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An electric bulb of volume $250 c m^{3}$ was sealed off during manufacture at a pressure $10^{- 3} m m$ of mercury at $27^{0} C$ . The number of air molecules contained in the bulb. $(R = 8.31 J / m o l - k)$

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$7.83 \times 10^{15}$

$6.54 \times 10^{15}$

$5.67 \times 10^{15}$

$8.21 \times 10^{15}$

answer is A.

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Detailed Solution

$P = h d g; n = \frac{P V}{R T} = 1.3 \times 10^{- 8} m o l e$
No.of molecules $= n \times N$
N=Avogadro number

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