At what temperature is the rms velocity of a hydrogen molecule equal to that of an oxygen molecule a

Q: Why do we have to convert the temperature to Kelvin? Can't we just use Celsius?

In all gas laws and kinetic theory formulas, temperature represents the average kinetic energy of the molecules. Kinetic energy is directly proportional to absolute temperature (Kelvin). Zero Kelvin means zero kinetic energy, but zero Celsius does not. Therefore, you must always use Kelvin.

Q: Why does hydrogen need such a low temperature to have the same speed?

Because hydrogen is much lighter than oxygen. At the same temperature, lighter molecules move much faster than heavier ones. To force the fast, light hydrogen molecules to slow down and match the speed of the heavier oxygen molecules, you have to drastically reduce their temperature.

Q: What are the molar masses of oxygen and hydrogen?

Oxygen gas exists as a diatomic molecule ($O_2$), so its molar mass is $16 \times 2 = 32\text{ g/mol}$. Similarly, hydrogen gas is $H_2$, so its molar mass is $1 \times 2 = 2\text{ g/mol}$.

Q: Does the value of the gas constant $R$ matter in this calculation?

No, it doesn't. Because we are equating the velocities of two gases, the term $3R$ appears on both sides of the equation and cancels out completely. You don't need to plug in its numerical value.

Q: Would the answer change if the question asked for average velocity instead of rms velocity?

No, the final temperature would be exactly the same. The formula for average velocity is $\sqrt{\frac{8RT}{\pi M}}$. If you equate the average velocities, the $\frac{8R}{\pi}$ terms will cancel out, leaving you with the exact same ratio: $\frac{T_O}{M_O} = \frac{T_H}{M_H}$.

Question

At what temperature is the rms velocity of a hydrogen molecule equal to that of an oxygen molecule at $47^{\circ}	ext{C}$ ?

Options:

(A) 80 K

(B) -73 K

(C) 3 K

(D) 20 K

Sigma Percentile · Accepted Answer

### The Race of the Molecules: Hydrogen vs. Oxygen

Imagine you are at a racetrack. On one lane, you have a massive, heavy truck representing an oxygen molecule ($O_2$). On the other lane, you have a tiny, lightweight sports car representing a hydrogen molecule ($H_2$). If both vehicles are given the exact same amount of kinetic energy (which is what temperature represents), the lightweight sports car is going to zoom off much faster than the heavy truck.

This is the core intuition behind the Kinetic Theory of Gases. But what if we want them to move at the exact same speed? We would have to drastically reduce the energy of the sports car. Let's see how the math perfectly mirrors this physical reality.

### The Master Equation

The speed of a gas molecule is best described by its root mean square (rms) velocity. The formula for this is a beautiful relationship between temperature and mass:

$$v_{rms} = \sqrt{\frac{3RT}{M}}$$

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

### Equating the Speeds

The problem asks us to find the temperature at which the hydrogen molecules will have the exact same rms velocity as the oxygen molecules at $47^{\circ}	ext{C}$. First, we must always convert Celsius to Kelvin.

$$T_O = 47 + 273 = 320	ext{ K}$$

Now, we set the rms velocities equal to each other:

$$\sqrt{\frac{3RT_O}{M_O}} = \sqrt{\frac{3RT_H}{M_H}}$$

By squaring both sides, the square roots vanish. The $3R$ terms on both sides also cancel out perfectly, leaving us with a highly elegant and simple ratio:

$$\frac{T_O}{M_O} = \frac{T_H}{M_H}$$

### The Final Calculation

Now, we simply substitute the known values. The molar mass of oxygen ($O_2$) is $32	ext{ g/mol}$, and the molar mass of hydrogen ($H_2$) is $2	ext{ g/mol}$.

$$\frac{320}{32} = \frac{T_H}{2}$$

Simplifying the left side gives us $10$.

$$10 = \frac{T_H}{2}$$

Multiplying both sides by $2$, we arrive at our final answer:

**$$T_H = 20	ext{ K}$$**

Think about what this means physically! Oxygen is at a warm $320	ext{ K}$, while hydrogen has to be cooled down to a freezing $20	ext{ K}$ just to slow it down enough to match the speed of the sluggish oxygen molecules. The math confirms our intuition perfectly.

Animated Solution for Physics - Thermodynamics: At what temperature is the rms velocity of a hydrogen molecule equal to that of an oxygen molecule at $4 7^{\circ} C$ ?

Select Answer:

Visualized Solution

Visualizing the Setup

The RMS Velocity Formula

Equating the Velocities

Simplifying the Equation

Substituting the Values

Final Calculation

Physical Interpretation

The Sigma Insight: Kinetic Theory of Gases

The Race of the Molecules

The Master Equation

Equating the Speeds

The Final Calculation

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Animated Solution for Physics - Thermodynamics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } At what temperature is the rms velocity of a hydrogen molecule equal to that of an oxygen molecule at 47∘C ?