Question

If ${10}^{22}$ gas molecules each of mass ${10}^{-26}kg$ collide with a surface (perpendicular to it) elastically per second over an area $1m^2$ 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. ${10}^{3}N/m^2$
• C. ${10}^{4}N/m^2$
• D. $2N/m^2$

Answer 1

The correct option is D $2N/m^2$

Pressure is defined as normal force per unit area.
Force is calculated as change in momentum per unit time.
Magnitude of change in momentum per collision $=2mv$
$Pressure=\frac{Force}{Area}=\frac{N\left(2mv\right)}{1}$

$=\frac{{10}^{22}\times 2\times {10}^{-26}\times {10}^4}{1}$
$=2N/m^2$