Question:
If $10^{22}$ gas molecules each of mass $10^{-26}$ $\mathrm{kg}$ collide with a surface (perpendicular to it) elastically per second over an area $1 \mathrm{~m}^{2}$ with a speed $10^{4}$ $\mathrm{m} / \mathrm{s}$, the pressure exerted by the gas molecules will be of the order of :
Correct Option: 1
Solution:
(1)
Rate of change of momentum during collision
$=\frac{m v-(-m v)}{\Delta t}=\frac{2 m v}{\Delta t} N$
so pressure
$P=\frac{N \times(2 m v)}{\Delta t \times A}$
$\frac{=10^{22} \times 2 \times 10^{-26} \times 10^{4}}{1 \times 1}=2 \mathrm{~N} / \mathrm{m}^{2}$
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