Two spherical flasks having total volume `V_(0) = 1.0 L` containing air are connected by a tube diameter `d =6` mm and length `lambda = 1` m. A small droplet of mercury contained in the tube is at its middle at `0^(@)C`. By what distance do the mercury droplets move if the flask `1` is heated by `2^(@)C` while flask `2` cooled by `2^(@)C`. Ignore any expansion of flask wall.

Answer

`0.259`

Explanation

Let `n_(1)` = no. of moles in vessel -`1`
`n_(2)`= no. of moles in vessel -`2`
`P_(1)`= initial pressure in both vessels
`P_(2)`= final pressure om both vessels.
Initially volume of vessel-`1` = volume of vessel-`2`
`(n_(1)RT_(0))/(P_(1)) = (n_(2)RT_(0))/(P_(1)) implies n_(1) = n_(2)`
Finally , `V_(1) = (n_(1)RT_(1))/(P_(2))` & `V_(2)=(n_(2)RT_(2))/(P_(2)) = ((n_(1)R)/(P_(2)))T_(2)`
`(V_(1))/(V_(2)) = (T_(1))/(T_(2)) = (275)/(271)` & `V_(1) + V_(2) = V_(0) + lambdaA`
Displacemet of mercury droplets `DeltaL = (V_(1)-V_(2))/(A) = (4)/(546)((V_(0) + lambdaA)/(A)) = 0.26m`