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 ω about an axis perpendicular 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)
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Question

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 ω about an axis perpendicular 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)

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G1 Physics · Accepted Answer

Consider a thin gas element at distance x from the rotation axis and of thickness dx. If its cross-sectional area is A, the pressure force difference is A dP. The element requires centripetal force dm ω²x. Therefore, A dP = dm ω²x. Since dm = ρA dx, we get dP = ρ ω²x dx. For an ideal gas, ρ = PM/(RT), where M is the molar mass. Therefore, dP/P = [Mω²/(RT)]x dx. Integrate from A to B: ∫ from P(A) to P(B) of dP/P = [Mω²/(RT)] ∫ from 0 to l of x dx. Hence, ln[P(B)/P(A)] = Mω²l²/(2RT). Therefore, P…