A horizontal tube of length `I` closed at both ends contains an ideal gas of molecular weight `M`, The tube is rotated at a constant angular velocity `omega` about a vertical axis passing through an end. Assuming the temperature to be uniform and constant, show that
`p_(2)=p_(1)e^((Momega^(2)l^(2))/(2RT),`
where `p_(2)` and `p_(1)` denote the pressures at the free end and the fixed end respectively.

Explanation

Consider an element of the gas between `x` and `x + dx` from `O`.
Let `p` and `p + dp` be the pressures as shown.
The force acting on the element towards the centre
`= (p + dp - p)A = dp A`,
`A`: area of `X`-section
This elements is moving in circle of radius `x`
`dpA = d m omega^(2) x` (i)
`pV = nRT`
`pAd x = (d m)/(M)RT`
`dm = (MpA)/(RT)dx` (ii)
From (i) and (ii)
`dp A = (MpA)/(RT)dx omega^(2) x`
`underset(p_(1)) overset(p_(2))(int) (dp)/(p) = (M omega^(2))/(RT) underset(0)overset(l)(int) x dx`
`1n (p_(2))/(p_(1)) = (M omega^(2) l^(2))/(2RT)`
`p_(2) = p_(1) e^((M omega^(2) l^(2))/(RT))`