Which will have a greater volume when the following gases are compared at S.T.P?
(a) $1.2 l N_{2}$ at $25^{o}$ C and $748$ mm Hg
(b) $1.25 l O_{2}$ at S.T.P.

O_{2}

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N_{2}

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both have same volume

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can not be calculated

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Solution

The correct option is A $O_{2}$
(a) $1.2 L N_{2}$ at $25^{o} C$ and $748 m m H g$

$P_{1} = 0.9842 a t m; P_{2} = 1 a t m$

$V_{1} = 1.2 L : V_{2} = ?$

By Boyle’s Law,

$P_{1} V_{1} = P_{2} V_{2}$

$V_{2} = \frac{P_{1} V_{1}}{P_{2}} = \frac{0.9842 \times 1.2}{1} = 1.181 L$ at STP
(b) $1.25 L O_{2}$ at STP

As we can see, volume of $N_{2}$ at STP is much lesser than volume of $O_{2}$ at STP. Therefore answer is $O_{2}$

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