Question

The temperature of an ideal gas is increased from $120K$ to $480K$. If at $120K$, the root mean square speed of gas molecules is v, then at $80k$, it will be

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

Answer 1

The correct option is B $2v$

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

$\Rightarrow \frac{v}{2}=\frac{1}{2}$

$\Rightarrow v_2=2v$

Hence, the correct option is $B$