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 pe1pendicular 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 is same at all points in the tube)

Consider a small gas element at distance r from the axis at end A.

Because the tube rotates with angular speed ω\omegaω, that gas element needs centripetal force.

Required centripetal force on volume element dV=A dr is

$$dmω^2r$$

where

$$dm=ρAdr$$

So force balance due to pressure difference gives

$$dP⋅A=ρAdrω^2r$$

Cancel A:

$$dP=ρω^2rdr$$

For an ideal gas,

$$\rho=\frac{MP}{RT}$$

(M is molar mass, one mole gas)

Substitute:

$$dP=\frac{MP}{RT}ω^2rdr$$

Rearrange:

$$\frac{dP}{P}=\frac{M}{RT}ω^2rdr$$

Integrate from end A to end B:

At

$$r=0,\quad P=P_A$$

At

$$r=l,\quad P=P_B$$

So

$$\int_{P_A}^{P_B}\frac{dP}{P}=\frac{M\omega^2}{RT}\int\ rdr$$

$$\ln\frac{P_B}{P_A}=\frac{M\omega^2}{RT}\cdot\frac{l^2}{2}l$$

Which is same as

$$P_{B}=P_{A} exp(M \omega^{2}l^{2}/2RT)$$