About 0.014 kg nitrogen is enclosed in a vessel at temperature of `27^(@)C` How much heat has to be transferred to the gas to double the rms speed of its molecules ? `(R = 2 cal//mol K)`

Explanation

Number of moles in `0.014 kg` of nitrogen `n=(0.014xx10^(3))/(28)= 1/2 mol e`
As Nitrogen is a diatomic gas,
`:. C_(v)= 5/2R=5/2xx2= 5 cal. Mol e^(-1)K^(-1)`
As `r.m.s` speed of molecules is directly proprtional to square root of absolute temperature of the gas
`(v_(2))/(v_(2))= sqrt((T_(2))/(T_(1)))= 2, T_(2)= 4T_(1)`
`:. DeltaT= T_(2)-T_(1)=4T_(1)-T_(1)=3T_(1)=3xx300= 900K`
`DeltaQ= nC_(v)DeltaT=1/2xx5xx900= 2250cals`