Question

Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at $0^o$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 ${62}^o$C. What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible.

• A. $P=103.75$ cm of Hg.
• B. $P=83.75$ cm of Hg.
• C. $P=93.75$ cm of Hg.
• D. $P=3.75$ cm of Hg.

Answer 1

Answer: (B)

$P=83.75$ cm of Hg.

Solution:

At first , at 273 K

the mole in each bulb = $\frac{n_{total}}{2}=\frac{1}{2}\frac{P.2.V}{RT}=\frac{76V}{273\times R}$

on heating second bulb , some mole of gas are transferred to another bulb until the pressure of both bulb become same .

$\therefore n_1=\frac{PV}{273R}$ and $n_2=\frac{PV}{335R}$

$n_{total}=n_1+n_2$

$2\times \frac{76V}{273R}=\frac{PV}{273R}+\frac{PV}{335R}\Rightarrow P$

= $83.75$ cm of Hg.

Hence the correct option: B