Question

Find the number of molecules in 1 $c{m}^3$ of an ideal gas at $0^0$'C' and at a pressure of ${10}^{-5}$ mm of mercury.

Answer 1

V = 1 $c{m}^3$, T = $0^0$ C,
P= ${10}^{-}5$ mm of Hg
n = $\frac{PV}{RT}=\frac{\rho gh\times v}{RT}$

= $\frac{13.6\times 980\times {10}^{-6}\times 1}{8.31times273}$

= $5.874\times {10}^{-13}$

No. of molecules

= N $\times$ n

= $6.023\times {10}^{23}\times 5.874\times {10}^{-13}$

= $35.384\times {10}^{10}$

=$3.538\times {10}^{11}$