Question

The number of gas molecules striking per second per square metre of the top surface of a table placed in a room at $20^{\circ }C$ and $1$ atmospheric pressure is of the order of $(k_B=1.4\times 10^{-23}J{K}^{-1}$ and the average mass of an air molecule is $5\times 10^{-27}kg$ )

• A. $10^{27}$
• B. $10^{23}$
• C. $10^{25}$
• D. $10^{29}$

Answer 1

Answer: (A)

Using, $v_{\text{rms }}=\sqrt{\frac{3kT}{m}}$ and
$p=N\times 2m{v}_{\text{rms }}$
We have,
$N=\frac{p}{2m\cdot v_{\text{rms }}}=\frac{p\times \sqrt{m}}{2m\sqrt{3kT}}=\frac{(p/2)}{\sqrt{3mkT}}$
$=\frac{1.01\times 10^5}{2\times \sqrt{\left(3\times 5\times 10^{-27}\times 1.4\times 10^{-23}\times 293\right)}}$
$=6.4\times 10^{27}$