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The average kinetic energy of gas molecu...

The average kinetic energy of gas molecule at `27^(@)C` is `6.21xx10^(-21)` J. Its average kinetic energy at `127^(@)C` will be

A

`52.2xx10^(-21)J`

B

`5.22xx10^(-21)J`

C

`10.35xx10^(-21)J`

D

`11.35xx10^(-21)J`

Text Solution

Verified by Experts

The correct Answer is:
C
Average kinetic energy, `E=(3)/(2)k_(B)T`
where `k_(B)` is Boltzmann constant and T is the absolute temperature.
`therefore (E_(2))/(E_(1))=(T_(2))/(T_(1))=(273+227)/(273+27)=(500)/(300)`
or `E_(2)=(5)/(3)xx6.21xx10^(-21)J=10.35xx10^(-21)J`
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