A closed tube filled with ideal gas is rotating with angular speed \(\omega\) about an axis passing through end \(A\). Find the pressure at the other end \(B\).
(\(M\) is the molar mass of the gas, \(\ell\) is the length of the tube and \(T\) is the temperature of the gas.)
Given pressure at \(A\) is \(P_A\).
Show Hint
For gases in rotation:
Pressure increases exponentially with \(r^2\)
Use \( \dfrac{dp}{p} = \dfrac{M\omega^2 r}{RT}\,dr \) for isothermal conditions
Concept:
For a gas in steady rotation with angular speed \(\omega\), pressure varies with radial distance \(r\) due to the centrifugal effect.
For an ideal gas at uniform temperature \(T\):
\[
\frac{dp}{p} = \frac{M\omega^2 r}{RT}\,dr
\]
Step 1: Set up the pressure variation.
Here, the axis of rotation passes through end \(A\).
Thus,
\[
r_A = 0,\qquad r_B = \ell
\]
Step 2: Integrate between ends \(A\) and \(B\).
\[
\int_{P_A}^{P_B} \frac{dp}{p}
= \frac{M\omega^2}{RT}\int_{0}^{\ell} r\,dr
\]
\[
\ln\!\left(\frac{P_B}{P_A}\right)
= \frac{M\omega^2}{RT}\left[\frac{r^2}{2}\right]_0^{\ell}
= \frac{M\omega^2 \ell^2}{2RT}
\]
Step 3: Exponentiate to find \(P_B\).
\[
P_B = P_A \exp\!\left(\frac{M\omega^2 \ell^2}{2RT}\right)
\]
\[
\boxed{P_B = P_A\,e^{\dfrac{\omega^2 \ell^2 M}{2RT}}}
\]
