A horizontal tube of length l closed at both ends, contains an ideal gas of molecular weight M. The tube is rotated at a constant angular velocity ω 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 the fixed end respectively, then choose the correct relation

$\frac{p_{2}}{p_{1}} = e^{\frac{{Mω}^{2} l^{2}}{2^{RT}}}$

$\frac{p_{1}}{p_{2}} = e^{\frac{M {ωl}^{2}}{RT}}$

$\frac{p_{1}}{p_{2}} = e^{\frac{ω l M}{3 RT}}$

$\frac{p_{2}}{p_{1}} = e^{\frac{M^{2} ω^{2} l^{2}}{3 RT}}$

Solution

$\frac{p_{2}}{p_{1}} = e^{\frac{{Mω}^{2} l^{2}}{2^{RT}}}$

Consider the diagram

Consider the elementary part of thickness dx at a distance x from the axis of rotation, then force on this part

Adp = (dm)ω² x ......(i)

where, dm = mass of element

ω = angular velocity

Now, from ideal gas equation

pV = nRT

R - Avagadro constant

V = Volume

n = Number of moles of gas

pA dx = $\frac{dm}{M} RT$

⇒ dm = $\frac{MpA}{RT} dx$ ......(ii)

From Eqs. (i) and (ii)

Adp = $\frac{M p A}{R T} ω^{2} x dx$

0(minimum) to l (maximum) is the limit

$\int_{p_{1}}^{p_{2}} \frac{dp}{p} = \int_{0}^{l} \frac{M ω^{2}}{R T} x dx$

⇒ $\ln (p) = \frac{{Mω}^{2}}{RT} {[\frac{x^{2}}{2}]}_{0}^{l}$

⇒ ln $\frac{p_{2}}{p_{1}} = \frac{{Mω}^{2} l^{2}}{2 RT}$

⇒ $\frac{p_{2}}{p_{1}} = e^{\frac{M ω^{2} l^{2}}{2 RT}}$

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