1.

At what temperature will the average speed of CH_(4) molecules have the same value as O_(2) has at 300 K

Answer»

1200K
150 K
600 K
300 K

Solution :`(V_(av)CH_(4))/(V_(ab)O_(2))=sqrt((T_(CH_(4)))/(T_(O_(2))).(M_(O_(2)))/(M_(CH_(4))))=1`
`(T_(CH_(4)))/(300).(32)/(16)=1, T_(CH_(4))=150K`


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