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.

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
$PV=nRT$

$$\begin{gathered} PV=nRT \\ \Rightarrow n=\frac{PV}{RT} \\ \Rightarrow n=\frac{\rho ghV}{RT} \\ \Rightarrow n=\frac{{10}^{-6}\times 13600\times 10\times 250\times {10}^{-6}}{8.314\times 300} \\ \mathrm{Now},\mathrm{number}\mathrm{of}\mathrm{molecules}=nN \\ =\frac{{10}^{-6}\times 13600\times 10\times 250\times {10}^{-6}}{8.314\times 300}\times 6\times {10}^{23} \\ =8\times {10}^{15} \end{gathered}$$