1. The root mean square speed of a गैस particle is given by

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

where:

• $T$ = absolute temperature
• $m$ = mass of one particle
• $k$ = Boltzmann constant

2. Initially, the gas consists of oxygen molecules $\mathrm{O_2}$.

So,

$v_1=\sqrt{\frac{3kT}{m_{\mathrm{O_2}}}}$

3. Now two changes happen:

• temperature is doubled: $T \to 2T$
• oxygen molecules dissociate into atomic oxygen, so mass per particle becomes half:

$m_{\mathrm{O}}=\frac{m_{\mathrm{O_2}}}{2}$

4. Therefore the new rms speed is

$v_2=\sqrt{\frac{3k(2T)}{m_{\mathrm{O_2}}/2}}$

5. Simplify:

$v_2=\sqrt{\frac{6kT}{m_{\mathrm{O_2}}/2}}$

$v_2=\sqrt{\frac{12kT}{m_{\mathrm{O_2}}}}$

6. Compare with the initial speed:

$v_1=\sqrt{\frac{3kT}{m_{\mathrm{O_2}}}}$

Hence,

$\frac{v_2}{v_1}=\sqrt{\frac{12kT/m_{\mathrm{O_2}}}{3kT/m_{\mathrm{O_2}}}}=\sqrt{4}=2$

So,

$v_2=2v_1$

7. Option check:

• A: same $\rightarrow$ incorrect
• B: doubles $\rightarrow$ correct
• C: half $\rightarrow$ incorrect
• D: four times $\rightarrow$ incorrect

Therefore, the rms velocity becomes double.