A cylindrical tube A B of length l , closed at both ends contains...

A cylindrical tube

A B

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

A B

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)
A cylindrical tube A B of length l , closed at both ends contains an ideal gas of 1 mol having molecular weight M . The

A cylindrical tube A B of length l , closed at both ends contains an ideal gas of 1 mol having molecular weight M . The

Options

A $P_{B}=P_{A}$
B $P_{B}=P_{A} \exp \left(M \omega^{2} l^{2} / 3 R T\right)$
C $P_{B}=P_{A} \exp \left(M \omega^{2} l^{2} / R T\right)$
D $P_{B}=P_{A} \exp \left(M \omega^{2} l^{2} / 2 R T\right)$

Correct answer

D. $P_{B}=P_{A} \exp \left(M \omega^{2} l^{2} / 2 R T\right)$

Step-by-step solution

A[(P + dP) - P] = (dm)(\omega^2 x)

dP = \dfrac{(dm)}{A} \omega^2 x

dP = \dfrac{(\rho)(A)(dx)\omega^2 x}{A}

also

[PM = \rho RT]

\rho = \dfrac{PM}{RT}

dP = \left( \dfrac{PM}{RT} \right) \omega^2 x \, dx

\int_{P_A}^{P_B} \dfrac{dP}{P} = \dfrac{\omega^2 M}{RT} \int_{0}^{\ell} x \, dx

\ln \left( \dfrac{P_B}{P_A} \right) = \dfrac{\omega^2 \ell^2 M}{2RT}

P_B = P_A e^{\dfrac{M\omega^2 \ell^2}{2RT}}

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