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 \(P_A\).
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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 tube rotating with angular speed \(\omega\), filled with an ideal gas, the pressure at point \(B\) is related to the pressure at point \(A\) by taking into account the centrifugal force acting on the gas. The centrifugal force causes a pressure gradient from point \(A\) to point \(B\). The pressure difference between the two ends of the rotating tube can be found using the equation: \[ \frac{P_B}{P_A} = \exp \left( \frac{\omega^2 \ell^2 M}{2RT} \right) \] Where: - \(\omega\) is the angular speed, - \(\ell\) is the length of the tube, - \(M\) is the molar mass of the gas, - \(R\) is the universal gas constant, - \(T\) is the temperature of the gas. Step 1: Relate the pressures at points \(A\) and \(B\). The pressure at point \(B\) is related to the pressure at point \(A\) as: \[ P_B = P_A \exp \left( \frac{\omega^2 \ell^2 M}{2RT} \right) \] Hence, the final expression for the pressure at point \(B\) is: \[ P_B = P_A e^{\frac{\omega^2 \ell^2 M}{2RT}} \] Step 2: Conclusion. Thus, the pressure at the other end \(B\) is given by: \[ P_B = P_A e^{\frac{\omega^2 \ell^2 M}{2RT}} \] The correct answer is: \[ \boxed{\frac{\omega^2 \ell^2 M}{2RT} \cdot P_A e} \]
