In the given figure a glass tube lies horizontally with the middle `20cm` containing mercury. The two ends of the tube contains air at `27^(@)C` and at a pressure `76 cm` of mercuy. Now the air column on one side in maintained at `0^(@)C` and the other side is maintained at `127^(@)C`. Find the new length of the air column on the cooler side. Neglect the changes in the volume of mercury and of the glass.

Answer

`l = (21840)/(673)cm`

Explanation

At equilibrium pressure on both side is same, for initial condition `PV = n_(1)RT` and `PV = n_(2)RT`
here `n_(1) = n_(1)` and `n_(2) = n_(2)`
for `n_(1) = n'_(1) rArr (pA xx 40)/(R xx 300) = (p'A xx x)/(R xx 273) ….(1)`
for `n_(2) = n'_(2) rArr (pA xx 40)/(R xx 300) = (p'A xx (80 -x))/(R xx 400) ...(2)`
form (1) & (2)
`(x)/(273) = (80-x)/(400) rArr x = (80 xx 273)/(673) = (21840)/(673) cm`