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An electric bulb of volume $250 c m^{3}$ was sealed off during manufacture at a pressure of $10^{- 3} m m$ of mercury at $27 ° C$ . Compute the number of air molecules contained in the bulb. Given that $R = 8.31 J / m o l - K$ and $N_{A} = 6.02 \times 10^{23}$ per mol.

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$8 \times 10^{15}$

$45 \times 10^{15}$

$95 \times 10^{15}$

$9 \times 10^{15}$

answer is A.

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Detailed Solution

p V=n R T

$∴ n = \frac{p V}{R T} = \frac{(ρ g h) V}{R T} = \frac{(13.6 \times 10^{3}) (9.8) (10^{- 6}) (250 \times 10^{- 6})}{8.31 \times 300} = 1.33 \times 10^{- 8} ∴ Number of molecules = (n) N_{A} = (1.33 \times 10^{- 8}) (6.02 \times 10^{23})$

= $8 \times 10^{15}$

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