Question

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

• A. 83.75 cm of Hg
• B. 63.75 cm of Hg
• C. 183.75 cm of Hg
• D. 89.75 cm of Hg

Answer 1

The correct option is A 83.75 cm of Hg

According to gas equation PV = nRT, initially the moles in the two bulbs will be -
$n_1=n_2=\frac{PV}{RT}=\frac{76\times V}{R\times 273}$ ...(1)
Now when one of the bulb is placed in a water bath at ${62}^{\circ }$C, redistribution of gas will take place and the system will acquire a new pressure say $P^{\prime }$. In this situation number of moles in the two bulbs will be
$n_1^{\prime }=\frac{P^{\prime }V}{R\times 273}and{n}_2^{\prime }=\frac{P^{\prime }V}{R\times 335}$ (2)
Now as no gas leaves or enter the system, the mass of system remains constant, i.e.
$n_1+n_2=n_1^{\prime }+n_2^{\prime }$
So from eqn. (1) and (2)
$2\times \frac{76V}{R\times 273}=\frac{P^{\prime }V}{R}[\frac{1}{273}+\frac{1}{335}]$
i.e., $P^{\prime }=2\times 76\times \frac{335}{608}=83.75\text{cm of Hg}$