If the rms speed of oxygen molecules at 0°C is 160 m/s, find the rms speed of hydrogen molecules at

Q: Why do we use molar mass instead of the mass of a single molecule?

You can actually use either! The formula $v_{\text{rms}} = \sqrt{\frac{3kT}{m}}$ uses the mass of a single molecule ($m$) and the Boltzmann constant ($k$). The formula $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ uses molar mass ($M$) and the gas constant ($R$). Since we are taking a ratio, the constants cancel out, and the ratio of molar masses is identical to the ratio of single molecule masses.

Q: Do I need to convert the temperature from Celsius to Kelvin for this problem?

In this specific problem, because we are taking a ratio of speeds at the *same* temperature, the temperature term cancels out entirely. However, if you were calculating the absolute speed from scratch, you **must** always convert temperature to Kelvin!

Q: Why is the speed of hydrogen so much higher if they are at the same temperature?

Temperature is a measure of average kinetic energy ($\frac{1}{2}mv^2$). Since both gases are at $0^\circ\text{C}$, they have the same average kinetic energy. Because hydrogen is much lighter (smaller $m$), it must have a much higher velocity ($v$) to balance the equation and have the same energy as the heavier oxygen.

Q: What if the gases were at different temperatures?

If the temperatures were different, they wouldn't cancel out in our ratio. The ratio formula would become $\frac{v_1}{v_2} = \sqrt{\frac{T_1 M_2}{T_2 M_1}}$. You would simply plug in the respective temperatures (in Kelvin!) along with the molar masses.

Q: Is RMS speed the exact speed every single molecule is moving at?

No! Gas molecules move at a wide variety of speeds, constantly colliding and changing velocities according to the Maxwell-Boltzmann distribution. The RMS speed is a specific type of statistical average that directly relates to the kinetic energy of the gas.

Question

If the rms speed of oxygen molecules at $0^\circ	ext{C}$ is $160	ext{ m/s}$, find the rms speed of hydrogen molecules at $0^\circ	ext{C}$.

Options:

(A) 640 m/s

(B) 40 m/s

(C) 80 m/s

(D) 332 m/s

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Gases

Imagine two identical containers sitting side by side. One is filled with oxygen ($O_2$), and the other with hydrogen ($H_2$). Both are kept at the exact same temperature of $0^\circ	ext{C}$.

Even though the macroscopic temperature is identical, the microscopic world inside these containers is vastly different. Temperature is simply a measure of the average translational kinetic energy of the gas molecules.

Because both gases are at the same temperature, their molecules possess the exact same average kinetic energy. However, kinetic energy depends on both mass and velocity ($K = \frac{1}{2}mv^2$).

### The Master Equation

To understand how fast these molecules are zipping around, we use the formula for the Root Mean Square (RMS) speed:

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

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

Since $3$, $R$, and $T$ are identical for both our oxygen and hydrogen samples, they act as constants in this scenario. This reveals a beautiful, inverse relationship: the RMS speed is inversely proportional to the square root of the molar mass.

$$v_{	ext{rms}} \propto \frac{1}{\sqrt{M}}$$

### The Ratio of Speeds

We can set up a mathematical ratio to compare the two gases directly. By dividing the RMS speed of oxygen by the RMS speed of hydrogen, the constants cancel out perfectly:

$$\frac{(v_{	ext{rms}})_{O_2}}{(v_{	ext{rms}})_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}}$$

Notice how the masses are flipped on the right side of the equation. This is the mathematical manifestation of our inverse relationship!

### Final Calculation

Now, we substitute the known values into our elegant ratio. We are given that the RMS speed of oxygen is $160	ext{ m/s}$. The molar mass of hydrogen ($H_2$) is $2	ext{ g/mol}$, and the molar mass of oxygen ($O_2$) is $32	ext{ g/mol}$.

$$\frac{160}{(v_{	ext{rms}})_{H_2}} = \sqrt{\frac{2}{32}}$$

Simplifying the fraction inside the square root:

$$\frac{160}{(v_{	ext{rms}})_{H_2}} = \sqrt{\frac{1}{16}}$$

Taking the square root of $\frac{1}{16}$ gives us exactly $\frac{1}{4}$.

$$\frac{160}{(v_{	ext{rms}})_{H_2}} = \frac{1}{4}$$

Cross-multiplying yields our final answer:

$$(v_{	ext{rms}})_{H_2} = 160 	imes 4 = 640	ext{ m/s}$$

**The hydrogen molecules are moving four times faster than the oxygen molecules!** This makes perfect physical sense. Because hydrogen is sixteen times lighter than oxygen, it must move four times as fast to maintain the exact same average kinetic energy at the same temperature.

Animated Solution for Physics - Thermodynamics: If the rms speed of oxygen molecules at $0^{\circ} C$ is $160 m/s$ , find the rms speed of hydrogen molecules at $0^{\circ} C$ .

Select Answer:

Visualized Solution

$Given Data$

$Formula for v_{rms}$

$Ratio of Speeds$

$Substituting Values$

$Calculating (v_{rms})_{H_{2}}$

$Conclusion$

The Sigma Insight: Kinetic Theory of Gases

The Setup

The Master Equation

The Ratio of Speeds

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; } If the rms speed of oxygen molecules at 0∘C is 160 m/s, find the rms speed of hydrogen molecules at 0∘C.

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Data

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Speeds

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating (vrms​)H2​​

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