A cylindrical tube $A B$ of length $l$, closed at both ends contains an ideal gas of 1 mol having mo

Question

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)

Professor PC Thomas Classes · Answer

Consider a small gas element of thickness $dx$ at a distance $x$ from the axis. The mass of this element is $dm = \rho A dx$. The pressure difference provides the centripetal force: $A dP = (dm) x \omega^2 = \rho A \omega^2 x dx \Rightarrow dP = \rho \omega^2 x dx$. Using the ideal gas equation $\rho = \frac{PM}{RT}$, we get $dP = \frac{PM}{RT} \omega^2 x dx \Rightarrow \frac{dP}{P} = \frac{M\omega^2}{RT} x dx$. Integrating from $x=0$ to $x=l$ gives $\ln\left(\frac{P_B}{P_A}\right) = \frac{M\omega^2 l^2}{2RT} \Rightarrow P_B = P_A \exp\left(\frac{M\omega^2 l^2}{2RT}\right)$.