Question

Air is filled at ${60}^{\circ }\text{C}$ in a vessel of open mouth. The vessel is heated to a temperature $T$, so that $\frac{1}{4}\text{th}$ part of the air escapes. Assuming the volume of the vessel remains constant, the value of $T$ is

• A. ${80}^{\circ }\text{C}$
• B. ${44}^{\circ }\text{C}$
• C. ${333}^{\circ }\text{C}$
• D. ${171}^{\circ }\text{C}$

Answer 1

The correct option is D ${171}^{\circ }\text{C}$

Mass of the air left in the vessel $M-\frac{M}{4}=\frac{3M}{4}$
(as $\frac{1}{4}\text{th}$ part of air escapes)
Given, initial temperature $T_1={60}^{\circ }\text{C}=333\text{K}$

$\because PV=nRT$ [Ideal gas equation ]
$PV=\left(\frac{M}{\text{Molecular Mass}}\right)RT$
Pressure and volume of the gas remain constant. [because vessel is open to atmosphere, pressure will be the same as atmospheric pressure]
i.e $MT=$ constant.
$\therefore \frac{T_2}{T_1}=\frac{M_1}{M_2}=\left(\frac{M}{\frac{3M}{4}}\right)=\frac{4}{3}$
$\Rightarrow T_2=\frac{4}{3}\times T_1=\frac{4}{3}\times 333=444\text{K}={171}^{\circ }\text{C}$