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

The number of air molecules contained in the bulb is  $\bold{8.18 \times 10^{15}}$

Explanation:

Given:

Electric bulb Volume , V = 250 cc

Temperature at which fabrication takes place, T = 27  + 273  = 300 K

Mercury height, $h = 10^{-3} mm$

Mercury Density ,$\rho = 13600 \text{kgm}{-3}$

Avogadro constant, $N = 6 \times 10^{23} mol^{-1}$

Solution:

We know that  

Pressure  $P =\rho gh$------------------------------(1)

Using the ideal equation for gas we get

$n = \frac{PV}{RT}$-----------------------------------(2)

Substituting (1) in (2), we get

$n=\frac{\rho \mathrm{gh} V}{R T}$

Now substituting the given values,

$n=\frac{10^{-6} \times 13600 \times 10 \times 250 \times 10^{-6}}{8.31 \times 300}$

$n=\frac{34 \times 10^{-6}}{2493}$

$n=1.3638 \times 10^{-8}$

Number of Molecules   = $n \times N$

$=1.3638 \times 10^{-8} \times 6 \times 10^{23}$

$=8.18 \times 10^{15}$

Answer 2

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{yellow}{Answer}}}}$

$8.18 \times {10}^{15}$

hope it help