The average velocity of an ideal gas at `0^(@)C` is `0.4 ms^(-1)`. If the temperature of gas is increased to `546^(@)C` its average velocity will be

Options

A. `0.8 m s^(-1)`
B. `1.6 ms^(-1)`
C. `0.346 ms^(-1)`
D. `0.69 ms^(-1)`

Answer

D

Explanation

Apply formula
`bar(v) = (8RT//pi M)^(1//2)` and compare.
`((bar(v)_(T) = 273)/(bar(v)_(T) = 819))^(2) = (T_(1))/(T_(2))`
or `((0.4)/(bar(v)))^(2) = (273)/(819) " or " (0.16)/(bar(v)^(2)) = (273)/(819)`
or `(bar(v))^(2) = (0.16 xx 819)/(273) = 0.48`
or `bar(v) = sqrt(0.48) = 0.69 ms^(-1)`