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 $\left(k_{B}=1.4 \times 10^{-23}\, JK ^{-1}\right.$ 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, $v_{\text{rms }}=\sqrt{\frac{3 k T}{m}}$ and
$p=N \times 2 m v_{\text{rms }}$
We have,
$N =\frac{p}{2 m \cdot v_{\text{rms }}}=\frac{p \times \sqrt{m}}{2 m \sqrt{3 k T}}=\frac{(p / 2)}{\sqrt{3 m k T}} $
$=\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)}} $
$=6.4 \times 10^{27}$