Q62 / 89 · KVPY 2017 (5 Nov SB/SX) · Physics
The number of gas molecules striking per second per square meter of the top surface of a table placed in a room at $20^{\circ} \mathrm{C}$ and 1 atmospheric pressure is of the order of $\left(\mathrm{k}_{\mathrm{B}}=1.4 \times 10^{-23} \mathrm{~J} / \mathrm{K}\right.$, and the average mass of an air molecules is $5 \times 10^{-27} \mathrm{~kg}$ )
- Correct answerA$10^{27}$
- B$10^{23}$
- C$10^{25}$
- D$10^{29}$
Solution
$\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}=\sqrt{\frac{3 \mathrm{kT}}{\mathrm{M}_{0}}}$
$\mathrm{P}=\dot{\mathrm{N}} \times 2 \mathrm{mV}_{\mathrm{rms}}$
$1.01 \times 10^{5}=\dot{\mathrm{N}} \times 2 \times 5 \times 10^{-27} \times \mathrm{V}_{\mathrm{rms}}$
$\begin{aligned} \dot{N} &=\frac{1.01 \times 10^{5} \sqrt{5 \times 10^{-27}}}{2 \times 5 \times 10^{-27} \sqrt{3 \times 1.4 \times 10^{-23} \times 293}} \\ &=6.43 \times 10^{27} \end{aligned}$
Hence, the correct answer is option: 'A'.