Air is filled at 60∘C in a vessel of open mouth. The vessel is heated to a temperature T, so that 14th part of the air escapes. Assuming the volume of the vessel remains constant, the value of T is
A
80∘C
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B
44∘C
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C
333∘C
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D
171∘C
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Solution
The correct option is D171∘C Mass of the air left in the vessel M−M4=3M4 (as 14th part of air escapes) Given, initial temperature T1=60∘C=333K
∵PV=nRT [Ideal gas equation ] PV=(MMolecular Mass)RT Pressure and volume of the gas remain constant. [because vessel is open to atmosphere, pressure will be the same as atmospheric pressure] i.e MT= constant. ∴T2T1=M1M2=⎛⎜
⎜
⎜⎝M3M4⎞⎟
⎟
⎟⎠=43 ⇒T2=43×T1=43×333=444K=171∘C