Question

An 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.

Use R=8.314J K^-1 mol^-1

Answer 1

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

\[PV = nRT\]

\[PV   =   nRT\] 

\[ \Rightarrow n   = \frac{PV}{RT}\] 

\[ \Rightarrow n = \frac{\rho gh V}{RT}\] 

\[ \Rightarrow n   = \frac{{10}^{- 6} \times 13600 \times 10 \times 250 \times {10}^{- 6}}{8 . 314 \times 300}\] 

\[\text { Now,   number  of  molecules }  = nN\] 

\[ = \frac{{10}^{- 6} \times 13600 \times 10 \times 250 \times {10}^{- 6}}{8 . 314 \times 300} \times 6 \times  {10}^{23} \] 

\[ = 8 \times  {10}^{15}   \]