Question

The root mean square speed of molecules of nitrogen gas is $v$ at a certain temperature. When the temperature is doubled, the molecules dissociate into individual atoms. The new rms speed of the atom is :

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

Answer 1

Answer: (B)

$2v$

$V_{rms}=\sqrt{\frac{3RT}{M}}$

$V_{rms}^{‘}=\sqrt{\frac{3RT}{M^{‘}}}$

$T^{‘}=2T$

$M^{‘}=M/2$

$V^′_{rms}=\sqrt {\dfrac {3R\times 2T}{M/2}}

$V_{rms}^{\prime }=2V_{rms}$

If $V_{rms}=v$ then the new rms speed of atom is $2v$

Hence, the correct option is $\text{B}$