The KE of `N` molecule of `O_(2)` is `x` joules at `-123^(@)C`. Another sample of `O_(2)` at `27^(@)C` has a `KE` of `2x` joules. The latter sample contains

Options

A. `N` molecules of `O_(2)`
B. `2N` molecules of `O_(2)`
C. `N//2` molecules of `O_(2)`
D. `N//4` molecules of `O_(2)`

Answer

A

Explanation

`KE=(3)/(2)RT, T= -123+273= +150 K`
`(3)/(2)xxRxx150=(3)/(2)xx8.314xx75`
`= x J = 225 xx 8.314 = x J`
At `27^(@)C=27+223= 300K`
`KE for = 2xJ=(3)/(2)xx8.314xx300`
`N` molecules
`:. xJ= 3xx8.314xx75`
In both the cases `xJ` corresponds to `N` molecules.