Calculate the root mean square speed total and average translational kintic energy in joule of the molecules in 9 methane at `27^(@)C` .

Explanation

`T =27 + 273 = 300 K`
`R =8.314 xx 10^(7)` erg
`u_("rms")` for `CH_(4)`
`= sqrt(((3RT)/(M))) = sqrt(((3 xx 8.314 xx 10^(7) xx 300)/(16)))`
` =6.84 xx 10^(4) cm sec^(-1)`
Now `KE//mol` of `CH_(4) = (1)/(2) Mu^(2)`
` = (1)/(2) xx 16 xx (6.84 xx 10^(4))^(2) = 374.28 xx 10^(8) erg mo1^(-1)`
`:. KE` for `(1)/(2)` mole `CH_(4) = (374.28 xx 10^(8))/(2) erg`
`= (374.28 xx 10^(8))/(2 xx 10^(7))` joule = 1871. 42 joule
Average kinetic energy
`= (K.E//mo1)/(Av.no) = (374.28 xx 10^(8))/(6.023 xx 10^(23))`
`= 62. 14 xx 10^(-15) erg = 62.14 xx 10^(-22)` joule .