A horizontal tube of length l closed at both ends, contains the ideal has 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. If p1 and p2 denote the pressure at free and fixed end respectively, then choose the correct relation.
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The correct relation between the pressure is (p2 = p1) eM(ω^2l^2) / 2RT
Explanation:
- As this element is going in a circle of radius X.
- The force acting in the element towards the centre is Adp.
- Where A is the cross sectional area.
Adp=(dm)(ω2)x ...(1)
Where dm=mass of the element.
Using pV=nRT on this element.
pA.dx = dm / MRT
dm = MpA / RT.dx
Putting in equation (1)
Adp = MpA / RT(ω^2)x.dx
OR
∫ p2 - p1 dp / p = ∫l-0 m(ω^2) RT x.dx
In p2 / p1 = M(ω^2)l^2 / 2RT
(p2 = p1) eM(ω^2l^2) / 2RT
Thus the correct relation between the pressure is (p2 = p1) eM(ω^2l^2) / 2RT
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