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Q.
A gas at $300 \,K$ has pressure $4 \times 10^{-10}\, N\, m^{-2}$. If $k_B = 1.38 \times 10^{-23}\, J \,K^{-1}$, the number of molecules per $cm^3$ is of the order of
Kinetic Theory
Solution:
Using, $PV = nRT = nN\left(\frac{R}{N}\right) T = nNk_{B}T$
$\frac{nN}{V} = \frac{P}{k_{B}T} =$ number of molecules per $m^{3}$
$\therefore \quad$ Number of molecules/cm$^{3} = \frac{P}{k_{B}T}\times10^{-6}$
$= \frac{4 \,\times\,10^{-10}\,\times\,10^{-6}}{1.38\,\times\,10^{-23}\,\times\,300} = 10^{5}$