Physics: Kinetic Theory

Question

Kinetic theory

a. What is the average kinetic energy of a monatomic gas at a temperature of 354 K?

b. If this is Ne gas (neon has 20 nucleons) what is the root-mean-squared speed?

c. What temperature would be required to triple the speed of these particles?

Answer 1

By Kinetic Theory Of Gases, Pressure Times Volume Equals (2/3)(Total Number Of Particles Present) Times (0.5m×(v²)_bar) Which In Turn Equals (Total Number Of Particles Present) Times Boltzmann's Constant Times Kelvin Temperature. That is, PV = (2/3)N × (0.5m(v²)_bar) = NkT.

Reduce (2/3)N × (0.5m(v²)_bar) = NkT to T = (2/3k)×(0.5m(v²)_bar).

Atomic Mass or M for Neon equals 20.18E-3 Kilograms Per Mole.

Avogadro's Number N_A equals 6.022136736E23 Particles Per Mole.

Boltzmann's Constant k equals 1.38065812E-23 Joules Per Kelvin.

Also, m equals Atomic Mass/Avogadro's Number, equal here to 20.18E-3 Kilograms Per Mole/6.022136736E23 Atoms Per Mole or 3.350970077E-26 Kilograms Per Atom.

The Average Translational Kinetic Energy sought is equal to (0.5m(v²)_bar).

For T = 354 Kelvin, write 354 = (2/3/1.38065812E-23)(0.5m×(v²)_bar) which will equate

(0.5m(v²)_bar) to 7.331294617E-21 Joules.

Root-Mean-Square Speed or v_rms is obtained from T = (2/3k)×(0.5m(v²)_bar), rewritten as

(3kT/m) = (v²)_bar. Then √(v²)_bar = √[(3×1.38065812E-23×354) ÷ (20.18E-3/6.022136736E23)] which reduces to 661.4850894 Meters Per Second.

To triple the speed of the particles or atoms, equate 3(661.4850894) to √[(3kT)/(M/N_A)] or 3(661.4850894) =

√[(3×1.38065812E-23×T) ÷ (20.18E-3/6.022136736E23)] which gives the value of T sought as 3186 Kelvin.