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_______.
Write numerical value (upto 2 digit only) after decimal point only as for 0.19, write 19)

Answer 1

As 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, answer is 60.