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Air is filled at $60^{\circ} \mathrm{C}$ in a vessel of open mouth. The vessel is heated to a temperature $t^{\circ} \mathrm{C}$ so that $\frac{1}{4}$ th of the air is escaped from the vessel. Assuming air as ideal gas and the volume of the vessel remaining constant, then the value of ' $t$ ' is
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Verified Answer
The correct answer is:
$171^{\circ} \mathrm{C}$
Initial temperature, $T_1=60^{\circ} \mathrm{C}=60+273=333 \mathrm{~K}$
Mass of the air left in the vessel, $M_2=M-\frac{M}{4}$
$$
=\frac{3 \mathrm{M}}{4}
$$
Here pressure and volume remain constant from ideal gas equation
$$
\begin{aligned}
& P V=n R T \\
& P V=\left(\frac{M}{m}\right) R T
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{MT}=\text { constant } \\
& \mathrm{M}_1 \mathrm{~T}_1-\mathrm{M}_2 \mathrm{~T}_2 \\
& \mathrm{~T}_2=\frac{\mathrm{M}_1 \mathrm{~T}_1}{\mathrm{M}_2}=\frac{\mathrm{M} \times 333}{3 \mathrm{M}} \times 4 \\
& =444 \mathrm{~K} \\
& \Rightarrow \mathrm{t}=\mathrm{T}_2=171^{\circ} \mathrm{C}
\end{aligned}
$$
Mass of the air left in the vessel, $M_2=M-\frac{M}{4}$
$$
=\frac{3 \mathrm{M}}{4}
$$
Here pressure and volume remain constant from ideal gas equation
$$
\begin{aligned}
& P V=n R T \\
& P V=\left(\frac{M}{m}\right) R T
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{MT}=\text { constant } \\
& \mathrm{M}_1 \mathrm{~T}_1-\mathrm{M}_2 \mathrm{~T}_2 \\
& \mathrm{~T}_2=\frac{\mathrm{M}_1 \mathrm{~T}_1}{\mathrm{M}_2}=\frac{\mathrm{M} \times 333}{3 \mathrm{M}} \times 4 \\
& =444 \mathrm{~K} \\
& \Rightarrow \mathrm{t}=\mathrm{T}_2=171^{\circ} \mathrm{C}
\end{aligned}
$$
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