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JEE MAIN 2019
08-04-2019 S1
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:
Select the correct option:
A
$10^8 \mathrm{~N} / \mathrm{m}^2$
B
$10^3 \mathrm{~N} / \mathrm{m}^2$
C
$10^{16} \mathrm{~N} / \mathrm{m}^2$
D
$10^4 \mathrm{~N} / \mathrm{m}^2$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & P=\frac{N \times(2 \mathrm{mv})}{\Delta \mathrm{r} \times \mathrm{A}} \\ & =\frac{10^{22} \times 2 \times 10^{-26} \times 10^4}{1 \times 1}=2 \mathrm{~N} / \mathrm{m}^2\end{aligned}$
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