Question

Consider a mixture of gas molecules of type $A,B$ and $C$ having masses, $m_A<m_B<m_C$. The ratio of their root-mean-square speeds at normal temperature and pressure is :

• A. $v_A=v_B\ne v_C$
• B. $\frac{1}{v_A}<\frac{1}{v_B}<\frac{1}{v_C}$
• C. $\frac{1}{v_A}>\frac{1}{v_B}>\frac{1}{v_C}$
• D. $v_A=v_B=v_C=0$

Answer 1

The correct option is B $\frac{1}{v_A}<\frac{1}{v_B}<\frac{1}{v_C}$

$v_{\text{rms}}=\sqrt{\frac{3k_{B}T}{m}}$

At the same, normal temperature and pressure,

$v_{\text{rms}}\propto \frac{1}{\sqrt{m}}$

$\because m_A<m_B<m_C$

$\Rightarrow v_A>v_B>v_C$

$\Rightarrow \frac{1}{v_A}<\frac{1}{v_B}<\frac{1}{v_C}$

Hence, option $\left(D\right)$ is the correct answer.