Two vessels are filled with ideal gases A and | vedclass

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(D) Initial state for gas $A$: $P_1 = 8 \ atm$, $V_1 = 12 \ L$. Total volume after opening the stop cock: $V_{total} = 12 \ L + 8 \ L = 20 \ L$.Since

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$4.8, 2.0$

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(D) Initial state for gas $A$: $P_1 = 8 \ atm$, $V_1 = 12 \ L$. Total volume after opening the stop cock: $V_{total} = 12 \ L + 8 \ L = 20 \ L$.Since the temperature is constant, we use Boyle's Law $(P_1V_1 = P_2V_2)$ for each gas independently.For gas $A$: $8 \ atm 	imes 12 \ L = P_A 	imes 20 \ L \implies P_A = \frac{96}{20} = 4.8 \ atm$.For gas $B$: $5 \ atm 	imes 8 \ L = P_B 	imes 20 \ L \implies P_B = \frac{40}{20} = 2.0 \ atm$.Thus, the partial pressures of $A$ and $B$ are $4.8 \ atm$ and $2.0 \ atm$ respectively.

Two vessels are filled with ideal gases $A$ and $B$ and are connected through a pipe of zero volume as shown in the figure. The stop cock is opened and the gases are allowed to mix homogeneously and the temperature is kept constant. The partial pressures of $A$ and $B$ respectively (in $atm$) are:

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