A thermally isolated cylindrical closed vessel of height 8 m is kept vertically . It is divided into two equal parts by a diathermic (perfect conductor) frictionless partition of mass 8.3 kg . Thus the partition is held initially at a distance of 4 m from the top , as shown in the schematic figure below . Each o the two parts of the vessel contians 0.1 mole of an ideal gas at temperature 300 K. The partition is now released and moves without any gas leaking from one part of the vessel to the other . When equilibrium is reached, the distance of the parition form one part of the vessel to the otehr . When equilibrium is reached the distacne of the partition from the top (in w) will be _______(take the accelearation due to gravity = 10 `ms^(-2)` and the universal gas constant ` = 8.3 J ` mol^(-1) K^(-1)`).

Answer

6

Explanation

Initially `PA(4) = 0.1 xx R xx 300……….` (i) for both part
Left final distance of pistion from the top is y .
` 0.1 xx R xx 300 = P_(1) xx A. y = P_(2) A(8-y)………………(ii)`
`P_(1) A + 8.3 g= P_(2)A`
`30 R + 8.3g = 30R`
` y = 8 - y`
`30+10=30`
` y = 8- y`
` 3 + y = 3`
` y = 8 - y`
`24 + 8y - 3y - y^(2) = 3y`
` y^(2) - 2y - 24 = 0 , y = (2 pm 4 + 96)/( 2 xx 1) = ( 2 pm 10)/(2) = 6m `