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(B) Initial mass of air is $M_1 = M$ at temperature $T_1 = 60 + 273 = 333 \ K$.Since the vessel is open,the pressure remains constant (atmospheric pre

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$171$

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(B) Initial mass of air is $M_1 = M$ at temperature $T_1 = 60 + 273 = 333 \ K$.Since the vessel is open,the pressure remains constant (atmospheric pressure).According to the ideal gas law,$PV = nRT = \frac{M}{m}RT$,where $m$ is the molar mass of air.Since $P$,$V$,and $m$ are constant,we have $M_1 T_1 = M_2 T_2$.After heating,one-fourth of the air escapes,so the remaining mass is $M_2 = M - \frac{M}{4} = \frac{3M}{4}$.Substituting the values: $M 	imes 333 = \frac{3M}{4} 	imes T_2$.$T_2 = 333 	imes \frac{4}{3} = 444 \ K$.Converting back to Celsius: $T = 444 - 273 = 171^{\circ}C$.

$A$ vessel with an open mouth contains air at $60^{\circ}C$. When the vessel is heated up to temperature $T$,one-fourth of the air goes out. The value of $T$ is ..... $^{\circ}C$

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