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?
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
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
Can I type my own questions and you guys will help to answer it?
To solve this question, we'll use the concept of the average kinetic energy of gas particles, which is directly related to their temperature. The equation for average kinetic energy is given by:
KE = (3/2) kT
Where KE represents the average kinetic energy, k is the Boltzmann constant, and T is the temperature.
Now, since we're comparing the average velocities of argon and radon gases, we can use the relationship between the average kinetic energy and velocity:
KE = (1/2) m v^2
Where m is the mass of an individual gas particle, and v is the average velocity.
Given that the temperature of argon gas is 400 K, we'll calculate the average velocity of argon using its temperature:
KE_argon = (3/2) k T_argon
We'll set the velocity of radon as v_radon and since it's twice that of argon, we have:
v_radon = 2 v_argon
Now we can compare the average kinetic energies of argon and radon:
(1/2) m_argon v_argon^2 = (1/2) m_radon v_radon^2
Substituting the relationship v_radon = 2 v_argon, we get:
(1/2) m_argon v_argon^2 = (1/2) m_radon (2 v_argon)^2
(1/2) m_argon v_argon^2 = 4 m_radon v_argon^2
m_argon = 4 m_radon
Since the masses of argon and radon gas particles depend on their atomic masses, the atomic mass of argon (Ar) is around 40 g/mol, and the atomic mass of radon (Rn) is around 222 g/mol.
Using the equation m_argon = 4 m_radon, we have:
40 g/mol = 4 (222 g/mol)
40 g/mol = 888 g/mol
This equation is not true, which means our initial assumption that the average velocity of radon is twice that of argon is incorrect.
Therefore, we cannot determine the temperature of radon gas based on the given information.