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 $1 m ^{2}$ with a speed $10^{4} m / s$, the pressure exerted by the gas molecules will be of the order of

• A. $10^{4} N / m ^{2}$
• B. $10^{8} N / m ^{2}$
• C. $10^{3} N / m ^{2}$
• D. None of the above

Answer 1

Answer: (D)

Momentum imparted to the surface in one collision,
$\Delta p=\left(p_{i}-p_{f}\right)=m v-(-m v)=2 m $
Force on the surface due to $n$ collision per second,
$F =\frac{n}{t}(\Delta p)=n \Delta p (\because t=1 s )$
$ =2 m n v { from Eq. (i) }]$
So, pressure on the surface,
$p=\frac{F}{A}=\frac{2 m n v}{A}$
Here, $m=10^{-26} kg , n=10^{22} s ^{-1}, v=10^{4} ms ^{-1}$, $A=1 m ^{2}$
$\therefore$ Pressure, $p=\frac{2 \times 10^{-26} \times 10^{22} \times 10^{4}}{1}=2 N / m ^{2}$
So, pressure exerted is of order of $10^{\circ}$.