I have around 200 barometers (altitude up to 200 m above sea-level) of which some in direct sunlight that can become relatively hot. I am investigating whether there is any relation between measured air pressure and temperature.

Since some barometers also measure temperature, I was able to make the following plot:

enter image description here

Air pressure is measured in cmH2O and temperature in °C.

Based on 0.5-quantile (i.e. median) regression (blue line in the plot) I get the following coefficients

                  Value       Std. Error  t value     Pr(>|t|)   
(Intercept)        1035.69548     0.02500 41423.62580     0.00000
TEMPERATURE_VALUE    -0.06811     0.00115   -59.40118     0.00000

which suggest a drop of 0.068 cmH20 per unit °C of temperature increase.

Note that quantile regression is used to mitigate the effects of outliers caused by broken barometers.

Now my question is whether this result can be backed by theory? Most obvious starting point is the ideal gas law which can be written in the following specific form:

$$ P = \rho \frac{R^*}{M} T $$

where $R^*$ is the universal gas constant and $M$ the molar mass of Earth's air.

Assuming constant $\rho$ and using $T_b$ = 288.15 K, $R^*$ = 8.3144598 J/(mol·K) and $M$ = 0.0289644 kg/mol one gets:

$$ \rho \frac{R^*}{M} (T_b + 1) - \rho \frac{R^*}{M} T_b = 351.65 \;\text{Pa} = 3.59 \; \text{cmH2O} $$

which is clearly wrong (in magnitude and sign).

Furthermore, the confusing bit is the fact that $\rho$ also depends on $T$. Thus, as the air gets hotter, it expands and changes the density. This results in upward convection which in turn causes local air flow increase. Both the density and flow speed impact pressure according to Bernoulli's principle.

Which brings me to my question: what is the theoretical correct way to compute the effect of temperature on air pressure around sea-level?

PS I couldn't find any relevant paper on this topic, so paper suggestions are also welcome.


If I assume that $\rho$ depends on $T$ as well, I end up here. Assuming dry air, the density calculation is again based on the ideal gas law, and as such doesn't bring any new information. If I take this information into consideration, then

$$ \rho (T_b + 1) \frac{R^*}{M} (T_b + 1) - \rho (T_b)\frac{R^*}{M} T_b = 0 \;\text{Pa} $$

meaning that increase in temperature will reduce the density, resulting in zero change in pressure.

If instead of dry air, I use the density formula for humid air (90 %) and Buck equation for saturation vapor pressure of water, I end up with -0.0004 Pa.

  • $\begingroup$ Are your sensors shielded from direct sunlight? If not, the temperature measurements will be skewed as temperature will read possibly much higher when exposed to direct insolation. $\endgroup$ Commented Mar 3, 2020 at 15:16
  • $\begingroup$ @planetmaker, they are not shielded. Most sensors also measure temperature, and looking at that data, there are no real extremes: here is a link. Units are °C. For Belgium, these values seem OK. Part of analysis is to ascertain whether the above effect on air pressure due to temperature can be attributed to barometers being exposed to heat. $\endgroup$ Commented Mar 3, 2020 at 15:29
  • $\begingroup$ Additionally, if you take data from all times, you average-out the effects of annual variation. You might want to consider to bin data by month and do a separate analysis for those groups. Maybe also reduce temperature and pressures by the exepected monthly means to make datasets again comparable. Using monthly bins allows you to work out the influence on this due to differences in insolation $\endgroup$ Commented Mar 3, 2020 at 15:30

1 Answer 1


If you know about some broken barometers, you might possibly remove them from the sampling which is a better way to mitigate their effect.

It might be a good thing to test some barometers. Keep them at constant air pressure and see how their readings change at different temperatures. Make sure that the change in pressure with temperature is not dominated by the effect on the barometers themselves.

How much are your barometers affected by airflow? That is also testable in the lab.

Eyeballing your plot, at first sight it looks unremarkable except that there is a bulge of low pressure at relatively low temperature.

At a second look, if the right number of low-pressure observations were transferred from the 0 to 15 degree area to the -5 to 10 degree area, it might look like there's no effect at all.

So maybe the question to ask is: Why do a fraction of low-temperature/low-pressure observations come out with higher temperatures, when that does not happen with low-temperature/high-pressure observations?

I could speculate that there might be different amounts of sunlight when a cold front comes in than when a warm front comes, but that's only speculation.

  • $\begingroup$ Indeed, the bulge. A partial explanation is the higher variance during not-summer time. See here. During this time, many observations fall in the lower pressure range, while the higher range stays stable during the year. For humidity, see here. It has a similar seasonality pattern, but can not explain the bulge during not-summer. Humidity change from 75 to 90 % would only decrease the pressure by 0.001 cmH2O. $\endgroup$ Commented Mar 3, 2020 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.