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Q. A flat plate is moving normal to its plane through a gas under the action of a constant force $F$. The gas is kept at a very low pressure. The speed of the plate $v$ is much less than the average speed $u$ of the gas molecules. Which of the following options is/are true?

JEE AdvancedJEE Advanced 2017Kinetic Theory

Solution:

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Consider area $A$ of the plate moving towards right (see figure). In next dt time, only those particles of the gas can collide with the plate (on its right face) as are within $( v + u ) dt$ distance from it. Number of these particles $= n ( u + y )$ dt $A$, where $n$ is number of gas particles per unit volume. Only half of these particles will collide with the plate, the other half will be moving away from it.
$\therefore $ Number of particles, colliding with the plate from right, in dt time $=\frac{1}{2} n(u+v)$ dt $A$ Upon collision each particle (of mass $m$ ) imparts momentum of $2 m ( u + v )$ to the plate towards left, as they are moving with a velocity $u+v$ relative to the plate.
The force applied on the right face of the plate by the collisions $=\frac{\text { momentum imparted }}{\text { time }}$
$=\frac{\left[\frac{1}{2} n(u+v) d t A\right][2 m(u+v)]}{d t}=m n A(u+v)^{2}$
$\therefore $ Pressure on the right face of plate $=\frac{\text { Force }}{\text { Area }}=\frac{m n A(u+v)^{2}}{A}=m n(u+v)^{2}$
Similarly, pressure on its left plate $= mn ( u - v )^{2}$ [The particles on the left, move in with a velocity u-v relative to the plate].
And, the difference of pressure $=m n\left[(u+v)^{2}-(u-v)^{2}\right]=4\, mn\, uv\, \propto$ uv With passage of time, $v$ will increase (or decrease) such that the resistive force (due to this pressure difference) balances $F$ : Also, the resistive force is proportional to v