Question 94: Centre of Mass and Rotational Motion

Question

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)

Accepted Answer

As the tube rotates, the gas molecules are pushed outward due to the rotation. This outward push (centrifugal effect) creates a pressure gradient; the pressure must increase as we move from the axis of rotation A towards the outer end B to provide the necessary centripetal force to keep the gas in circular motion.
Let the cross-sectional area of the tube be S . Assume a small element of thickness dx be at a distance x from the axis of rotation A .
Let the pressure on the inner face be P and the pressure on the outer face be $\mathrm{P}+\mathrm{dP}$.
Force pushing outward on the inner face $=\mathrm{P} \cdot \mathrm{S}$
Force pushing inward on the outer face $=(\mathrm{P}+\mathrm{dP}) \cdot \mathrm{S}$
The net inward force provided by the pressure difference is:
$$ \mathrm{F}_{\mathrm{net}}=(\mathrm{P}+\mathrm{dP}) \cdot \mathrm{S}-\mathrm{P} \cdot \mathrm{~S}=\mathrm{S} \cdot \mathrm{dP} $$
This net inward force acts as the centripetal force required to keep this small mass of gas moving in a circle of radius x with angular velocity $\omega$.
The mass of this small element (dm) is its volume (S Ã¢ÂÂ dx) multiplied by the local density ( $\rho$ ) of the gas :
$$ \mathrm{dm}=\rho \cdot \mathrm{S} \cdot \mathrm{dx} $$
At equilibrium,
$$ S \cdot d P=(\rho \cdot S \cdot d x) \cdot x \cdot \omega^2 $$
$\Rightarrow \mathrm{dP}=\rho \cdot \omega^2 \cdot \mathrm{x} \cdot \mathrm{dx} \ldots(\mathrm{i})$
It is given that it's an ideal gas at a constant temperature T .
From the ideal gas equation:
$$ \mathrm{PV}=\mathrm{nRT} $$
The number of moles $n=\frac{m}{M}$, where $m$ is the mass and $M$ is the molar mass (molecular weight).
$$ \mathrm{PV}=\left(\frac{\mathrm{m}}{\mathrm{M}}\right) \mathrm{RT} $$
$$\Rightarrow $$ $$ \mathrm{PM}=\left(\frac{\mathrm{m}}{\mathrm{~V}}\right) \mathrm{RT} $$
Since density $\rho=\frac{\mathrm{m}}{\mathrm{V}}$, we get :
$$ \mathrm{PM}=\rho \mathrm{RT} $$
$$\Rightarrow $$ $$ \rho=\frac{\mathrm{PM}}{\mathrm{RT}} $$
Now, substitute this expression for $\rho$ into equation (i) :
$$ \mathrm{dP}=\left(\frac{\mathrm{PM}}{\mathrm{RT}}\right) \omega^2 \mathrm{xdx} $$
$$ \frac{\mathrm{dP}}{\mathrm{P}}=\left(\frac{\mathrm{M} \omega^2}{\mathrm{RT}}\right) \mathrm{xdx} $$
At end $\mathrm{A}(\mathrm{x}=0)$, the pressure is $\mathrm{P}_{\mathrm{A}}$.
At end $\mathrm{B}(\mathrm{x}=\mathrm{l})$, the pressure is $\mathrm{P}_{\mathrm{B}}$.
$$ \int\limits_{\mathrm{P}_{\mathrm{A}}}^{\mathrm{P}_{\mathrm{B}}} \frac{1}{\mathrm{P}} \mathrm{dP}=\left(\frac{\mathrm{M} \omega^2}{\mathrm{RT}}\right) \int\limits_0^1 \mathrm{x} \mathrm{dx} $$
$$\Rightarrow $$ $[\ln (\mathrm{P})]_{\mathrm{P}_{\mathrm{A}}}^{\mathrm{P}_{\mathrm{B}}}=\left(\frac{\mathrm{M} \omega^2}{\mathrm{RT}}\right)\left[\frac{\mathrm{x}^2}{2}\right]_0^1$
$$\Rightarrow $$ $\ln \left(\mathrm{P}_{\mathrm{B}}\right)-\ln \left(\mathrm{P}_{\mathrm{A}}\right)=\left(\frac{\mathrm{M} \omega^2}{\mathrm{RT}}\right)\left(\frac{\mathrm{l}^2}{2}-0\right)$
$$\Rightarrow $$ $\ln \left(\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}}\right)=\frac{\mathrm{M} \omega^2 \mathrm{l}^2}{2 \mathrm{RT}}$
$$\Rightarrow $$ $\frac{P_B}{P_A}=e^{\left(\frac{M \omega^2 l^2}{2 R T}\right)} $
$\Rightarrow P_B=P_A e^{\left(\frac{M \omega^2 l^2}{2 R T}\right)}$
Hence, the correct option is (C).

Answer

${P}_{{B}}={P}_{{A}} {e}^{\left(\frac{M \omega^2 {l}^2}{3 R T}\right)}$

Answer

$P_B=P_A$

Answer

${P}_{{B}}={P}_{{A}} {e}^{\left(\frac{{M} \omega^2 {l}^2}{2 {R T}}\right)}$

Answer

$P_B=P_A e^{\left(\frac{M \omega^2 l^2}{R T}\right)}$

Centre of Mass and Rotational Motion

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