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(B) At equilibrium, the pressure $P_1$ and $P_2$ on both sides of the partition must be equal $(P_1 = P_2)$, and the temperature $T$ is the same for b

Vedclass · Accepted Answer

$1.8$

Vedclass · Answer

(B) At equilibrium, the pressure $P_1$ and $P_2$ on both sides of the partition must be equal $(P_1 = P_2)$, and the temperature $T$ is the same for both gases.Let $A$ be the cross-sectional area of the container. Let $x$ be the distance of the partition $P$ from the right wall $B$. Then the distance from the left wall $A$ is $(5 - x)$.The volume of the left compartment is $V_1 = A(5 - x)$ and the volume of the right compartment is $V_2 = Ax$.Using the ideal gas equation $PV = nRT = (m/M)RT$, where $m$ is the mass and $M$ is the molar mass:For the left side: $P_1 V_1 = (m/32)RT$For the right side: $P_2 V_2 = (m/18)RT$Since $P_1 = P_2$ and $T$ is constant, we have $\frac{V_1}{V_2} = \frac{m/32}{m/18} = \frac{18}{32} = \frac{9}{16}$.Substituting the volumes: $\frac{A(5 - x)}{Ax} = \frac{9}{16} \implies \frac{5 - x}{x} = \frac{9}{16}$.$16(5 - x) = 9x \implies 80 - 16x = 9x \implies 25x = 80 \implies x = 3.2 \,m$.The distance from the left wall $A$ is $5 - x = 5 - 3.2 = 1.8 \,m$.

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