Thermal conductivity and kinetic theory

Consider three horizontal planes in the gas each of area A. The heat conducted downwards through A per second is then -kAdθ/dx

However, each second a mass of gas m at a temperature θ₁ crosses A moving downwards and a mass of gas m at a temperature θ₂ crosses A moving upwards.

Now: m = ρcA/6
θ₁ = θ + λdθ/dz and θ₂ = θ + λdθ/dz

Therefore, since heat = mcθ , the net transfer of heat downwards is:
–[ρcA/6]λ[2dθ/dz]C_v.

But this must equal –kAdθ/dz and therefore:

k = 1/3ρcλC_v and since η = 1/3[ρcλ]   k = ηC_v