# 2.4: The higher the temperature, the thicker the layer

## A surprising way to relate the distance between two pressure surfaces to the temperature of the layer between them.

Consider a column of air between two pressure surfaces. If the mass in the column is conserved, then the column with the greater average temperature will be less dense and occupy more volume and thus be higher. But the pressure is related to the weight of the air above the column and so the upper pressure surface rises. If the temperature of the column is lower, then the pressure surface at the top of the column will be lower.

We can look at this behavior from the point-of-view of hydrostatic equilibrium.

$\frac{\mathrm{d} p}{d z}=-\rho g=-\frac{p g}{R_{d} T}$

If the temperature is greater, then the change in p with height is less, which means that any given pressure surface is going to be higher.

The difference between any two pressure surfaces is called the thickness. We can show that the thickness depends only on temperature:

$d z=-\frac{d p}{p} \frac{R_{d}}{g} T$

Integrate both sides:

$\int_{z_{1}}^{z_{2}} d z=-\int_{p_{1}}^{p_{2}} \frac{d p}{p} \frac{R_{d}}{g} T$

or

$z_{2}-z_{1}=\frac{R_{d}}{g} \ln \left(\frac{p_{1}}{p_{2}}\right) \overline{T}$

where T is the average temperature of the layer between p1 and p2. So, the thickness is actually a measure of the average temperature in the layer.

$$\overline{\mathrm{T}}=\frac{10^{*}\left(z_{2}-z_{1}\right)}{(287 / 9.8) \ln \left(p_{1} / p_{2}\right)}=\frac{10^{*}(560-0)}{(287 / 9.8) \ln (1000 / 500)}=276 K$$