{0.04}{R} = \frac{0.06}{R}$After opening the valve,the gas occupies both compartments. Since the vessels are thermally insulated,the temperature in each compartment remains at its initial value ($400 \ K$ and $800 \ K$ respectively).$n_{total} = \frac{P_f V_A}{R T_A} + \frac{P_f V_B}{R T_B}$$\frac{0.06}{R} = P_f \left[ \frac{2}{400R} + \frac{4}{800R} ight]$$0.06 = P_f \left[ 0.005 + 0.005 ight]$$0.06 = P_f 	imes 0.01$$P_f = 6 \ atm$

Hydrogen gas is present in compartment $A$ and compartment $B$ as shown in the following figure. If the valve is opened,then the final pressure in the compartments is ............. $atm$. [Assume the temperature of each vessel remains constant at its initial value]

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