Question
An open mouthed vessel is filled with air at $60^{\circ} C$. To what temperature should the vessel be heated so that $\frac{1}{4}$ part of air comes out?
✓
Answer
Suppose initially there are n molecules in the vessel while the absolute temperature of gas is $T_1 K$. On heating till $T _2 K \frac{1}{4}$ molecules come out hence the reamining molecule :
$\begin{aligned}& n^{\prime}=n-\frac{1}{4} n=\frac{3}{4} n \\\because \quad nT_1 & =n^{\prime} T_2 \Rightarrow \frac{n}{n^{\prime}}=\frac{T_2}{T_1} \\\frac{3}{\frac{3}{4} n} & =\frac{T_2}{273+60}=\frac{T_2}{333} \\\Rightarrow \quad \frac{4}{3} & =\frac{T_2}{333} \quad \therefore T_2=\frac{4 \times 333}{3} \Rightarrow T_2=444 K \\\therefore\quad T_2 & =444-273=171^{\circ} C\end{aligned} $
$\begin{aligned}& n^{\prime}=n-\frac{1}{4} n=\frac{3}{4} n \\\because \quad nT_1 & =n^{\prime} T_2 \Rightarrow \frac{n}{n^{\prime}}=\frac{T_2}{T_1} \\\frac{3}{\frac{3}{4} n} & =\frac{T_2}{273+60}=\frac{T_2}{333} \\\Rightarrow \quad \frac{4}{3} & =\frac{T_2}{333} \quad \therefore T_2=\frac{4 \times 333}{3} \Rightarrow T_2=444 K \\\therefore\quad T_2 & =444-273=171^{\circ} C\end{aligned} $
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