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Question

Air is made up mostly of N₂ (MW = 28) and O₂ (MW = 32). If these molecules share a container at the same temperature and pressure, which ones will generally be moving faster? Explain.

Expert Solution

Step 1

The Root Mean Square Velocity

The root mean squared velocity of a molecule of a gas is defined as the square root of the average velocity square of every molecule.

If $u_{1}, u_{2}, u_{3}, . . ., u_{n}$ be the velocity of $n$ molecules then the rms velocity is given as

$C = \sqrt{\frac{u_{1}^{2} + u_{2}^{2} + u_{3}^{2} + . . . . + u_{n}^{2}}{n}}$

The pressure exerted by the random movement of molecules of a gas whose density is $ρ$ is

$p = \frac{1}{3} ρ C^{2}$

Let us consider that one mole of a gas occupies $V$ volume, therefore for one mole the pressure exerted by the gas is

$p = \frac{1}{3} ρ C^{2} = \frac{1}{3} \frac{M}{V} C^{2}$

$M$ the molecular weight of the gas. We get

$p V = \frac{1}{3} M C^{2}$

Then by ideal gas law equation for one mole

$p V = \frac{1}{3} M C^{2} = R T \Rightarrow C = \sqrt{\frac{3 R T}{M}}$

$R$ the universal gas constant and $T$ is the temperature of the gas.

Step 2

Let us consider the temperature of the container be $T$ and one mole each of $N_{2}$ and $O_{2}$ occupies the container.

Given, the molecular weight of $N_{2} = M_{N} = 28$ and

the molecular weight of $O_{2} = M_{O} = 32$

Therefore rms velocity of $N_{2}$

$C_{N} = \sqrt{\frac{3 R T}{M_{N}}} = \sqrt{\frac{3 R T}{28}}$

and rms velocity of $O_{2}$

$C_{O} = \sqrt{\frac{3 R T}{M_{O}}} = \sqrt{\frac{3 R T}{32}}$

Taking the ratio

$\frac{C_{N}}{C_{O}} = \frac{\sqrt{\frac{3 R T}{28}}}{\sqrt{\frac{3 R T}{32}}} = \sqrt{\frac{32}{28}} = 1.07$

This gives

$C_{N} = 1.07 C_{O}$

This shows that rms velocity of nitrogen is greater and 1.07 times that of oxygen.

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which ones will generally be moving faster?