As I was reading my textbook, I noted that it stated that the atmosphere becomes less dense as height increases. I am unable to understand this. I understand that pressure decreases as height increases, but shouldn't the density remain constant?

  • $\begingroup$ Do you know the interpretation of temperature in terms of probability of finding a particle in low energy states? Low temperature means almost surely in the lowest energy state $\endgroup$ – AHusain Oct 6 '18 at 23:07
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    $\begingroup$ I'm sorry, I don't understand how that relates to the question (I think I'm misreading your comment btw) $\endgroup$ – Dude156 Oct 6 '18 at 23:09
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    $\begingroup$ What does the ideal gas law predict about how the pressure affects the density? $\endgroup$ – Chet Miller Oct 6 '18 at 23:22
  • $\begingroup$ Consider a single molecule. If you assume a uniform temperature, then low to the ground is lower energy so higher probability of being there. Do this for all the molecules and you get a statement about density. But temperature is not uniform, but the qualitative result remains. $\endgroup$ – AHusain Oct 6 '18 at 23:26
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    $\begingroup$ What I am describing is how you would derive the ideal gas law from first principles instead of taking it as simply empirical. $\endgroup$ – AHusain Oct 6 '18 at 23:28

The barometric formula predicts the atmospheric density with Newtonian gravity and fluid dynamics.

So basically you need three things:

  1. gravity

  2. pressure

  3. temperature

You are asking the right question, because for example water does not behave like that, as you go deeper in the ocean, the density of the water only slightly increases. Whereas in the atmosphere, this effect, that the density decreases as you go higher is more dominant.

Once you figure out what the difference is between the ocean and the atmosphere, and why one has a big density change with altitude and why the other does not, you will understand why the atmosphere is less dense at high altitude.

The incompressibility of fluids is caused by a rather small distance between water molecules, and when you are trying to compress them, they react with a rather big force (Van der Vaals and EM) that's not letting them closer.

This is not the case with the atmosphere. This gas is rather compressible, since the molecules are farther away, and they are not in constant contact with each other.

So now you would think it is only gravity that causes higher density closer to Earth. But the reality is that gravity (and curvature) directly only plays a little role here, because the difference in the strength of gravity almost the same at the surface as at the top of the atmosphere.

So why is the atmosphere less dense at the top? To understand this, imagine compression springs instead of gas molecules. If you look at the molecules at the top, they are not really compressed. As you go lower closer to the surface, those springs are really compressed. Why? Because of pressure from the other springs weight from above.

So although gravity does not significantly change with the altitude inside the atmosphere, the compression of the springs will linearly change, become more compressed as you go down. It is pressure.

Same thing happens with water in the ocean. The deeper you go, the more pressure. But why does water not get really denser? Because it is rather incompressible. The atmosphere is rather compressible and that is why it gets denser at the bottom.

See this question of mine:

Ocean density vs atmosphere density


From the ideal gas law, the density of a gas is described by the equation: $$\rho=\frac{PM}{RT}$$where $\rho$ is the density, P is the pressure, M is the molar mass, R is the universal gas constant, and T is the absolute temperature. So, at constant temperature, the density varies in direct proportion to the pressure. In the lower atmoephere, the temperature decreases somewhat with elevation, but the pressure variation easily wins out.


As Chester Miller said in the first comment you have that the Law of Ideal Gases holds true, hence you can write (where $R_d$ is the gas constant for dry air) \begin{equation} \rho=\frac{p}{R_dT} \label{eq:idealgas} \end{equation} But since it's the Atmosphere what we're dealing with, it's also true that the Barometric Equation holds, which states (verify it yourself) \begin{equation} p(z)=p_0e^{-\frac{z-z_0}{H}} \end{equation} Where with the subscript $\square_0$ I indicate the starting height and pressure, and with $H$ I indicated the $e$-folding height $$H=\frac{R_dT_0}{g_0}$$. Plugging it inside the Ideal Gas Law we get \begin{equation} \rho(z)=\frac{p_0}{R_dT}e^{-\frac{z-z_0}{H}} \end{equation} Which, without much doubt, is a monotonically decreasing function of height, q.e.d. $\square$


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