The molar specific heats of an ideal gas at constant volume and constant pressure are respectively 4.98 and 6.96 cal `mol^(-1) K^(-1)`. If the molecular weight of the gas be 32, then calculate the root means square speed of the molecule of the gas at `120^@ C`. (1 cal = 4.2 J)

Explanation

`C_(p)-C_(V)=R`
`R=6.96-4.98=1.98" cal "mol^(-1)K^(-1)`
`R=1.98xx4.2=8.316" J "mol^(-1)K^(-1)`
`(1)/(2)M overline(V^(2))=(3)/(2)RT`
`overline(V^(2))=(3RT)/(M)`
`V_(rms)=sqrt(overline(V^(2)))=sqrt((3RT)/(M))` ltbRgt Here, M=32g `mol^(-1)`=32`xx10^(-3)" kg "mol^(-1) and T=120^(@)C+273=393K`
`thereforeV_(rms)=sqrt((3xx8.316xx393)/(32xx10^(-3))=553.53ms^(-1)`.