0.014 kg of ntirogen is enclosed in a vessel at a temperature of `27^(@)C`. How much heat has to be transferred to double the rms speed of its molecules.

Explanation

Number of molecules in 0.014 kg of nitrogen
`n=(0.014xx10^(3))/(27)=0.0005xx10^(3)=0.5=(1)/(2)` mole.
Molecular specific heat
`C_(v)=(5)/(2)R,C_(v)=(5)/(2)R=(5)/(2)xx2=(10)/(2)=5` cal / molecule k
[`therefore` Nitrogen diatomic molecule `S_(0)=(5)/(2)`]
Ratio of volume `(V_(2))/(V_(1))=sqrt((T_(2))/(T_(1))),T_(2)=4T_(1)`
`DeltaT=T_(2)-T_(1)=4T_(1)-T_(1)=3T_(1)`
`=3xx300=900K`
`DeltaQ=nC_(V)DeltaT=(1)/(2)xx5xx900`
`therefore` Heat transfer `DeltaQ=2250` cal