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(A) Initially,both parts $A$ and $B$ have equal volumes $(V)$ and equal pressures $(P = 1 \ atm)$.When the partition is removed,the total volume becom

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Pressure remains unchanged in both $A$ and $B$.

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(A) Initially,both parts $A$ and $B$ have equal volumes $(V)$ and equal pressures $(P = 1 \ atm)$.When the partition is removed,the total volume becomes $2V$.According to Boyle's Law,$P_1V_1 = P_2V_2$.For the total system,the final pressure $P_f$ is calculated as: $P_f = \frac{P_A V + P_B V}{2V} = \frac{1 	imes V + 1 	imes V}{2V} = 1 \ atm$.Since the gases mix and occupy the total volume,the partial pressure of each gas decreases to $0.5 \ atm$,but the total pressure of the mixture remains $1 \ atm$ throughout the container.

$A$ container is divided into two equal parts $A$ and $B$ containing $H_2$ and $O_2$ respectively,each at $1 \ atm$ pressure. If the wall separating the two sections is removed,what happens to the pressure?

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