Haryana PMT Haryana PMT Solved Paper-2010

  • question_answer Air is filled at \[60{}^\circ C\] 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 the vessel remaining constant, the value of T is

    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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