Entropy of an ideal gas
Consider
is its molar entropy.
Thermodynamic derivation
By the ^Quasistatic and Energy of an ideal gas we have
Δ π = π π π π π + π π π for a quasistatic process with a fixed number of moles. The heat added per mole is then
Δ π = π π π π + π π π£ Applying the Ideal gas law
and dividing by temperature gives π π£ = π π π π = Δ π π = π π π π π + π π π£ π = π ( π π l n β‘ π + π l n β‘ π£ + Λ πΆ ) where
is an undetermined constant. Note the unitful logarithms are dealt with by the constant. Λ πΆ