Rotation of adiabatic tube

Suppose, in figure 2, pressure and density in the compartment at distance $y$ are $p$ and $\rho$ respectively. Also, suppose that the initial density of the gas is $\rho_0$ .

As the process is adiabatic [for diatomic gas, heat capacity ratio $\gamma\ = \frac{7}{5}$ ], $p_0 (\Delta x) ^ \gamma = p (\Delta y) ^\gamma$ $\rightarrow \frac{\Delta y}{\Delta x} = (\frac{p_0}{p})^{\frac{1}{\gamma}} ~~~...~~~...~~~...~~~(i)$ Again, the number of molecules in the compartment is same in both figures. Therefore, $\frac{\rho _0}{\rho} = \frac{\Delta y}{\Delta x}$ Using eqn (i), $\rho = (\frac{p}{p_0})^{\frac{1}{\gamma}} \cdot \rho_0 ~~~...~~~...~~~...~~~(ii)$ Let the cross-sectional area of the tube is $A_0$ . So, the net force on the compartment we are talking about is $[p(y+\Delta y) - p(y)]A_0$ . And this is equal to the centripetal force. Therefore, $[p(y+\Delta y) - p(y)]A_0 =(\rho A_0 \Delta y) \omega ^2 y$ $\rightarrow \Delta p =\rho \omega ^2 y \Delta y$ Using eqn (2), $\Delta p = (\frac{p}{p_0})^{\frac{1}{\gamma}} \cdot \rho_0 \omega ^2 y \Delta y$ $\rightarrow p^{-\frac{1}{\gamma}} \Delta p = (\frac{\rho_0 \omega ^2}{p_0 ^{\frac{1}{\gamma}}}) y \Delta y$ Let the pressure at the center compartment be $p_1$ in fig 2. Integrating both sides, $\int ^{p} _{p_1} p^{-\frac{1}{\gamma}} \,dp = \frac{\rho_0 \omega ^2}{p_0 ^{\frac{1}{\gamma}}} \int ^y _0 y \,dy$ $\rightarrow \frac{p^{-\frac{1}{\gamma}+1} - {p_1}^{-\frac{1}{\gamma}+1}}{-\frac{1}{\gamma} + 1 } = (\frac{\rho_0 \omega ^2}{2p_0 ^{\frac{1}{\gamma}}}) y^2$ $\rightarrow (\frac{p}{p_0})^{-\frac{1}{\gamma}+1} \cdot p_0 ^{-\frac{1}{\gamma}+1} - {p_1}^{-\frac{1}{\gamma}+1} = (-\frac{1}{\gamma} + 1 )(\frac{\rho_0 \omega ^2}{2p_0 ^{\frac{1}{\gamma}}}) y^2$ Using eqn. 1, $(\frac{\Delta y}{\Delta x})^{-\gamma(\frac{\gamma - 1}{\gamma})} \cdot p_0 ^{\frac{\gamma - 1}{\gamma}} - {p_1}^{\frac{\gamma - 1}{\gamma}} = (\frac{(\gamma - 1)\rho_0 \omega ^2}{2 \gamma p_0 ^{\frac{1}{\gamma}}}) y^2$ $\rightarrow \frac{dy}{dx} = [(\frac{(\gamma - 1)\rho_0 \omega ^2}{2 \gamma p_0}) y^2 + (\frac{p_1}{p_0})^{\frac{\gamma -1}{\gamma}}]^{-\frac{1}{\gamma-1}}$ So we have, $A = \frac{(\gamma - 1)\rho_0 \omega ^2}{2 \gamma p_0}$ $n = -\frac{1}{\gamma-1}$ Therefore, $nA = - \frac{\rho_0 \omega ^2}{2 p_0 \gamma }$ Using ideal gas equation, $\frac{\rho _0}{p_0} = \frac{M}{RT_0}$ Therefore, $nA= -\frac{M\omega^2}{2RT_0\gamma }$