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# 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 - Physics

Sum

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

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#### 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 × 1023 mol−1
Pressure $\left( P \right)$ is given by

P = $\rho gh$

Using the ideal gas equation, we get

$PV = nRT$

$PV = nRT$

$\Rightarrow n = \frac{PV}{RT}$

$\Rightarrow n = \frac{\rho gh V}{RT}$

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

$\text { Now, number of molecules } = nN$

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

$= 8 \times {10}^{15}$

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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 2 Kinetic Theory of Gases
Q 6 | Page 34
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