Solution

Heat flow is defined by

$$J =\frac{\Delta E}{A\cdot t}$$

We know that the change of the energy is given by

$$\Delta E = \frac{nR\Delta T}{\gamma-1}$$

Where $n$ is the number of moles

We know that Heat flow then

$$J = \frac{nR\Delta T}{(\gamma-1)A\cdot t}$$

We know that the time is given by because (the free path of atoms is much longer than $\delta$) furthermore $\lambda \ll \delta$ therefore the average collision between the molecules will be $\delta$:

$$t = \frac{\delta}{v}$$

Likewise:

$$J = \frac{nR\Delta Tv}{(\gamma-1)A\cdot\delta}$$

Now we have to:

$$A\cdot \delta=V$$

So from the previous equations we have to:

$$J = \frac{nRΔTv}{(\gamma-1)V}$$

Another way to write this equation is:

$$J = \frac{n'K\Delta Tv}{(\gamma-1)}$$

Where $n'$ is the number of the atoms per unit volume:

$$v=\sqrt{\frac{3R(T+\Delta T)}{\mu}}$$

How $\Delta T \ll T$ is much less than this

$$v\approx\sqrt{\frac{3RT}{\mu}}$$

Introducing this velocity into the equation for the flow we have to

$$J = \frac{n'KΔT}{(\gamma-1)}\sqrt{\frac{3RT}{\mu}}$$