Question

The temperature of an ideal gas is increased from $120$ K to $480$ K. If at $120$ K the root-mean square velocity of the gas molecules is v, at $480$ K it becomes.

• A. $4$v
• B. $2$v
• C. $v/2$
• D. $v/4$

Answer 1

The correct option is B $2$v

The root mean square velocity is given by $v_{rms}=\sqrt{\frac{3RT}{M}}$ so. $\frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}$

Now $T_1=120K,T_2=480K,v_1=v$

so ,

$\frac{v_1}{v_2}=\sqrt{\frac{120}{480}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

$\Rightarrow \frac{v_1}{v_2}=\frac{1}{2}$

$\Rightarrow v_2=2v$

Hence the correct option is B.