Question

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 $\omega$ about a vertical axis passing through an end. Assuming the temperature to be uniform and constant. If ${\rho }_1$ and ${\rho }_2$ denote the pressure at free and the fixed end respectively, then choose the correct relation.

• A. $\frac{{\rho }_2}{{\rho }_1}=e^{\frac{M{\omega }^2{l}^2}{2RT}}$
• B. $\frac{{\rho }_1}{{\rho }_2}=e^{\frac{M\omega l^2}{RT}}$
• C. $\frac{{\rho }_1}{{\rho }_2}=e^{\frac{\omega lM}{3RT}}$
• D. $\frac{{\rho }_2}{{\rho }_1}=e^{\frac{M^2{\omega }^2{l}^2}{3RT}}$

Answer 1

The correct option is A $\frac{{\rho }_2}{{\rho }_1}=e^{\frac{M{\omega }^2{l}^2}{2RT}}$

Rotation create centrifugal acceleration($r{\omega }^2$) which mimic gravity.

Consider a thin slice of tube at distance $r$ and $r+dr$, which has pressure difference $d\rho$.

Use the pressure difference equation in gravity by replacing gravity by centrifugal acceleration.

$d\rho =\sigma r{\omega }^{2}dr$, $\sigma$ is mass density.

Ideal gas law : $M=\frac{\sigma RT}{\rho }$, which gives, $d\rho =\frac{M\rho {\omega }^2}{RT}rdr$.

Integrating gives, $\rho \left(r\right)=e^{\frac{M{\omega }^2{r}^2}{2RT}}$=$\frac{{\rho }_2}{{\rho }_1}$