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)

Solution

Physics sir · 3 steps

Each slice of gas needs a centripetal push, and only the pressure difference across it can supply it.

1. Force on a slice. Take a thin slice of gas of thickness dx at distance x from A, with tube cross-section S. It moves in a circle, so the net inward force on it must be dm ω² x. Pressure is higher on the outer face, so S dP = dm ω² x.
2. Density of the gas. The slice has mass dm = ρ S dx. For an ideal gas ρ = PM/RT, so dP = PM/RT ω² x dx. Divide by P to separate the variables.

   ∫_(P_A)^(P_B)dP/P = Mω²/RT∫₀ˡ x dx ⇒ ln P_B/P_A = (Mω² l²)/2RT

3. So. P_B = P_A e^(Mω² l²/2RT). The 1/2 comes from ∫₀ˡ x dx = l²/2, so the answer is (c).

The idea to keep. For a gas in a rotating tube, balance S dP = ρ S ω² x dx with ρ = PM/RT. Because ρ depends on P, you get dP/P, and the pressure grows exponentially outward.

Answer from the NTA final answer key. Solution by PrepWiser.