Question

The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K the root
square velocity of the gas molecules is V, then at 480 K it becomes​

Answer 1

Explanation:

$\sf\large\underline\purple{Answer:-}$

$\displaystyle \sf \star\frac{v1}{v2} = \frac{ \sqrt{ \frac{3rt1}{m} } }{ \sqrt{ \frac{3rt2}{m} } } = \frac{ \sqrt{t1} }{ \sqrt{t2} } \\ \\ \displaystyle \sf \implies \: v2 = 2v1$