Question

If ${10^{22}}$ gas molecules each of mass ${10^{ - 26}}kg\;$ collided with a surface (perpendicular to it) elastically per second over an area $1\,{m^2}$ with a speed ${10^4}m/s$ , the pressure exerted by the gas molecules will be of the order of
A) ${10^8}\,N/{m^2}$
B) ${10^4}\,N/{m^2}$
C) ${10^3}\,N/{m^2}$
D) ${10^{16}}\,N/{m^2}$

Answer 1

Hint : In this solution, we will calculate the force exerted by the gas molecules on the surface using the change in the momentum of the gas molecules. Then we will use the relation of pressure, force, and area to determine the pressure exerted by the gas on it.

Formula used: In this solution, we will use the following formula:
Force due to change in momentum: $F = \dfrac{{\Delta P}}{{\Delta t}}$ where $\Delta P$ is the change in momentum in time $\Delta t$
Pressure due to force exerted: $P = \dfrac{F}{A}$ where $A$ is the area on which the force is exerted.

Complete step by step answer
If the gas molecules collide with the surface, they will impart momentum to the surface. The force exerted by the molecules can be calculated by Newton’s second law as the rate of change of momentum with respect to time as
$F = \dfrac{{\Delta P}}{{\Delta t}}$
When the molecules collide with the surface, they will have an opposite velocity. So, the momentum transferred will be $\Delta P = mv - (m( - v)) = 2mv$
So, the force transferred per second will be
$\dfrac{{\Delta F}}{{\Delta t}} = 2mv$
So, the pressure exerted on the surface will be
$P = \dfrac{{2mv}}{A}$
Which gives us the order of the pressure for $N$ molecules as
$P = \dfrac{{{{10}^{22}} \times 2 \times {{10}^{ - 26}} \times {{10}^4}}}{1}$
$\Rightarrow P = 2\,N/{m^2}$
So the correct answer is not present in the options.

Note
The force calculated by us is only an approximate estimation as in a real gas the gas molecules will not have a fixed velocity. Instead, there will be a range of different velocities of gas molecules which will exert different amounts of force on a surface. The assumption that the gas molecules will have an opposite velocity after the collision is also an approximation and is not true in general but it can help us in finding the order of pressure.