1.

0.0014kg of nitrogen is enclosed in a vessel at a temperature of 27^(@)C. How much heat has to be transferred to the gas to double the rms speed of its molecules.

Answer»

Solution :Number of mole in `0.014kg` of Nitrogen
`n=(0.0014xx10^(3))/(28)=(1)/(20)` mole
`C_(v)=(5)/(2)R=(5)/(2)xx2=5` cal/mole `k`
`(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)/(20)xx5xx900=225cal`


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