Question

Two glass bulbs of equal volumes are connected by a narrow tube and are filled with a gas at $0^{\circ }\text{C}$ and a pressure of $76\text{cm}$ of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at ${62}^{\circ }\text{C}$. What is the new value of the pressure inside the bulbs ? Consider the volume of the connecting tube is negligible.

• A. $75.75\text{cm Hg}$
• B. $86.50\text{cm Hg}$
• C. $70\text{cm Hg}$
• D. $83.75\text{cm Hg}$

Answer 1

The correct option is D $83.75\text{cm Hg}$

Let $V$ be the volume and $n$ be the no. of moles of gas in each bulb.

For left bulb, equation of state is $PV=nRT$.

Substituting the given data, we get

Initially,
$76\times V=nR\times 273.....\left(1\right)$

Let $x$ moles of gas shift from the high temperature side to low temperature side. Finally, let $P^{\prime }$ be the new pressure.


Then, for left bulb:
$P^{\prime }\times V=(n+x)R\times 273....\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get

$\frac{P^{\prime }}{76}=\frac{n+x}{n}......\left(3\right)$

For right bulb

Initially:

$76\times V=nR\times 273......\left(4\right)$

Finally:

$P^{\prime }\times V=(n-x)R\times 335.....\left(5\right)$

From $\left(4\right)$ and $\left(5\right)$,

$\frac{P^{\prime }}{76}=\frac{n-x}{n}\times \frac{335}{273}......\left(6\right)$

From $\left(3\right)$ and $\left(6\right)$,

$\frac{n+x}{n}=\frac{n-x}{n}\times \frac{335}{273}$

$\Rightarrow n=\frac{608}{62}x......\left(7\right)$

Substituting the value of $\left(7\right)$ in $\left(3\right)$, we get

$\frac{P^{\prime }}{76}=1+\frac{62}{608}$

$\Rightarrow P^{\prime }=\frac{670}{608}\times 76=83.75\text{cm Hg}$