Question

${10}^{22}$ particles each of mass ${10}^{-26}kg$are striking perpendicularly on a wall of area $1m^2$ with speed ${10}^{4}m{s}^{-1}$in $1s$ The pressure on the wall if the collision is perfectly elastic is:

• A. $2N{m}^{-2}$
• B. $4N{m}^{-2}$
• C. $6N{m}^{-2}$
• D. $8N{m}^{-2}$

Answer 1

The correct option is A

$2N{m}^{-2}$

Step 1. Given data

Number of particles, $n={10}^{22}$

Mass of each particle, $m={10}^{-26}kg$

Area of the wall, $A=1m^2$

Speed of the particles, $v={10}^{4}m{s}^{-1}$

Time, $t=1s$

Step 2. Finding the force on the particle, $F$

By using the formula of force,

$\mathrm{Force}\left(\mathrm{F}\right)=\frac{\mathrm{change}\mathrm{in}\mathrm{momentum}\left(∆\mathrm{p}\right)}{\mathrm{time}\left(\mathrm{t}\right)}$

When the molecules collide with the surface, they will have an opposite velocity. So, the momentum transferred will be,

$∆\mathrm{p}=\mathrm{mv}-\left(\mathrm{m}\left(-\mathrm{v}\right)\right)$

$∆\mathrm{p}=2\times \mathrm{m}\times \mathrm{v}$

$\mathrm{Force}\left(\mathrm{F}\right)=\frac{2\times \mathrm{mass}\left(\mathrm{m}\right)\times \mathrm{velocity}\left(\mathrm{v}\right)\times \mathrm{number}\mathrm{of}\mathrm{particles}\left(\mathrm{n}\right)}{\mathrm{time}\left(\mathrm{t}\right)}$ [$\mathrm{Momentum}=\mathrm{mass}\times \mathrm{velocity}$]

$F=\frac{2\times {10}^{-26}\times {10}^4\times {10}^{22}}{1}$ [From the given values]

$F=2N$

Step 3. Finding the pressure, $P$

By using the formula of pressure,

$\mathrm{Pressure}\left(\mathrm{P}\right)=\frac{\mathrm{Force}\left(\mathrm{F}\right)}{\mathrm{Area}\left(\mathrm{A}\right)}$

$P=\frac{2}{1}$ [from the above values ]

$P=2N{m}^{-2}$

Therefore, the pressure exerted is $2N{m}^{-2}$.

Hence, the correct option is $\left(A\right)$.