1.

Calculate for hydrogen at `27^(@)` (i) KE of one gram mole of the gas (ii) KE of one gram of the gas (iii) root mean square velocity of the molecule. Given, molecule wt. Of hydrogen = 2.

Answer» Here, `T=27^(@C = (27 + 273) K`
`=300 K,M=2`gram,
(i) KE of one gram mole
`=3/2 RT = 3/2 xx 8.31 xx 300 = 3.74 xx 10^(3)J`
(ii) KE of one gram `= (3.74 xx 10^(3))/(2)`
`1.87 xx 10^(3)J`
(ii) `upsilon-(rms) = sqrt((3RT)/(M)) = sqrt((3 xx 8.31 xx 300)/(2))`
`= 61.15 m//s`


Discussion

No Comment Found

Related InterviewSolutions