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(B) Initial temperature $T_1 = 60^{\circ} C = 60 + 273 = 333 \ K$.Let the initial mass of air be $M$. After heating, $\frac{1}{4}$th of the air escape

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$171$

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(B) Initial temperature $T_1 = 60^{\circ} C = 60 + 273 = 333 \ K$.Let the initial mass of air be $M$. After heating, $\frac{1}{4}$th of the air escapes, so the remaining mass is $M_2 = M - \frac{M}{4} = \frac{3M}{4}$.Since the vessel is open, the pressure $P$ remains constant (equal to atmospheric pressure), and the volume $V$ of the vessel is constant.From the ideal gas equation $PV = nRT = \frac{M}{m}RT$, where $m$ is the molar mass of air.Since $P, V, R,$ and $m$ are constant, we have $M_1 T_1 = M_2 T_2$.Substituting the values: $M 	imes 333 = \frac{3M}{4} 	imes T_2$.$T_2 = \frac{333 	imes 4}{3} = 111 	imes 4 = 444 \ K$.Converting back to Celsius: $t = 444 - 273 = 171^{\circ} C$.

Air is filled at $60^{\circ} C$ in a vessel of open mouth. The vessel is heated to a temperature $t^{\circ} C$ so that $\frac{1}{4}$th of the air escapes from the vessel. Assuming air as an ideal gas and the volume of the vessel remains constant, the value of '$t$' is: (in $^{\circ} C$)

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