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The average translational energy and the rms speed of molecules in a sample of oxygen gas at `300K` are `6.21xx10^(-21)J` and `484m//s`, respectively. The corresponding values at `600K` are nearly (assuming ideal gas behaviour)

A

`12.42xx 10^(-21) J, 968` m/s

B

`8.78 xx 10^(-21) J, 684` m/s

C

`6.21 xx 10^(-21) J, 968` m/s

D

`12.42 xx 10^(-21) J, 684` m/s

Text Solution

Verified by Experts

The correct Answer is:
D

`E=(3NKT )/(2 ),V_(rms ) = sqrt((3KT )/( m))`
`T_(1)= 300 K, T_(2)= 600 K`
`(E _(2))/(E_(1))=(T_(2))/(T_(1))=2 implies E_(2) =2E _(1)`
`E_(2)=2xx 6.21 xx10 -21 = 12.42 xx 10 -21 J`
`(V_(2))/(V_(1))=sqrt((T_(2))/(T_(1)))=sqrt(2) implies V_(2) = 484 xx sqrt(2) `
` V_(2) = 684 m//s`
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