A cylindrical tube AB of length l, closed at | vedclass

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(A) Consider a small element of gas of length $dx$ at a distance $x$ from the axis of rotation $A$. Let $A$ be the cross-sectional area of the tube.Th

Vedclass · Accepted Answer

$P_{B}=P_{A} \exp(M\omega^{2}l^{2}/2RT)$

Vedclass · Answer

(A) Consider a small element of gas of length $dx$ at a distance $x$ from the axis of rotation $A$. Let $A$ be the cross-sectional area of the tube.The net force on this element providing the centripetal acceleration is $(P+dP)A - PA = (dm) \omega^2 x$.$AdP = (dm) \omega^2 x$.Since $dm = \rho A dx$, we have $dP = \rho \omega^2 x dx$.Using the ideal gas law $PM = \rho RT$, we have $\rho = \frac{PM}{RT}$.Substituting $\rho$, we get $dP = \left(\frac{PM}{RT}\right) \omega^2 x dx$.Rearranging the terms, $\frac{dP}{P} = \frac{M \omega^2}{RT} x dx$.Integrating from $x=0$ to $x=l$ and $P=P_A$ to $P=P_B$:$\int_{P_A}^{P_B} \frac{dP}{P} = \frac{M \omega^2}{RT} \int_0^l x dx$.$\ln\left(\frac{P_B}{P_A}\right) = \frac{M \omega^2}{RT} \left[\frac{x^2}{2}\right]_0^l = \frac{M \omega^2 l^2}{2RT}$.Therefore, $P_B = P_A \exp\left(\frac{M \omega^2 l^2}{2RT}\right)$.

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