Question

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

• A. 80 cm of Hg
• B. 87.2 cm of Hg
• C. 83.8 cm of Hg
• D. 78.5 cm of Hg

Answer 1

The correct option is C

83.8 cm of Hg

We have, $PV=nRT\Rightarrow n=\left(\frac{PV}{RT}\right)$.

As total number of gas molecules in the bulbs would remain constant, we have $n_1+n_2=n_1\prime +n{\prime }_2$

$\frac{P_{0}V}{R{T}_0}+\frac{P_{0}V}{R{T}_0}=\frac{P.V}{RT}+\frac{PV}{R{T}_0}$

$\Rightarrow \frac{2P_0}{T_0}=P(\frac{1}{T}+\frac{1}{T_0}).$

Hence the new pressure $P=\left(\frac{2T}{T+T_0}\right).P_0.$

Here, $P_0=76cms$ of Hg (Initial pressure), $T_0=0^{\circ }C=273K.$

Given, $T={62}^{\circ }C=335K$

$\Rightarrow P=\left[\frac{2\times 335}{(273+335)}\right]\times 76=83.75cms$ of Hg.