What is the average velocity of atoms in 1.00 mol argon at 275k? For m use 0.0399 kg

Answer;

414.0 m/s

Explanation;

From kinetic theory for an ideal gas ...

Mean KE of a particle; ½mv² = 3kT/2

For 1 mol of particles (L); Total KE = L½mv² = 3RT/2

(Lm) ½v² = 3RT/2

Lm = molar mass 0.0399kg, R = molar gas constant 8.30 J / (K. mol),

T = Kelvin temp 275K

v² = 3RT / Lm

v² = 3 x 8.30 x 275 / 0.0399

= 1.716^5

v (rms) = √1.716^5

v (rms) = 414.0 m/s

414.0 m/s

Explanation:

A p E x