Question

The rms speed of oxygen molecules in a container is $V$. If the temperature is doubled and oxygen molecules dissociate into oxygen atoms, the rms speed will become :

• A. $V$
• B. $\sqrt{2}V$
• C. $2V$
• D. $4V$

Answer 1

The correct option is C $2V$

Suppose, mass of each atom of oxygen is $m$
then, molar mass of oxygen gas will be $2m$.
We know that, $V_{rms}=\sqrt{\frac{3RT}{M}}\text{Initially,}V=\sqrt{\frac{3RT}{2m}}..\left(1\right)$
After temperature is doubled and oxygen molecules dissociate,
$V_{rms}=\sqrt{\frac{3R\times 2T}{m}}...\left(2\right)$
On dividing equation $\left(2\right)$ by equation $\left(1\right)$,
$\frac{V_{rms}}{V}=\sqrt{\frac{\frac{6RT}{m}}{\frac{3RT}{2m}}}\Rightarrow V_{rms}=2V$