Question

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

Answer 1

Given $T_1=60+273=333K$
Let
$M_1=M,$
As ${\frac{1}{4}}^{th}$ of air escapes,
$M_2=M-\frac{M}{4}=\frac{3M}{4}$
If pressure and volume of the gas remains constant, then $MT=$ constant
$\therefore \frac{T_2}{T_1}=\frac{M_1}{M_2}=\left(\frac{M}{\frac{3M}{4}}\right)=\frac{4}{3}$
$\Rightarrow T_2=\frac{4}{3}\times T_1=\frac{4}{3}\times 333=444K={171}^{\circ }C$

Why this question? To understand why the gas escapes out, That is actually to maintain pressure. If temperature is raised, the gas molecules start vibrating with higher amplitude, so they need more space. So, to maintain pressure, the gas expands itself. Caution to use temperature in Kelvin scale.