Question

Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at $0^{\circ }C$ and a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at ${62}^{\circ }C$. What will be the new pressure (in cm of Hg) in the tube? (Answer upto two decimal places)

Answer 1

Let the right bulb be placed in water bath at ${62}^{\circ }C$ and the left bulb be placed in melting ice at $0^{\circ }C$.

Let molecules be tranfered from right bulb to left bulb.
For left bulb, using $PV=nRT$,
Initially, $76\times V=nR\times 273$
Finally, $P^{\prime }\times V=(n+x)R\times 273$
$\therefore \frac{P^{\prime }}{76}=\frac{(n+x)}{n}$ $\cdots \left(i\right)$

Now consider the right bulb.
Initially, $76\times V=nR\times 273$
Finally, $P^{\prime }\times V=(n-x)R\times (273+62)$
or $\frac{P^{\prime }}{76}=\frac{(n-x)}{n}\times \frac{335}{273}$ $\cdots \left(ii\right)$

From (i) and (ii),
$\frac{n+x}{n}=\frac{n-x}{n}\times \frac{335}{273}$
or $273(n+x)=335(n-x)$
or $273x+335x=335n-273n$
or $608x=62n$ or $n=\frac{608}{62}x$ $\cdots \left(iii\right)$

Put (iii) in (i) to obtain $P^{\prime }$
$\therefore \frac{P^{\prime }}{76}=\frac{n+x}{n}=\frac{n}{n}+\frac{x}{n}=1+\frac{x}{n}$
or
$P^{\prime }=76(1+\frac{62}{608})=\frac{76\times 670}{608}{P}^{\prime }=83.75cm\text{of Hg}$

$\therefore$ New pressure $83.75cm\text{of Hg}$