The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature

Q: Why doesn't the molar mass affect the kinetic energy?

According to the equipartition of energy, each degree of translational freedom contributes $\frac{1}{2}kT$ to the energy, regardless of the particle's mass. Since all molecules have 3 translational degrees of freedom, their average translational kinetic energy is always $\frac{3}{2}kT$.

Q: Does this mean heavier molecules move at the same speed as lighter ones?

No! Kinetic energy is $\frac{1}{2}mv^2$. If a heavy molecule and a light molecule have the same kinetic energy, the lighter molecule must be moving faster to compensate for its smaller mass.

Q: What about rotational and vibrational kinetic energies?

The question specifically asks for *translational* kinetic energy, which is always $\frac{3}{2}kT$. If it asked for *total* kinetic energy, we would need to consider the degrees of freedom (which is 5 for diatomic gases like $\text{O}_2$ and $\text{N}_2$ at room temperature), making it $\frac{5}{2}kT$. Even then, it would be the same for both since they are both diatomic.

Q: Is the total kinetic energy of 1 mole of O2 and 1 mole of N2 the same?

Yes. For 1 mole of gas, the total translational kinetic energy is $\frac{3}{2}RT$. Since $R$ is a universal constant and $T$ is the same, the total energy is identical.

Q: What if the gases were at different temperatures?

If the temperatures were different, the kinetic energies would be directly proportional to their absolute temperatures ($KE \propto T$). You would use the ratio $\frac{KE_1}{KE_2} = \frac{T_1}{T_2}$ to find the answer.

Question

The average translational kinetic energy of $	ext{O}_2$ (molar mass 32) molecules at a particular temperature is $0.048	ext{ eV}$. The translational kinetic energy of $	ext{N}_2$ (molar mass 28) molecules in eV at the same temperature is

Options:

(A) 0.0015

(B) 0.003

(C) 0.048

(D) 0.768

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Gases

Imagine you are standing in a laboratory, looking at two identical containers. One is filled with Oxygen gas ($	ext{O}_2$) and the other with Nitrogen gas ($	ext{N}_2$). The problem states that both containers are kept at the exact same temperature.

We are given a specific piece of data: the average translational kinetic energy of an Oxygen molecule is $0.048	ext{ eV}$. The question then asks us to find the average translational kinetic energy of a Nitrogen molecule. It also helpfully (or perhaps deceptively) provides the molar masses of both gases: $32$ for Oxygen and $28$ for Nitrogen.

### The Master Equation: Kinetic Theory Unveiled

To solve this, we need to dive into the Kinetic Theory of Gases. What is the fundamental formula for the average translational kinetic energy of a single gas molecule?

According to the theory, it is given by the elegant equation:

$$KE_{	ext{avg}} = \frac{3}{2}kT$$

Here, $k$ is the Boltzmann constant, and $T$ is the absolute temperature of the gas.

### The Trap: Molar Mass Distraction

Look closely at the master equation. Do you see any variable representing mass? No! The average translational kinetic energy depends **strictly and only** on the absolute temperature $T$.

It is completely independent of the molar mass, the size of the molecule, or the chemical nature of the gas. The examiner provided the molar masses ($32$ and $28$) specifically to bait you into performing unnecessary calculations. This is a classic trap in competitive exams!

### The Final Verdict

Since both the Oxygen and Nitrogen gases are at the exact same temperature, their molecules must possess the exact same average translational kinetic energy.

Therefore, without any complex math, we can confidently state that the kinetic energy of the Nitrogen molecules is also **$0.048	ext{ eV}$**.

**A Quick Thought Experiment:** What if the question had asked for the *Root Mean Square (RMS) speed* instead? The formula for RMS speed is $v_{	ext{rms}} = \sqrt{\frac{3RT}{M}}$. Because mass $M$ is in the denominator, the lighter Nitrogen molecules would actually be zipping around faster than the heavier Oxygen molecules, even though their average kinetic energies are identical. Always read carefully to see whether the question asks for energy or speed!

Animated Solution for Physics - Thermodynamics: The average translational kinetic energy of $O_{2}$ (molar mass 32) molecules at a particular temperature is $0.048 eV$ . The translational kinetic energy of $N_{2}$ (molar mass 28) molecules in eV at the same temperature is

Select Answer:

Visualized Solution

Visualizing the Setup

The Master Equation

Analyzing Dependencies

Applying the Logic

Final Answer

The Way Forward

The Sigma Insight: Kinetic Theory of Gases

The Setup

The Master Equation

The Trap

The Final Verdict

Similar Questions

The average translational energy and the rms speed of molecules in a sample of oxygen gas at $300$ K are $6.21 \times 1 0^{- 21}$ J and $484$ m/s respectively. The corresponding values at $600$ K are nearly (assuming ideal gas behaviour)

The average translational kinetic energy of $N_{2}$ gas molecules at ........... $^{\circ} C$ becomes equal to the KE of an electron accelerated from rest through a potential difference of $0.1 V$ . [Given, $K_{B} = 1.38 \times 1 0^{- 23} J/K$ ]

A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300 K. The ratio of the average rotational kinetic energy per $O_{2}$ molecule to per $N_{2}$ molecule is

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Let $\overset{v}{ˉ}$ , $v_{rms}$ and $v_{p}$ respectively denote the mean speed, root mean square speed and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature $T$ . The mass of a molecule is $m$ . Then,

A $15 g$ mass of nitrogen gas is enclosed in a vessel at a temperature $2 7^{\circ} C$ . Amount of heat transferred to the gas, so that rms velocity of molecules is doubled is about (Take, $R = 8.3 J /K -mole$ )

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

From the following statements concerning ideal gas at any given temperature $T$ , select the correct one (s).

One kg of a diatomic gas is at a pressure of $8 \times 1 0^{4} Nm^{- 2}$ . The density of the gas is $4 kgm^{- 3}$ . What is the energy of the gas due to its thermal motion?

Select Answer:

Visualized Solution

Visualizing the Setup

The Master Equation

Analyzing Dependencies

Applying the Logic

Final Answer

The Way Forward

The Sigma Insight: Kinetic Theory of Gases

The Setup

The Master Equation

The Trap

The Final Verdict

Similar Questions

.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.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } From the following statements concerning ideal gas at any given temperature T, select the correct one (s).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } One kg of a diatomic gas is at a pressure of 8×104 Nm−2. The density of the gas is 4 kgm−3. What is the energy of the gas due to its thermal motion?

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

From the following statements concerning ideal gas at any given temperature $T$ , select the correct one (s).

One kg of a diatomic gas is at a pressure of $8 \times 1 0^{4} Nm^{- 2}$ . The density of the gas is $4 kgm^{- 3}$ . What is the energy of the gas due to its thermal motion?