In the figure, chamber A contains a gas, a mo | vedclass

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(D) Let $M$ be the mass of chamber $B$ and $m$ be the mass of each ball. The total downward force exerted by the chamber and the balls on the gas is $

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$1/n$

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(D) Let $M$ be the mass of chamber $B$ and $m$ be the mass of each ball. The total downward force exerted by the chamber and the balls on the gas is $F = (M + nm)g$. If $A$ is the cross-sectional area of the chamber, the pressure $P$ of the gas is given by $P = F/A = (M + nm)g / A$. When one ball is removed, the new force is $F^{\prime} = (M + (n-1)m)g$. The new pressure $P^{\prime}$ is $P^{\prime} = F^{\prime} / A = (M + (n-1)m)g / A$. The difference in pressure is $P - P^{\prime} = (M + nm)g/A - (M + nm - m)g/A = mg/A$. Thus, $(P - P^{\prime}) / P = (mg/A) / ((M + nm)g/A) = m / (M + nm)$. Assuming the mass of the chamber $M$ is negligible compared to the total mass of the balls (or that the problem implies the weight is primarily due to the balls), we have $M \approx 0$. Then, $(P - P^{\prime}) / P = m / (nm) = 1/n$.

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