Question

Find the number of air molecules present in an electric bulb

Answer 1

An electric bulb of volume 250 cm3 which is sealed off during manufacture at pressure of 10-3 mm of hg at 25C. Number of air molecules present in the bulb is - Using the formula PV =nRT
here h = 10-3 mm = 10-3x 10-3m
density of mercury = 13.6 x 103 kg/m3  and value of g = 10m/s2
  so pressure = hpg = 10-3x 10-3x 13.6 x 103 x 10 = 0.136 atm
now volume  = 250 cm3 = 250 x (10-2)3 = 250 x 10-6 m3
now using the formula
n =  PV/RT . =0.136 ×250×10−68.31×(273+27) =34×10−62493 =0.0136×10−6 hence number of air molecules n ×6.02×10230.0136×10−6×6.02×1023=8.21×1015