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?

• A. The velocity of atomic oxygen remains same
• B. The velocity of atomic oxygen doubles
• C. The velocity of atomic oxygen becomes half
• D. The velocity of atomic oxygen becomes four times

Answer 1

The Correct Option is B

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 

1. 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 \(\sqrt{2}\).
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.

Answer 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.