Air is filled at 60^oC in a vessel of open mouth. The vessel is heated to a temperature T so that 1/4th part of air escapes. Assuming the volume of vessel remaining constant, the value of T is : 

171^oC

\(M_1 = M ; T_1 = 60 + 273 = 333 K\)

\(M_2 = M - \frac{M}{4} = \frac{3M}{4}\) ... [as 1/4th part of air escapes]

If pressure and volume of the gas remain constant, then : 

MT = constant

\(\Rightarrow \frac{T_2}{T_1} = \frac{M_1}{M_2} = \frac{M}{3M/4} = frac{4}{3}\)

\(T_2 = \frac{4}{3} T_1 = \frac{4}{3}*333\)

\(T_2 = 444 K\) = 171^oC