 Solution

ID: past-exam-of-the-theoretical-mechanics-course-of-the-university-of-oxford/2010/b/4/a/ii/solution

Continuity gives

D lo g ρ / D t = - \nabla \cdot u

. Using

p = ρRT

D lo g T / D t = - (R / c_{v}) \nabla \cdot u

. Therefore

\frac{D l o g p}{D t} = \frac{D l o g ρ}{D t} + \frac{D l o g T}{D t} = - γ \nabla \cdot u, γ = 1 + R / c_{v} .

Consequently

\frac{D S}{D t} = c_{v} (\frac{D lo g p}{D t} - γ \frac{D lo g ρ}{D t}) = 0.

Each smooth parcel evolves isentropically under these adiabatic inviscid equations; the argument does not cover entropy-producing shocks.

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