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At what temperature the molecule of nitrogen will have same rms velocity as the molecule of oxygen at `127^(@)C` ?A. `457^(@)C`B. `273^(@)C`C. `350^(@)C`D. `77^(@)C`

Answer» Correct Answer - D
The root mean square velocity , `v_" rms"sqrt((3RT)/(M))`
where , R is gas constant , T the absolute temperature ans M the molecular weight.
Given , `M_(N_(2))=28,M_(O_(2))=32` ,
`T_(O_(2))=127^(@)C=127+273=400K`
`therefore" " (v_(O_(2)))/(v_(N_(2)))=sqrt((T_(O_(2)))/(M_(O_(2)))xx(M_(N_(2)))/(T_(N_(2))))=sqrt((400)/(32)xx(28)/T_(N_(2)))=1`
`rArr" " T_(N_(2))=350K=77^(@)C`


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