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At what temperature is the K.E. Of a gas...

At what temperature is the `K.E`. Of a gas molecules half that of its value at `27^(@)C`

A

`13.5^(@)` C

B

`150^(@)` C

C

75 K

D

`-123 K`

Text Solution

Verified by Experts

The correct Answer is:
C
K.E.=`(3)/(2)RT,(E_(1))/(E_(2))=sqrt(T_(1)/T_(2)), `
`(1)/(2)==sqrt(T_(1)/(300))T_(1)=75` K
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