Five grams of helium having rms speed of molecules `1000 m//s` and `24 g` of oxygen having rms speed of `1000 m//s` are introduced into a thermally isolated vessel. Find the rms speeds of helium and oxygen individually when thermal equilibrium is attained. Neglect the heat capacity of vessel.

Explanation

For helium, `V = sqrt((3 RT)/(M))`
`1000 = sqrt((3RT_(1))/(4 xx 10^(-3))) implies 3 RT_(1) = 4000`
For oxygen,
`1000 = sqrt((3RT_(2))/(32 xx 10^(-3))) implies 3 RT_(2) = 32000`
By internal energy conservation `U_(1) + U_(2) = U_(1)^(1) + U_(2)^(1)`
As we know `U = (f)/(2) nRT ,`
`(3)/(4) xx (5)/(4) RT_(1) + (5)/(2) xx (24)/(32) RT_(2) = ((3)/(2) xx (5)/(4) + (5)/(2) xx (24)/(32)) RT`
`T = (T_(1) + T_(2))/(2)`
After mixing rms speed of helium,
`v_(1)^(1) = sqrt((3 RT)/(rms)) = sqrt((3 R)/(2)(T_(1) + T_(2))/(mw))`
`v_(1)^(1) = 15000 sqrt2 m//s`
For oxygen,
`v_(2)^(1) = sqrt((3 RT (T_(1) + T_(2)))/(2 mw)) = sqrt((4000 + 32000)/(2 xx 32 xx 10^(-3)))`
`v_(1)^(2) = 750 m//s`