A) \[80{}^\circ C\]
B) \[444{}^\circ C\]
C) \[333{}^\circ C\]
D) \[171{}^\circ C\]
Correct Answer: D
Solution :
For open mouth vessel, pressure is constant. Volume is also given constant. Hence from \[pV=\mu RT\] \[=\left( \frac{m}{M} \right)RT\] \[\Rightarrow \] \[T\propto \frac{1}{m}\] \[\Rightarrow \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{{{m}_{2}}}{{{m}_{1}}}\] \[\because \] \[\frac{1}{4}th\] part escapes, so remaining mass in the vessel \[{{m}_{2}}=\frac{3}{4}{{m}_{1}}\] \[\Rightarrow \] \[\frac{(273+60)}{T}=\frac{3/4{{m}_{1}}}{{{m}_{1}}}\] \[\Rightarrow \] \[T=444K={{171}^{o}}C\]
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