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 × 10^{23} mol^{−1}, density of mercury = 13600 kg m^{−3} and *g* = 10 m s^{−2}.

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

#### Solution

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 × 10^{23} mol^{−1}

Pressure \[\left( P \right)\] is given by

*P* = \[\rho gh\]

Using the ideal gas equation, we get

\[PV = nRT\]

\[PV = nRT\]

\[ \Rightarrow n = \frac{PV}{RT}\]

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

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

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

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

\[ = 8 \times {10}^{15} \]