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 $480K$ it will be

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

Answer 1

Answer: (B)

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