Consider an ideal gas with following distribution of spedds.

(a) Calculate `V_("rms")` and henceT. `(m=3.0xx 10^(-26) kg)`
(b) If all the molecules with speed `1000m//s` escape from the system, calculate new `V_("rms")` and hence T.

Explanation

(a) We know that `V_("rms")^(2) = (underset(i)(Sigma)n_(i)V_(i)^(2))/(Sigman_(i))`
This is the rms speed for all molecules collectively.
Now , `V_("rms") = ((underset(i)(Sigma)n_(i)v_(i)^(2))/(Sigman_(i)))^(1/2)`
`=sqrt((n_(1)v_(1)^(2) +n_(2)v_(2)^(2)+n_(3)v_(3)^(2)+.........+n_(n)v_(n)^(2))/(n_(1)+n_(2) +n_(3)+............+n_(n)))`
`=sqrt((n_(1)v_(1)^(2)+n_(2)v_(2)^(2)+n_(3)v_(3)^(2)+n_(4)v_(4)^(2)+n_(5)v_(5)^(2))/(n_(1)+n_(2)+n_(3)+n_(4)+n_(5)))`
`=sqrt((10xx(200)^(2) +20xx(400)^(2)+40xx(600)^(2)+20xx(800)^(2)+10xx(1000)^(2))/(100))`
`=sqrt((1000xx(4+32+144 +128 +100)))`
`=sqrt(408 xx 1000) ~~ 639 m//s.`
Now, according to kinetic theory of gasses
`1/2 mv_("rms")^(2) =3/2 kg T" "[underset( m="mass of gaseous molecules")(k_(B) ="Boltzmann constant"]]`
`T =1/3(mv_("rms")^(2))/(K_(B)) =1/3 xx (30xx10^(-26) xx 4.08 xx 10^(5))/(1.38 xx 10^(-23))`
`=2.96 xx 10^(2) K =296 K`
(b) If all the molecules with speed 1000`m//s` escape ,then
`V_("rms")^(2) =(10xx (200)^(2) +20 xx(400)^(2) +40xx(600)^(2)+20 xx(800)^(2))/(90)`
`=(10 xx100^(2) xx (1xx4 +2xx16 +4xx36 +2 xx 64))/(90)`
`=10000 xx (308)/(9) =342 xx 1000 m^(2)//s^(2)`
`V_("rms") =584 m//s`
`"Again"" "T=1/3 (mv_("rms")^(2))/(K)`
`=1/3 xx (3xx10^(-26) xx 3.42 xx 10^(5))/(1.38 xx 10^(-23))`
`=2.478 xx 10^(2)`
`=247 .8 -248 K`
Note After escaping of molecules with speed of 1000 `m//s` the temperature in part (b) is 248 K whereas in part (a) before escaping of molecules the temperature was 297 K. thus evaprations facilitates cooling.