The temperature at which the average speed of the gas molecules is double to that at a temperature of `27^(@)C` is

Options

A. `54^(@)C`
B. `198^(@)C`
C. `327^(@)C`
D. `927^(@)C`

Answer

D

Explanation

`bar(v) = ((8RT)/(pi M))^(1//2) " or " bar(v) prop (T_(1))^(1//2)` for a gas
Now `(bar(v)_(1))/(bar(v)_(2)) = (T_(1)//T_(2))^(1//2)`
or `(v_(1))/(2v_(1)) = ((T_(1))/(T_(2)))^(1//2) = ((300)/(T_(2)))^(1//2)`
or `(1)/(4) = (300)/(T_(2)) " or " T_(2) = 1200 K`
`= 1200 - 273 = 927^(@)C`