The total pressure of a mixture of 8g of oxygen and 14g of nitrogen contained in a 11.2L vessel at $0^{o} C$ is

0.5 atm

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1 atm

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1.5 atm

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2 atm

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Solution

The correct option is C
1.5 atm

PV=nRT
V=11.2 lit, $n = \frac{14}{28} + \frac{8}{32}$
$= \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$
$P = \frac{3}{4} \times \frac{22.4}{273} \times \frac{300}{11.2} = 1.49 = 1.5 a t m$

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An enclosure of volume V contains a mixture of 8g of oxygen, 14g of nitrogen and 22g of carbon dioxide at absolute temperature T. The pressure of the mixture of gases is (R is universal gas constant)

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The total pressure of a mixture of 8g of oxygen and 14g of nitrogen contained in a 11.2L vessel at 0o C is

The correct option is C 1.5 atm PV=nRT V=11.2 lit, n=1428+832 =12+14=34 P=34×22.4273×30011.2=1.49=1.5atm

The total pressure of a mixture of 8g of oxygen and 14g of nitrogen contained in a 11.2L vessel at $0^{o} C$ is

The correct option is C
1.5 atm

PV=nRT
V=11.2 lit, $n = \frac{14}{28} + \frac{8}{32}$
$= \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$
$P = \frac{3}{4} \times \frac{22.4}{273} \times \frac{300}{11.2} = 1.49 = 1.5 a t m$