### 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 in altitude. But the pressure at the top of the column is related to the weight of the air above the column, which is constant, and so the upper pressure surface is higher in altitude. If the temperature of the column is lower, then the pressure surface at the top of the column will be lower in altitude.

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

$\frac{\text{d}p}{dz}=-\rho g=-\frac{pg}{{R}_{d}T}$

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:

$dz=-\frac{dp}{p}\frac{{R}_{d}}{g}T$

Integrate both sides:

$\underset{{z}_{1}}{\overset{{z}_{2}}{{\displaystyle \int}}}dz=-\underset{{p}_{1}}{\overset{{p}_{2}}{{\displaystyle \int}}}\frac{dp}{p}\frac{{R}_{d}}{g}T$

or

${z}_{2}-{z}_{1}=\frac{{R}_{d}}{g}\text{ln}\left(\frac{{p}_{1}}{{p}_{2}}\right)\overline{T}$

where *T* is the average temperature of the layer between *p _{1}* and

*p*. So, the thickness is actually a measure of the average temperature in the layer.

_{2}#### To Learn More

As some of you already know, you can use the thickness between different pressure surfaces to estimate the type of precipitation that will fall - snow, rain, or a mixture. You can check out these resources for some more information and example problems:

#### Check Your Understanding

Suppose that the 500 mb surface is at 560 dam (decameters, 10s of meters) and the 1000 mb surface is at 0 dam. What is the average temperature of the layer between 1000 mb and 500 mb?

**Click for answer.**

Rearrange equation 2.4 to get an expression in terms of the average temperature and then put all the numbers into the equation to find the average temperature of the layer. Make sure that all the units are correct.

$$\overline{\text{T}}=\frac{10*\left({z}_{2}-{z}_{1}\right)}{\left(\raisebox{1ex}{$287$}\!\left/ \!\raisebox{-1ex}{$9.8$}\right.\right)\mathrm{ln}\left(\raisebox{1ex}{${p}_{1}$}\!\left/ \!\raisebox{-1ex}{${p}_{2}$}\right.\right)}=\frac{10*\left(560-0\right)}{\left(\raisebox{1ex}{$287$}\!\left/ \!\raisebox{-1ex}{$9.8$}\right.\right)\mathrm{ln}\left(\raisebox{1ex}{$1000$}\!\left/ \!\raisebox{-1ex}{$500$}\right.\right)}=276\text{\hspace{0.17em}}K$$