A mixture of 2 moles of helium gas (atomic mass = 4u) and 1 mole of argon gas (atomic mass = 40u) is

Q: Why don't we need to convert the atomic mass from 'u' to 'kg'?

Because we are taking a ratio! Any conversion factor you multiply the numerator by will also be multiplied in the denominator, perfectly canceling out. As long as both masses are in the same units, you are good to go.

Q: Does the number of moles (2 moles of He and 1 mole of Ar) affect the rms speed?

Not at all. The rms speed of an individual molecule depends only on the temperature of the gas and its own molar mass. The number of moles just tells us the total quantity of gas, which is irrelevant for finding the speed of a single molecule.

Q: Why is the formula $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ used instead of $\sqrt{\frac{8RT}{\pi M}}$?

The question specifically asks for the 'rms speed' (root mean square), which corresponds to the $\sqrt{\frac{3RT}{M}}$ formula. The other formula, $\sqrt{\frac{8RT}{\pi M}}$, is for the 'average speed'. Always read the specific type of speed requested!

Q: What if the gases were kept at different temperatures?

Then you couldn't cancel out the temperature $T$ when taking the ratio. The full ratio would be $\sqrt{\frac{T_1 M_2}{T_2 M_1}}$. You would need to plug in both the temperatures and the masses.

Q: Is the kinetic energy of Helium greater than Argon because it moves faster?

No! The average translational kinetic energy per molecule is $\frac{3}{2}kT$, which depends *only* on temperature. Since both are at $300\text{ K}$, their average translational kinetic energies are exactly the same. Helium moves faster simply to compensate for its smaller mass.

Question

A mixture of 2 moles of helium gas (atomic mass = 4u) and 1 mole of argon gas (atomic mass = 40u) is kept at 300 K in a container. The ratio of their rms speeds $\left[ \frac{v_{	ext{rms(helium)}}}{v_{	ext{rms(argon)}}} ight]$ is close to

Options:

(A) 0.32

(B) 2.24

(C) 3.16

(D) 0.45

Sigma Percentile · Accepted Answer

### The Microscopic Race

Imagine you are shrunk down to the size of an atom, standing inside a sealed container. The temperature is a balmy $300	ext{ K}$.

All around you, a chaotic dance is taking place. You see two types of participants in this microscopic race: Helium atoms, which are tiny and nimble, and Argon atoms, which are large and sluggish.

Even though they are in the same room and experiencing the exact same temperature, they are not moving at the same speed. Why is that? Let's dive into the physics of the Kinetic Theory of Gases to find out!

### The Master Equation

To understand how fast these gas molecules are zipping around, we need to look at their root mean square (rms) speed.

The formula that governs this is:
$$v_{	ext{rms}} = \sqrt{\frac{3RT}{M}}$$

Here, $R$ is the universal gas constant, $T$ is the absolute temperature in Kelvin, and $M$ is the molar mass of the gas.

Notice something crucial here. Both the Helium and Argon gases are trapped in the same container, which means they share the exact same temperature $T$.

### The Inverse Relationship

Since $3$, $R$, and $T$ are all constants in this specific scenario, we can strip away the noise and look at the core relationship.

The rms speed is inversely proportional to the square root of the mass:
$$v_{	ext{rms}} \propto \frac{1}{\sqrt{M}}$$

This mathematical relationship perfectly explains our initial observation. The lighter the molecule, the faster it must move to maintain the same thermal equilibrium. Helium, being the featherweight, is going to outpace the heavyweight Argon.

### Setting Up the Ratio

The question asks for the ratio of the rms speed of Helium to that of Argon. Let's set up our equation carefully.

Because of the inverse relationship, when we divide the speed of Helium by the speed of Argon, the mass of Argon must go in the numerator!

$$\frac{v_{	ext{rms(He)}}}{v_{	ext{rms(Ar)}}} = \sqrt{\frac{M_{	ext{Ar}}}{M_{	ext{He}}}}$$

**Watch out for this trap!** Many students accidentally put the mass of Helium in the numerator and get the inverse of the correct answer.

### The Final Sprint

Now, it is just a matter of plugging in the numbers. We are given the atomic masses: $M_{	ext{Ar}} = 40	ext{ u}$ and $M_{	ext{He}} = 4	ext{ u}$.

Substitute these into our ratio:
$$\frac{v_{	ext{rms(He)}}}{v_{	ext{rms(Ar)}}} = \sqrt{\frac{40}{4}}$$

This simplifies beautifully:
$$\frac{v_{	ext{rms(He)}}}{v_{	ext{rms(Ar)}}} = \sqrt{10}$$

We know that $\sqrt{9} = 3$, so $\sqrt{10}$ must be slightly larger than $3$. Calculating it gives approximately **$3.16$**.

This matches option (c) perfectly!

### The Kinetic Energy Trap

Before we celebrate, let's think about a classic variation of this question. What if the examiner had asked for the ratio of their *average translational kinetic energies* instead of their speeds?

You might be tempted to think that because Helium is moving faster, it has more kinetic energy. But that is a trap!

The average translational kinetic energy of a gas molecule is given by $\frac{3}{2}kT$. It depends **only** on the temperature.

Since both gases are at $300	ext{ K}$, their average kinetic energies are exactly the same. The ratio would simply be $1:1$. Helium moves faster precisely to compensate for its lack of mass, ensuring its kinetic energy matches that of the heavier Argon!

Animated Solution for Physics - Thermodynamics: A mixture of 2 moles of helium gas (atomic mass = 4u) and 1 mole of argon gas (atomic mass = 40u) is kept at 300 K in a container. The ratio of their rms speeds $[\frac{v _{rms(helium)}}{v _{rms(argon)}}]$ is close to

Select Answer:

Visualized Solution

Visualizing the Gas Mixture

The $v_{rms}$ Formula

Proportionality with Mass

Setting up the Ratio

Substituting the Values

Final Calculation

The Kinetic Energy Trap

The Sigma Insight: Kinetic Theory of Gases

The Microscopic Race

The Master Equation

The Inverse Relationship

Setting Up the Ratio

The Final Sprint

The Kinetic Energy Trap

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Select Answer:

Visualized Solution

Visualizing the Gas Mixture

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vrms​ Formula

Proportionality with Mass

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Substituting the Values

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The Master Equation

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Setting Up the Ratio

The Final Sprint

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