The average kinetic energy (KE) per molecule of a gas is $\frac{3}{2}kT$. While finding this we do

$$ \text{ Average KE} =\frac{1}{2} M \frac{1}{N}\sum v^2=\frac{3}{2}kT$$ But why do we not add rotational kinetic energy here?


The expression you quote is for a ideal monatomic gas, and we get $C_v = 3/2$ for the three degrees of freedom. For ideal diatomic gases we do indeed have to count rotational degrees of freedom and we get $C_v = 5/2$. See the Wikipedia article on ideal gases for more info.


For any Ideal Gas:-

  • PV = 2/3E
  • PV = (γ-1)U

where U = E + R (Internal energy),

E = 1/2mvᵣₘₛ² (avg. translational Kinetic energy),

& R is rotational Kinetic energy.

For eg: A diatomic gas with γ = 7/5 has PV = 2/3E & PV = (7/5 - 1)U.

Therefore, 2/3E = 2/5U

OR E = 3/5U

CONCLUSION #1: Translational kinetic energy of a diatomic gas makes up 3/5th of it's internal energy.

From equation of state, PV = NKᵦT

Therefore, NKᵦT = 2/3E

OR E = 3/2NKᵦT

CONCLUSION #2: Avg. Translational Kinteic Energy Per Molecule (i.e. E/N = 3/2KᵦT) is related only to Temperature & is independent of pressure, volume or nature of gas.

FINAL CONCLUSION: Temperature depends on Avg. TRANSLATIONAL kinetic energy in ideal cases, whether mono, di or poly-atomic.


  • $\begingroup$ I don't think I've ever seen someone use $\beta$ as a replacement for the subscript $B$ in Boltzmann's constant. $\endgroup$
    – Kyle Kanos
    Mar 15 '19 at 20:02

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