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Gay Lussac's Law

16 gm of O ...

Question

16 gm of $O_{2}$ was filled in a container of capacity 8.21 lit. at 300 K. Calculate Pressure exerted by $O_{2}$

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Solution

$P v = n R T = \frac{m}{M} R T$
$P = \frac{m R T}{M V} = \frac{16 \times 0.0821 \times 300}{32 \times 8.2}$
$P = \frac{3.94.08}{262.72} = 1.5 a t m$

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