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Mixture

The pressure ...

Question

The pressure of a gas of volume $22.4$ l is $3$ atm at certain temperature. Then find the pressure of the gas of volume $44.8$ l at the same temperature.

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Solution

according to boyles law
$p_{1} v_{1} = p_{2} v_{2}$
$p_{1} = 3 a t m$
$v_{1} = 22.4 L$
$p_{2} =$
$v_{2} = 44.8 L$
$3 \times 22.4 = P_{2} \times 44.8$
$P_{2} = 1.5 a t m$ \\

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