Question:
What will be the effect on the root mean square velocity of oxygen molecules if the temperature is doubled and oxygen molecule dissociates into atomic oxygen?
What will be the effect on the root mean square velocity of oxygen molecules if the temperature is doubled and oxygen molecule dissociates into atomic oxygen?
Updated On: Apr 12, 2026
- The velocity of atomic oxygen remains same
- The velocity of atomic oxygen doubles
- The velocity of atomic oxygen becomes half
- The velocity of atomic oxygen becomes four times
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The Correct Option is B
Approach Solution - 1
To determine the effect on the root mean square (RMS) velocity of oxygen molecules when the temperature is doubled and oxygen molecules dissociate into atomic oxygen, we need to understand the formulas and concepts involved.
Concept and Formula
- The root mean square (RMS) velocity of a gas molecule is given by the formula: \(v_{rms} = \sqrt{\frac{3kT}{m}}\), where
- \(k\) is the Boltzmann constant,
- \(T\) is the absolute temperature,
- \(m\) is the mass of the gas particle (molecule or atom).
Step-by-step Analysis
- Effect of Doubling the Temperature:
- If the temperature is doubled, i.e., \(T \to 2T\), then the RMS velocity becomes: \(v'_{rms} = \sqrt{\frac{3k(2T)}{m}} = \sqrt{2} \cdot v_{rms}\).
- This implies that if only temperature is considered, the RMS velocity increases by a factor of \(\sqrt{2}\).
- Effect of Dissociation:
- An oxygen molecule (\(O_2\)) dissociates into two oxygen atoms (\(O\)). Thus, the mass of the atom is half of that of the molecule.
- The RMS velocity of atomic oxygen is then calculated as: \(v''_{rms} = \sqrt{\frac{3k(2T)}{m/2}} = \sqrt{4} \cdot v_{rms} = 2 v_{rms}\).
- This indicates that due to dissociation, each oxygen atom now moves with a velocity doubled from that of the original molecule's RMS velocity (not considering temperature change).
Conclusion
The overall effect of both doubling the temperature and dissociating the oxygen molecules into atoms results in a doubling of the RMS velocity of atomic oxygen.
Thus, the correct answer is: The velocity of atomic oxygen doubles.
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Approach Solution -2
The correct answer is (B) : The velocity of atomic oxygen doubles
As
\(v_{rms}=\sqrt{\frac{3RT}{M_0}}\)
T is doubled and oxygen molecule is dissociated into atomic oxygen molar mass is halved.
So,
\(v'_{rms}=\sqrt{\frac{3R×2T_0}{M_0/2}}=2v_{rms}\)
So velocity of atomic oxygen is doubled.
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Concepts Used:
Kinetic Molecular Theory of Gases
Postulates of Kinetic Theory of Gases:
- Gases consist of particles in constant, random motion. They continue in a straight line until they collide with each other or the walls of their container.
- Particles are point masses with no volume. The particles are so small compared to the space between them, that we do not consider their size in ideal gases.
- Gas pressure is due to the molecules colliding with the walls of the container. All of these collisions are perfectly elastic, meaning that there is no change in energy of either the particles or the wall upon collision. No energy is lost or gained from collisions. The time it takes to collide is negligible compared with the time between collisions.
- The kinetic energy of a gas is a measure of its Kelvin temperature. Individual gas molecules have different speeds, but the temperature and
kinetic energy of the gas refer to the average of these speeds. - The average kinetic energy of a gas particle is directly proportional to the temperature. An increase in temperature increases the speed in which the gas molecules move.
- All gases at a given temperature have the same average kinetic energy.
- Lighter gas molecules move faster than heavier molecules.
