Etotal = Etrans = RT² ( dlnq trans)v .....(1)
dT
We know that, qtrans= (2πmkT)³/₂ v .....(2)
h³
Take log on both sides of equation (2)
lnq trans= 3/2ln (2πmk) + 3/2 lnT + lnV -
3lnh.............. (3)
(∂lnq trans)v = 0 + 3/2 (1/T) + 0−0 = 3
∂T 2T
Substitute the value of the above differential in equation (1)
Etotal = Etrans = RT² 3 = 3/2 RT .......(4)
2T
This is the internal of a monoatomic gas depends only on temperature.
Heat capacity of monoatomic gas at constant volume is,
∈v = ( ∂E ) v = 3 R
∂T 2
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