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Question

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

Easy

A

$\frac{RT}{V}$
B

$\frac{3 RT}{2 V}$
C

$\frac{5 RT}{4 V}$
D

$\frac{7 RT}{5 V}$

Solution

Mass of oxygen = 8 g
Molecular mass of O₂=32 g
Number of moles of $O_{2} = \frac{8}{32} = \frac{1}{4} = 0.25$
Mass of nitrogen = 14 g
Molecular mass of nitrogen gas = 28 g
Number of moles of nitrogen gas $= \frac{14}{28} = 0.5 moles$
Mass of carbon dioxide = 22 g
Molecular mass of carbon dioxide = 44 g
Number of moles carbon dioxide $= \frac{22}{44} = 0.5 moles$
Total number of moles of the mixture is
$0.5 + 0.5 + 0.25 μ_{mix} = 1.25$
We know that $PV = μ_{mix} RT P_{mix} = \frac{μ_{mix} RT}{V} = \frac{1.25 RT}{V} = \frac{5}{4} \frac{RT}{V}$

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