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The KE of N molecule of O(2) is x joules...

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

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)`

Text Solution

Verified by Experts

The correct Answer is:
A

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