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. `108^(@)C`
C. `300^(@)C`
D. `327^(@)C`

Answer

D

Explanation

As, ` T_(1) = 27 + 273 = 300K`
` Rightarrow (KE_(2))/(KE_(1)) = T_(2)/T_(1) = T_(2)/300`
` Rightarrow (2KE_(1))/(KE_(1)) = T_(2)/300`
` Rightarrow T_(2) = 600K = 327^(@) C`