As to P=ρgh, would the height of the atmosphere be affected by an increase in nitrogen and oxygen mass or would the height remain constant? Thanks!

  • $\begingroup$ This question is ambiguous. As per my answer below, the equation quoted has nothing to do to answer the question. Please provide a clearer statement of the problem that you truly want to be addressed. Define the system. Describe the information that you already have about it. State what answers you think apply and explain what answers still cause you confusion. $\endgroup$ Aug 2 '18 at 18:15
  • $\begingroup$ There are slightly different questions in the title and the body of the question. To a first approximation we can say that air is mostly nitrogen and oxygen, so effectively you're asking "if I double the amount of air, what will happen to the depth and pressure of the atmosphere?". People have effectively answered this below. $\endgroup$ Aug 3 '18 at 12:34
  • $\begingroup$ @SemidiurnalSimon ... Duly noted. The ambiguity would be lifted by deciding whether the question is about pressure change (as per the title) or about height change (as per the contents). $\endgroup$ Aug 3 '18 at 16:42
  • $\begingroup$ True, but those things are closely linked... $\endgroup$ Aug 3 '18 at 22:18
  • $\begingroup$ The phrase "closely linked" is itself ambiguous. $\endgroup$ Aug 7 '18 at 13:36

The starting question is phrased as such ...

As to $P=\rho gh$, would the height of the atmosphere be affected by an increase in nitrogen and oxygen mass or would the height remain constant? Thanks!

Density is mass per volume.

You can only add mass to a system in one of two ways. One way is to keep density constant. In this case, volume must increase. Since volume is area times height, either area or height must increase. On the earth, the air on the surface cannot expand itself laterally. Area is constant. Therefore, we learn

—> ANSWER 1: As mass is added to air on the earth under the condition of constant density, the height of the air above the earth will increase.

Another way to add mass is to allow density to change. The intuitive first response is that, when we add mass, we expect density will increase. At the limit, this is the same as keeping volume constant. For this case, we learn

—> ANSWER 2: As mass is added to the air on the earth under the condition where density increases, the limiting case is one where the height of the air above the earth will not change.

Finally, we could argue that we want to allow volume to expand somewhat as mass is added. The result is an answer between the two above.

Note that air is compressible. Its density will change as a function of pressure and temperature. By the ideal gas law, $\rho = M n / V = M p / R T$ where molar mass, moles, volume, pressure, gas law constant, and temperature are used. The overall effect is that density is not a constant for the system of air above the earth.

Any other answer requires that we have clearer specifications of the system. We correspondingly learn, either the equation quoted is of no consequence to the question or the question is not phrased properly in respect to the equation quoted.

  • $\begingroup$ To better understand this relationship I think I should study scale height for planetary atmospheres. Thank you. $\endgroup$ Aug 2 '18 at 15:34
  • $\begingroup$ To better understand the relationship I would hope that you would first define the problem statement better. As it is, the question is rather vague and fraught with puzzles. Hence my answer with caveats noted. $\endgroup$ Aug 2 '18 at 18:05
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    $\begingroup$ I added a comment that air is compressible. $\endgroup$ Aug 2 '18 at 18:05

The surface pressure is just the weight of the atmosphere per unit area, so yes.

The density decreases with altitude so there is no well defined "top" to it. Assuming constant temperature (Isothermal-barotropic approximation) the density decreases exponentially with altitude.

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If it were doubled there would the same portion of mass above any given altitude (but more mass above any given altitude).

  • $\begingroup$ If a molecule is higher up, the weight is less, so I wonder what the calculations are. And I suspect that's what the OP was asking. $\endgroup$
    – Spencer
    Jul 29 '18 at 13:48
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    $\begingroup$ @Spencer Half of the atmosphere is below $5.5$ km. That is less than $1/1000$ of the Earth's radius, so the difference in weight is very small. $\endgroup$ Jul 29 '18 at 15:59
  • $\begingroup$ @Spencer: I would encourage the OP to clarify what is really being asked. $\endgroup$ Aug 2 '18 at 18:11

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