Question

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?

Answer 1

temp is a measure of the average kinetic energy of molecules.
KE molecule=1/2 mass*v^2
so consider Argon.
400K=k*40*v^2 or v^2=400/40k
now consider Radon
T=k*222*(v/2)^2=k*55.5 v^2

solving for T...
T=k*55.5*400/(k40)=555Kelvins

Answer 2

Another way is through the KM-Theory where 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

Answer 3

Can I type my own questions and you guys will help to answer it?