Question

The average velocity of an ideal gas molecule at ${27}^{\circ }C$ is $0.3m{s}^{-1}$. The average velocity at ${927}^{\circ }C$ will be

• A. $0.6m{s}^{-1}$
• B. $0.3m{s}^{-1}$
• C. $0.9m{s}^{-1}$
• D. $3.0m{s}^{-1}$

Answer 1

The correct option is A $0.6m{s}^{-1}$

We know, $\frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}$

$T_1={27}^{\circ }C=300K$

$T_2={927}^{\circ }C=1200K$

$V_1=0.3m{s}^{-1}$

$V_2=\sqrt{\frac{T_2}{T_1}}\times V_1=\sqrt{\frac{1200}{300}}\times 03.=0.6m{s}^{-1}$

So, the correct option is $A$