1.

The average kinetic energy of a gas molecule at 27^(@)C is 621xx10^(-1)J. The average kinetic energy of gas molecule at 227^(@)C will be

Answer»

`52.2xx10^(-21)J`
`5.22xx10^(-21)J`
`10.35xx10^(-21)J`
`11.35xx10^(-21)J`

Solution :`K_(1)=(3)/(2)kT_(1)`, `K_(2)=(3)/(2)kT_(2)` i.e. `(K_(1))/(K_(2))=(T_(1))/(T_(2))impliesK_(2)=(K_(1)T_(2))/(T_(1))=(621xx10^(-21))/(300)xx500=10.35xx10^(-21)J`


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