Avessel with open mouth contains air at $60^oC$. When the vessel is heated upto temperature $T$, one fourth of the air goes out. The value of $T$ is ..... $^oC$
Diffcult
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$M_{1}=M, T_{1}=60+273$

$=333 K, M_{2}$

$=M-\frac{M}{4}=\frac{3 M}{4}$

[As $1/ 4$ th part of air escapes]

If pressure and volume of gas remains constant then $\mathrm{MT}=$ constant

$\therefore \frac{T_{2}}{T_{1}}=\frac{M_{1}}{M_{2}}=\left(\frac{M}{3 M / 4}\right)=\frac{4}{3} \Rightarrow T_{2}$

$=\frac{4}{3} \times T_{1}=\frac{4}{3} \times 333=444 \mathrm{K}$

$=171^{\circ} \mathrm{C}$

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