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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 K

C

`150^(@)C`

D

`-123K`

Text Solution

Verified by Experts

The correct Answer is:
B
`(KE_(2))/(KE_(1))=(T_(2))/(T_(1))`
`(1)/(2)=(T_(2))/(300)" " thereforeT_(2)=(300)/(2)=150K`
`T_(2)=150K`.
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