Question

The root mean square(rms) speed of X₂ gas is x m/s at a given temperature. When the temperature is doubled, the X₂ molecules dissociated completely into atoms. The root mean square speed of the sample of gas then becomes (in m/s)

• A. \(\frac{x}{2}\)
• B. x
• C. 2x
• D. 4x

Answer 1

The Correct Option is C

The root mean square (rms) speed of gas molecules is given by:

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

Where:

• $v_{\text{rms}}$ is the root mean square speed
• $k$ is the Boltzmann constant
• $T$ is the temperature in Kelvin
• $m$ is the mass of one gas molecule

Now, suppose the gas is $X_2$ and its initial rms speed is $x$ m/s at temperature $T$.

When the temperature is doubled (i.e., $T \rightarrow 2T$), and the molecules completely dissociate into atoms, the mass of each particle becomes half (since each $X_2$ becomes two $X$ atoms).

So, in the new situation:

$v_{\text{rms}}' = \sqrt{\frac{3k(2T)}{m/2}} = \sqrt{\frac{6kT}{m/2}} = \sqrt{\frac{12kT}{m}} = \sqrt{4 \cdot \frac{3kT}{m}} = 2v_{\text{rms}}$

Therefore, $v_{\text{rms}}'$ becomes $2x$ m/s

Correct option: (C) 2x