How to calculate rate of change of temperature with pressure in Earth's troposhere?

Question

I was given a problem in a competitive exam from previous year:

Calculate the rate of change of temperature of the Earth's troposphere with pressure ($\frac{dT}{dP}$), given

• Scale height, $H$, where $P(\text{H}) = (\frac{1}{e}\cdot P_0) = 8\text{ km}$ (the geopotential height where the pressure equals to $\frac{1}{e}$ of the initial one)
• Lapse rate $= \gamma= 6.5 \frac{°C}{km}$ (change of temperature with height)

I tried using potential temperature equation and pressure profile equation, but I am not able to connect it.

Please help me or just give an advice. Thank you!

Answer 1

Wow, @John came with a simple answer.

Our goal is to find

$$\frac{dT}{dP}=\frac{dT}{dz}\cdot\frac{dz}{dP}$$

By the equation for pressure lapse: $$P=P_0\cdot e^{-\frac{z}{H}}$$ where $H=8 km=8000 m$ (scale height) and $z$ is the height above the surface.

We differentiate it once, to get: $$\frac{dP}{dz}=\frac{P_0}{H}e^{-\frac{z}{H}}$$

The inverse value is $$\frac{dz}{dP}=\frac{H}{P_0\cdot e^{-\frac{z}{H}}}$$

And what is $\frac{dT}{dz}$? It is simply the lapse rate $\gamma$.

Using that we restate the above:

$$\frac{dT}{dP}=\frac{dT}{dz}\cdot\frac{dz}{dP}=\frac{\gamma H}{P_0\cdot e^{-\frac{z}{H}}}$$ But not we have that misterious $z$ variable which we don't want. So, $$P=P_0 e^{-\frac{z}{H}}$$ and $$z=-H\ln\frac{P}{P_0}$$ We insert that in the $\frac{dT}{dP}$ equation and we get $$\frac{dT}{dP}=\frac{\gamma H}{P}$$ where $H=8000\rm\,m$, $\gamma=6.5\rm\,\frac{K}{km}$ and $P=100\rm\,kPa$. So, $$\frac{dT}{dP}=0.052\rm\,\frac{K}{hPa}$$ And that is (I believe) close enough.