The rms speeds of the molecules of hydrogen, oxygen and carbondioxide at the same temperature are v_

Q: Why did we take the mass of hydrogen as 2 and oxygen as 32? Aren't their atomic masses 1 and 16?

Great question! Hydrogen and oxygen exist as diatomic molecules ($H_2$ and $O_2$) in their natural gaseous state. Therefore, we must use their molecular masses (2 g/mol and 32 g/mol) rather than their atomic masses when applying the kinetic theory of gases.

Q: Does the pressure of the gas affect the rms speed?

No, it does not! As long as the temperature remains constant, changing the pressure of an ideal gas will cause a proportional change in its density, leaving the ratio of pressure to density (and thus the rms speed) completely unchanged.

Q: Why is the kinetic energy the same if their speeds are so different?

Average translational kinetic energy depends ONLY on the absolute temperature ($KE = \frac{3}{2}kT$). Even though hydrogen moves much faster, its mass is much smaller. The higher speed squared perfectly compensates for the lower mass, resulting in the exact same kinetic energy as the slower, heavier carbon dioxide molecules.

Q: What if the gases were at different temperatures?

If the temperatures were different, you could no longer just look at the molar mass. You would have to calculate and compare the ratio of $\frac{T}{M}$ for each gas to determine which one has the highest rms speed.

Q: Is rms speed the exact same thing as average speed?

No, they are slightly different statistical averages. The average speed is $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$, while the rms speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$. However, notice that both are proportional to $\frac{1}{\sqrt{M}}$, so the relative order of speeds between different gases will always be the same regardless of which average you use.

Question

The rms speeds of the molecules of hydrogen, oxygen and carbondioxide at the same temperature are $v_H$, $v_O$ and $v_{CO_2}$ respectively, then

Options:

(A) $v_H > v_O > v_{CO_2}$

(B) $v_{CO_2} > v_O > v_H$

(C) $v_H = v_O > v_{CO_2}$

(D) $v_H = v_O = v_{CO_2}$

Sigma Percentile · Accepted Answer

### The Need for Speed in Gases Imagine you are observing a microscopic race between different gas molecules inside a container. We have three competitors: Hydrogen, Oxygen, and Carbon Dioxide. They are all kept at the exact same temperature. The question is, who is moving the fastest on average? To answer this, we need to look at the root mean square (rms) speed of the molecules. ### The Master Equation The kinetic theory of gases provides us with a beautiful and elegant formula for the rms speed of gas molecules: $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ Here, $R$ is the universal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass of the gas. The problem explicitly states that all three gases are at the **same temperature**. Since $3$, $R$, and $T$ are all constants in this scenario, we can establish a direct proportionality: $$v_{rms} \propto \frac{1}{\sqrt{M}}$$ This mathematical relationship tells us a profound physical truth: **the heavier the molecule, the slower it moves**. It makes perfect intuitive sense! If you give the same amount of thermal energy to a heavy bowling ball and a light tennis ball, the lighter tennis ball will zip away much faster. ### The Heavyweight Championship Now, let's weigh our competitors. We must remember that these gases exist as molecules in their natural state, not as isolated atoms. * **Hydrogen ($H_2$):** Molar mass $M_{H_2} = 2 ext{ g/mol}$ * **Oxygen ($O_2$):** Molar mass $M_{O_2} = 32 ext{ g/mol}$ * **Carbon Dioxide ($CO_2$):** Molar mass $M_{CO_2} = 44 ext{ g/mol}$ Comparing their masses, the order is clear: $$M_{H_2} < M_{O_2} < M_{CO_2}$$ ### The Final Verdict Because the rms speed is inversely proportional to the square root of the molar mass, the order of their speeds will be the exact opposite of their masses. The lightest gas will be the fastest, and the heaviest gas will be the most sluggish. Therefore, Hydrogen takes the gold medal for speed, Oxygen takes silver, and Carbon Dioxide takes bronze. $$v_H > v_O > v_{CO_2}$$ This perfectly matches option (a). Always remember, at a given temperature, all gases have the same average translational kinetic energy, but their speeds vary drastically depending on how heavy they are!

Animated Solution for Physics - Thermodynamics: The rms speeds of the molecules of hydrogen, oxygen and carbondioxide at the same temperature are $v_{H}$ , $v_{O}$ and $v_{C O_{2}}$ respectively, then

Select Answer:

Visualized Solution

$v_{r m s}$ Formula

Constant Temperature

Molar Masses

Comparing Masses

Comparing Speeds

Kinetic Energy

The Sigma Insight: Kinetic Theory of Gases

The Need for Speed in Gases

The Master Equation

The Heavyweight Championship

The Final Verdict

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