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laws

Question

A vessel of volume 1660 ${cm}^{3}$ contains 0.1 mole of oxygen and 0.2 mole of nitrogen. If the temperature of the mixture is 300 K, find its pressure.

Moderate

A

$2.5 \times 10^{5} Pa$
B

${1.5 × 10}^{5} Pa$
C

$4.5 \times 10^{5} Pa$
D

$6.5 \times 10^{5} Pa$

Solution

We have pV = nRT $\Rightarrow p = \frac{nRT}{V}$
The partial pressure due to oxygen is
$p_{1} = \frac{(0.1 mol) (8.3 J / mol - K) (300 K)}{(1660 \times 10^{- 6} m^{3})} = 1.5 \times 10^{5} Pa$
Similarly, the partial pressure due to nitrogen is
$p_{2} = 3.0 \times 10^{5} Pa$
The total pressure in the vessel is $p = p_{1} + p_{2}$
= $(1.5 + 3.0) \times 10^{5} Pa = 4.5 \times 10^{5} Pa$

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