If the temperature of argon gas is 400 K and the average velocity of argon is twice that of radon what is the temperature of radon gas?
From KM-Theory => RMS velocity = 158(Sqr-Rt(T/M)) meters/sec
F.Wt(Ar) = 40 g/mol
F.Wt(Rn) ~ 222 g/mol
RMS V(Ar) = (158)[Sqr-Rt(400/40)] m/s
---------> = 500 m/s
RMS V(Rn) =0.5(500) m/s = 250 m/s
Solve RMS Velocity Equation for 'T'
=> (250/158)^2 = (T/222)
=> T = (222)(2.50) = 555K