The average velocity of an ideal gas molecule at `27^(@)C` is `0.3 m s^(-1)`. The average velocity at `927^(@)C` will be

Options

A. `0.6 m s^(-1)`
B. `0.3 m s^(-1)`
C. `0.9 m s^(-1)`
D. `3.0 m s^(-1)`

Answer

A

Explanation

`(v_(1))/(v_(2))=sqrt((T_(1))/(T_(2)))`
`T_(1)=273^(@)C=300K`
`T_(2)=927^(@)C=1200K`
`V_(1)=0.3 m s^(-1)`
`V_(2)=sqrt((T_(2))/(T_(1))xxV_(1)=sqrt(1200)/(300)xx0.3=0.6m s^(-1)`