1.

The r.m.s. speed of the molecules of a gas at 100^(@)C is v. The temperature at which the r.m.s. speed will be sqrt(3)v is

Answer»

`546^(@)C`
`646^(@)C`
`746^(@)C`
`846^(@)C`

Solution :The RMS speed of a GAS molecules is given by
`v_(rms)=sqrt((3RT)/(M))orv_(rms)propsqrt(T)therefore (v_(rms_(1)))/(v_(rms_(2)))=sqrt((T_(1))/(T_(2)))`
Here, `v_(rms_(1))=v,v_(rms_(2))=sqrt(3)v`
and `T_(1)=100^(@)C=100+273=373K`
`therefore ((v)/(sqrt(3)v))^(2)=((373)/(T_(2)))or(1)/(3)=(373)/(T_(2))`
`implies T_(2)=3xx373=1119K`
`T_(2)=1119-273=846^(@)C`.


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