How to calculate temperatures on my world given different pressures and altitudes

Question

The fictional world is Earth-like with 1g gravity and similar solar radiation to Earth but the inhabitants only live on very high plateaus at around 5000m where the pressure is already 3bar. How do I calculate the temperature at sea level with high humidity levels? And what features could I change or add to maximize this temperature whilst maintaining 20-30 degrees C at 5000m?

Answer 1

The spherical cow in a vacuum approach: 70-80°C

Ok, here we will use simplifying assumptions. Big ones:

• The air is an ideal gas and g is constant with altitude (those ones are fine)
• The atmospheric system is globally adiabatic (yeah, ok, mayyybe)
• The air is dry (I'm starting to sweat a bit)

Then we pull out the maths cause that's what we're here for. (what? you're here for an answer? Haha. I like your humor).

We will use the foundational relation of hydrostatics:

$$dP=-\rho(z)gdz$$

We combine it with the well known ideal gas law $PV=nRT$, a.k.a $P=\frac{\rho}{M}RT$, with $M$ the molar mass, and we get an equation that we will call equation (1):

$$\frac{dP}{dz}=-\frac{gM}{R}\frac{P}{T} \text{ (1)}$$

Cool, cool. Just one more equation to have a well-defined physical system. An adiabatic reversible process satisfies:

$$P^{1-\gamma}T^{\gamma}=constant$$

If you take the derivative of that thing with repsect to $z$, and you re-arrange everything to make it look fabulous, you get equation (2):

$$\frac{dP}{dz}=\frac{\gamma}{\gamma-1}\frac{P}{T}\frac{dT}{dz} \text{ (2)}$$

Equations (1) and (2) both tell us about the rate of change of pressure with altitude, so the right hand parts are equal:

$$-\frac{gM}{R}\frac{P}{T} = \frac{\gamma}{\gamma-1}\frac{P}{T}\frac{dT}{dz}$$ $$\frac{dT}{dz}=\frac{1-\gamma}{\gamma}\frac{gM}{R}$$

We demonstrate the remarkable result that, under our assumptions, the rate of change of temperature with altitude is constant and depends only on the nature of the gas ($M,\gamma$) and the gravity of your planet ($g$). So if your atmosphere is made of the same stuff, temperature falls with altitude exactly at the same rate as on earth!. You can calculate this rate to be $9.8°C/km$ from the above equation

So, quite simply, 5km lower, your atmoshpere should be about 50°C hotter, giving roughly temperatures of 70-80°C (which I will certainly not give in Farenheits cause y'all gotta start using units that make sense).

Note that your pressure specifications were not needed to get that result. They will just make pressure higher at sea-level.

For a more realistic approach:

I'd rather shoot myself.

Simply taking into account the moisture in the air already makes things a lot more complicated. Water content varies with altitude and locations. It reaches saturation, condenses in an exothermic transition, creates clouds which eventually will change everything anyway. With coarse approximations, it is still manageable by hand but definitely nastier. And there is no point anyway. There are so many other effects at play (winds, water currents, land topology, shading, geothermic activity, position on the globe, precipitations, etc...) that you can't reasonably hope to get an answer that is anything else than a very rough ballpark estimation (which is what I did above).

Just think that you are asking for a planetary answer on a fictional ill-defined planet, while our own planet has temperatures ranging from -40 to +50 at any given time at sea level, depending on the location. And we can't even reliably predict the weather of earth a week from now with all the data and the years of intimate experience with this planet. What you are asking is 100% currently impossible, especially since you give close to no relevant meteorological information.

Note that on earth, the experimental rate of change of temperature is closer to 7°C/km below 11km of altitude (hey, we were pretty close!), suggesting lower sea-level temperatures of about 55-65°C in your case. And those orders of magnitude are pretty much the best you can hope to get. They just should be plausible enough for a story.

If you are very attached to including the humidity in your approach, you could use different corrective terms for different humidity regimes. However, I keep thinking that it is not very significant to focus on that specific aspect while leaving aside all the other factors which could probably realistically offset the answer by dozens of degrees C. For the purposes of a story, I think no one would bat an eye whether you tell them your lapse rate is 4°C/km or 12°C/km. Both are scientifically plausible enough

The factors you can play with

Looking back at the parameters in our equation above, if you want to increase the temperature at sea level, you could:

• have a higher gravity
• have an atmopshere with a higher molar mass (take bigger molecules)
• Humidity is mostly going to decrease the rate (it is the main reason for the experimental lapse rate being 30% lower than the one we calculated). and also make it harder to heat the sea level (because you spend energy vaporizing water rather than heating the air). It has an adverse effect. If you want a big temperature difference, keep humidity low