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 $\omega$ 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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Solution
$$P \rightarrow \square \leftarrow^{c}(P + dP)$$
$${A\lbrack(P + dP) - P\rbrack = (dm)\left( \omega^{2}x \right) }{dP = \frac{(dm)}{A}\omega^{2}x }{dP = \frac{(\rho)(A)(dx)\omega^{2}x}{A} }$$also $\lbrack PM = \rho RT\rbrack$
$${\rho = \frac{\text{~PM~}}{\text{~RT~}} }{dP = \left( \frac{PM}{RT} \right)\omega^{2}xdx }{\int_{P_{A}}^{P_{B}}\mspace{2mu}\frac{dP}{P} = \frac{\omega^{2}M^{\mathcal{l}}}{RT}\int_{0}^{\mathcal{l}}\mspace{2mu} xdx }{\mathcal{l}n\left( \frac{P_{B}}{P_{A}} \right) = \frac{\omega^{2}\mathcal{l}^{2}M}{2RT} }{P_{B} = P_{A}e^{\frac{M\omega^{2}\mathcal{l}^{2}}{2RT}}}$$
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