Question 123 of 168

A cylindrical tube $A B$ of length $l$, closed at both ends contains an ideal gas of 1 mol having molecular weight $M$. The tube is rotated in a horizontal plane with constant angular velocity $\omega$ about an axis perpendicular to $A B$ and passing through the edge at end $A$, as shown in the figure. If $P_{A}$ and $P_{B}$ are the pressures at $A$ and $B$ respectively, then (Consider the temperature is same at all points in the tube)

A$P_{B}=P_{A}$B$P_{B}=P_{A} \exp \left(M \omega^{2} l^{2} / 3 R T\right)$C$P_{B}=P_{A} \exp \left(M \omega^{2} l^{2} / R T\right)$D$P_{B}=P_{A} \exp \left(M \omega^{2} l^{2} / 2 R T\right)$

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Answer: (D) $P_{B}=P_{A} \exp \left(M \omega^{2} l^{2} / 2 R T\right)$

$A[(P + dP) - P] = (dm)(\omega^2 x)$
$dP = \dfrac{(dm)}{A} \omega^2 x$
$dP = \dfrac{(\rho)(A)(dx)\omega^2 x}{A}$
also $[PM = \rho RT]$
$\rho = \dfrac{PM}{RT}$
$dP = \left( \dfrac{PM}{RT} \right) \omega^2 x \, dx$
$\int_{P_A}^{P_B} \dfrac{dP}{P} = \dfrac{\omega^2 M}{RT} \int_{0}^{\ell} x \, dx$
$\ln \left( \dfrac{P_B}{P_A} \right) = \dfrac{\omega^2 \ell^2 M}{2RT}$
$P_B = P_A e^{\dfrac{M\omega^2 \ell^2}{2RT}}$