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?

Question

Find net kinetic energy (maximum possible) associated with 20 diatomic molecules (Here k_B is Boltzmann constant and T is absolute temperature of diatomic gas).

Answer 1

To determine the effect on the root mean square (RMS) velocity of oxygen molecules when the temperature is doubled and the oxygen molecules dissociate into atomic oxygen, we must understand the relationship between temperature, molar mass, and RMS velocity.

The formula for the RMS velocity of a gas is given by:

v_{\text{rms}} = \sqrt{\frac{3kT}{m}}

Where:

v_{\text{rms}} is the root mean square velocity.
k is the Boltzmann constant.
T is the absolute temperature.
m is the molecular mass of the gas particles.

Initially, for an O₂ molecule, the RMS velocity is:

v_{\text{rms, O_2}} = \sqrt{\frac{3kT}{M_{\text{O}_2}}}

Suppose the temperature is doubled (\( T \to 2T \)) and the oxygen molecule dissociates into atomic oxygen. The molar mass of an O atom is half that of an O₂ molecule:

For O₂, M_{\text{O}_2} = 2 \times M_{\text{O}}

When the oxygen molecule is dissociated, the formula becomes:

v_{\text{rms, O}} = \sqrt{\frac{3k(2T)}{M_{\text{O}}}}

Substituting M_{\text{O}_2} = 2 \cdot M_{\text{O}}, we simplify:

v_{\text{rms, O}} = \sqrt{\frac{6kT}{M_{\text{O}}}}

Compared to the initial RMS velocity of O₂, the ratio becomes:

\frac{v_{\text{rms, O}}}{v_{\text{rms, O}_2}} = \frac{\sqrt{\frac{6kT}{M_{\text{O}}}}}{\sqrt{\frac{3kT}{M_{\text{O}_2}}}} = \sqrt{4} = 2

Therefore, the RMS velocity of atomic oxygen doubles. Hence, the correct answer is:

The velocity of atomic oxygen doubles

Answer 2