Question

A cylindrical tube \(AB\) 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 \(AB\) 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 to be same at all points in the tube) 

• A. \( P_B = P_A \exp\!\left(\dfrac{M\omega^2l^2}{RT}\right) \)
• B. \( P_B = P_A \)
• C. \( P_B = P_A \exp\!\left(\dfrac{M\omega^2l^2}{3RT}\right) \)
• D. \( P_B = P_A \exp\!\left(\dfrac{M\omega^2l^2}{2RT}\right) \)

Hint:

In rotating systems, pressure variation arises due to centrifugal force, analogous to pressure variation in a gravitational field.

Answer 1

The Correct Option is A

The tube is rotating with angular velocity \( \omega \) about point \(A\). A gas molecule at a distance \(x\) from \(A\) experiences a centrifugal force given by \[ dF = m\omega^2 x, \] where \(m\) is the mass of a molecule.
Step 1: Pressure gradient due to centrifugal force.
For equilibrium of a thin gas element of thickness \(dx\), \[ dP = \rho \omega^2 x \, dx, \] where \( \rho \) is the mass density of the gas.
Using the ideal gas relation, \[ \rho = \frac{MP}{RT}, \] we get \[ \frac{dP}{P} = \frac{M\omega^2}{RT} x\, dx. \] Step 2: Integrate from \(A\) to \(B\).
At \(A\), \(x=0\) and pressure is \(P_A\). At \(B\), \(x=l\) and pressure is \(P_B\). \[ \int_{P_A}^{P_B} \frac{dP}{P} = \frac{M\omega^2}{RT} \int_0^{l} x\, dx. \] \[ \ln\!\left(\frac{P_B}{P_A}\right) = \frac{M\omega^2}{RT}\left[\frac{l^2}{2}\right] \times 2. \] \[ \ln\!\left(\frac{P_B}{P_A}\right) = \frac{M\omega^2l^2}{RT}. \] Step 3: Final expression.
Taking exponential on both sides, \[ P_B = P_A \exp\!\left(\frac{M\omega^2l^2}{RT}\right). \] Final Answer: \[ \boxed{P_B = P_A \exp\!\left(\dfrac{M\omega^2l^2}{RT}\right)} \]