NEET-XII-Physics
24: Kinetic Theory of Gases
- #6An electric bulb of volume 250 cc was sealed during manufacturing at a pressure of 10-3 mm of mercury at 27°C. Compute the number of air molecules contained in the bulb. Avogadro constant = 6 × 1023 mol-1, density of mercury = 13600 kg m-3 and g = 10 m s-2.Ans : Given:
Volume of electric bulb, V = 250 cc
Temperature at which manufacturing takes place, T = 27 + 273 = 300 K
Height of mercury, h = 10-3 mm
Density of mercury, `` \rho `` = 13600 kgm-3
Avogadro constant, N = 6 × 1023 mol-1
Pressure `` \left(P\right)`` is given by
P = `` \rho gh``
Using the ideal gas equation, we get
`` PV=nRT``
`` PV=nRT``
`` \Rightarrow n=\frac{PV}{RT}``
`` \Rightarrow n=\frac{\rho ghV}{RT}``
`` \Rightarrow n=\frac{{10}^{-6}\times 13600\times 10\times 250\times {10}^{-6}}{8.314\times 300}``
`` ``
`` \,\mathrm{\,Now\,},\,\mathrm{\,number\,}\,\mathrm{\,of\,}\,\mathrm{\,molecules\,}=nN``
`` =\frac{{10}^{-6}\times 13600\times 10\times 250\times {10}^{-6}}{8.314\times 300}\times 6\times {10}^{23}``
`` =8\times {10}^{15}``
`` ``