Question

The average speed of nitrogen molecules in a gas is v. If the temperature is doubled and the $N_2$ molecule dissociate into nitrogen atoms, then the average speed will be

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

Answer 1

The correct option is C $2v$

$V_{av}=\sqrt{\frac{8RT}{\pi M}}$ $\left\{averagespeedof{N}_{2}atT\right\}$

$\Rightarrow V_{a{v}^{\prime }}=\sqrt{\frac{8R{T}^{\prime }}{\pi M^{\prime }}}$

$\Rightarrow$ Given that $T^{\prime }=2T$

$\Rightarrow m^{\prime }=\frac{M}{2}$

$\Rightarrow$ So, $V_{a{v}^{\prime }}=\sqrt{\frac{8R\left(2T\right)}{\pi \left(\frac{M}{2}\right)}}=2\sqrt{\frac{8RT}{\pi M}}$

$\Rightarrow V_{a{v}^{\prime }}=2V_{av}$

Hence, the answer is $2v.$