# If $10^{22}$ gas molecules each of mass $10^{-26}$ kg collide 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 : Option 1) $2\; N/m^{2}$ Option 2) $10^{4}\; N/m^{2}$ Option 3) $10^{16}\; N/m^{2}$   Option 4) $10^{8}\; N/m^{2}$

$\\V_{i}=10^{4}\; m/s\; \; (\widehat{l})\\\; \; \; \\V_{f}=10^{4}\; m/s\; \; (\widehat{-l})$

$\Delta V=2\times 10^{4}\; m/s$

For 1 particle

Momentum  = $\Delta P=m\Delta V$

change for n particle

$\Delta P=m.n.\Delta V$

For unit second

$Pressure=\frac{F}{A}=\frac{\Delta P}{A}=\frac{2mnV}{A}$

$Pressure=\frac{2\times10^{-26}\times10^{22}\times10^{4}}{1}$

$Pressure= 2\; N/m^{2}$

Option 1)

$2\; N/m^{2}$

Option 2)

$10^{4}\; N/m^{2}$

Option 3)

$10^{16}\; N/m^{2}$

Option 4)

$10^{8}\; N/m^{2}$

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