Question

The number of gas molecules striking per second per square meter of the top surface of a table placed in a room at ${20}^o$C and $1$ atmospheric pressure is of the order of($k_B=1.4\times {10}^{-23}J/K$, 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

The correct option is A ${10}^{27}$

Force exerted by the gas molecules on unit area of table $F=P_{o}A={10}^5\times 1={10}^5$ $N$
Average velocity of gas molecules $v=\sqrt{\frac{8k_{B}T}{\pi m}}$
where $T=20+273=293K$
$\therefore$ $v=\sqrt{\frac{8\times 1.4\times {10}^{-23}\times 293}{\pi \times 5\times {10}^{-27}}}=1.4\times {10}^{3}m/s$
Rate of change in momentum $\frac{\Delta P}{\Delta t}=nm(v-(-v\left)\right)=2mnv$
Using $F=\frac{\Delta P}{\Delta t}=2mnv$
Or ${10}^5=2\times 5\times {10}^{-27}\times n\times 1.4\times {10}^3$
$⟹n=7\times {10}^{27}$
Thus correct answer is option A.