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? Key Concept: If temperature is raised, the gas molecules start vibrating with higher amplitude and they need more space. So, to maintain pressure, the gas expands. Caution: Use temperature in Kelvin scale.