1.

The temperature at which the root mean square velocity of a molecules will be double of its value at 100^(@)C is

Answer»

`1492^(@)C`
`1219^(@)C`
`546^(@)C`
`273^(@)C`

Solution :`BARC=sqrt((3RT)/(M)),(T_(2))/(T_(1))=((barc_(2))/(barc_(1)))^(2)`
`T_(1)=(273+100)K`
`T_(2)=(273+100)(2)^(2)=1492K`
`=(1492-273)=1219^(@)C`


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