By the principle of hydrostatic balance, the atmospheric pressure $P$ at surface level is such that a column of air with cross-sectional area $A$ and mass $m$ satisfies


If the temperature of the column of air increases, I expect the pressure in general to increase, as the molecules move faster on average and so the collisions impart more force. In particular, the surface pressure increases. However, there is no change in the mass. Does this imply that the hydrostatic balance is not longer achieved and the entire colunn of air is supposed to rise indefinitely?

On the other hand, I also read (for example, articles on how sea/land breeze is generated) that the surface pressure for a warm column of air is lower (increase of higher pressure by molecular consideration). Which is actually correct?

  • $\begingroup$ The pressure doesn't increase. The specific volume increases. $\endgroup$ Mar 1, 2020 at 13:33

1 Answer 1


The atmospheric pressure does not change, when increasing the temperature. Instead the surface density decreases. The hydrostatic law is valid at all times, with only one total mass and one surface pressure.

Longer answer:
I think this is an interesting question, which is a nice example how some problems cannot be resolved with arguments, only with math.
For this, I did a quick re-derivation of your $PA=mg$, just not for a fluid column, but for the planet as a whole. We start with the hydrostatic law $$\frac{\partial P}{\partial r}=-\rho(r) g, \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1) $$ where we need to take the ideal gas law as closure relation $$P=\frac{\rho k_B T}{\mu} , \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)$$ with the mean molecular weight $\mu$, density $\rho$ and Temperature $T$. With $(2)$ and assuming an isothermal atmosphere i.e. $T(r) =T$ everywhere, now $(1)$ has the solution $$\rho(r)=\rho_0 \, \exp(-r/H),\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(3) $$ where $$H=\frac{k_B T}{\mu g}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(4)$$ is called the scale-height of the atmosphere. $H$ is a tremendously useful quantity, because on one hand it is a measure of the temperature of the atmosphere (large H means large T) but because of $(3)$ it also gives us the physical extent of the atmosphere (large H means very extended, but thin atmosphere).

Now that we have the density profile, we can go on to construct a relation between the total mass of the atmosphere and the scale-height, which will inform us what will change under a temperature change.
So next, we compute $$ M = \int dm = 4\pi\int^{\infty}_{R_{\oplus}} dr\, r^2 \rho(r) \approx 4\pi \rho_0 H R^2_{\oplus}\, , \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(5)$$

where $\rho_0$ is the atmospheric density on the surface, and $R_{\oplus}$ is the radius of the planet. Now $(5)$ is a result that we can work with. Assuming that the total Mass M doesn't change (how would it?) we can sort the constant and non-constant terms left and right: $$ \frac{M}{4\pi R^2_{\oplus}} = const. = H \rho_0, \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(6)$$ from which we can see, that when the temperature changes, the surface density must adjust accordingly. So this already looks like the surface density will adjust itself as to compensate the temperature increase and keep the pressure constant. But we can calculate what happens to the pressure from $(4)$ and then $(2)$: $$ H\rho_0 \stackrel{(4)}{=} \frac{k_B T}{\mu g} \rho_0 \stackrel{(2)}{=} \frac{P_0}{g}, \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(7)$$ so we see that in fact, the surface pressure remains constant, when changing the atmospheric temperature.

Your other question concerns a dynamic process that involves horizontal transport of mass, which is not captured by a simplistic model for the vertical balance, such as $(1)$. As stated before, still the vertical balance must hold largely over the planet. We have $\sim$ 1000 $hPa$ pressure balancing the gravitational pull of the atmospheric mass vertically. Horizontal motions of air, up to the largest hurricanes we can experience on this planet have horizontal pressure perturbations of 10-30 $hPa$. So even the strongest horizontal disturbances can only upset the vertical hydrostatic equilibrium by a few percent.


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