Question

An electric bulb of volume 250 cc was sealed during manufacturing at a pressure of ${10}^{-3}$ mm of mercury at ${27}^0$C. Compute the number of air molecules contained in the bulb. Avogadro constant= 6 $\times$ ${10}^{23}$ $mo{l}^{-1}$, density of mercury = 13600 kg $m^{-3}$and g = 10 m $s^{-2}$.

Answer 1

V = 250 cc = $250\times {10}^{-6}{m}^3$

P = ${10}^{-3mm}={10}^{-3}\times {10}^{-3}$ m

= $({10}^{-6}\times 13600\times 10)$

= $136\times {10}^{-3}$ Pascal

T = ${27}^0$ C = 300 k

n = $\frac{PV}{RT}$

= $\frac{136\times {10}^{-3}\times 250\times {10}^{-6}}{8.3\times 300}=1.36\times {10}^{-8}$

No. of molecules

= $1.36\times {10}^{-8}\times 6\times {10}^{23}$

= $8.17\times {10}^{15}$