Brilliant's question on air pressure

Question

I have always thought that as temperature increases the pressure does as well in an ideal gas, of course I know the air in the atmosphere isn't exactly an ideal gas but I would've thought that the correlation would work as well but I found a question about the pressure of warm and cooler air of the pockets of air in the atmosphere...

And given the explanation at the end...

But I quite can't understand the difference, one refers to the inside pressure and the other refers to the outside pressure of the pockets of air, I've quite never heard of this, may I ask of a clarification?

Answer 1

This has nothing to do with some difference between "inside" and "outside" pressure. That is not a thing. It is of course not true that hot air has a low "inside" pressure but a high "outside" pressure. And I don't think Brilliant's answer was intended to suggest that these were different concepts. But, to the extent that it did, it's misleading.

It is easy to explain why you can't just look at the ideal gas law and say "high pressure implies higher temperature". If the air were an ideal gas (and it is probably close enough for this purpose), we'd have

$$P = \frac{nRT}{V}.$$

Note that $n/V$ is proportional to the density of the gas. So if the density of a gas is constant, $P$ is proportional to $T$. But if it is not constant, that relationship does not hold. And there is no reason for the density to be constant in the atmosphere. So that line of reasoning is not helpful.

However, Brilliant's explanation is also unhelpful. It says that the pressure is determined by the amount of atmosphere overhead. It is of course true that the pressure of a static air column is exactly enough to support its weight. However, this does not explain why hot systems would have lower pressure than cold systems. After all, why shouldn't hot systems have the same amount of air in the air column as cold systems? Even if hot air is less dense, the air column could simply be taller -- there is no ceiling at the top of the atmosphere. In fact, the air column is taller in hot regions. For example, the 500 mb altitude is more than 1000 m higher at the equator than it is at the poles.

The reality is that atmospheric pressure is far more complex than "hot low, cold high" (or the reverse), and such a rule of thumb is much more wrong than it is right.

Answer 2

If you heat air in a closed container, the pressure will increase and if you heat it enough the container will eventually explode. If instead, you place a open topped cylinder full of air on a cooker stove, the air is free to expand so there is almost no increase in pressure in the open topped cylinder. The heated air in the cylinder expands and some of leaves the top of the cylinder. There is less air by mass left inside the cylinder. Therefore the density of the hot air left inside the open cylinder has decreased. Pressure on a surface is proportional to the weight of whatever is pressing down on it divided by the surface area of the surface. if the density has decreased , the weight also decreases. We can write the equation for the pressure acting on the surface as $mg/A$ where m is the mass of the air , g is the acceleration of gravity and A is the area of the surface. We can also write the equation as $(d\times V)/A$ where d is the density of the air and V is volume of the cylinder. Since the density has decreased and the volume and base area of the open cylinder has not changed, the pressure on the base area of the cylinder must reduce.

Answer 3

If the volume $V$ of an ideal gas is fixed then, yes, its pressure increases as its temperature increases because for fixed $V$ the ideal gas law

$pV = nk_B T$

tells us that

$p \propto T$

But in the atmosphere $V$ is not fixed. In addition, the Earth's atmosphere is far from being an ideal gas, due to the presence of water vapour. The link between pressure and temperature in the Earth's atmosphere is not as simple as the course suggests - according to Wikipedia:

High pressure systems in the temperate latitudes generally bring warm weather in spring and summer, when the amount of heat received from the Sun during daytime exceeds what is lost at night, and cold weather in fall and winter when the amount of heat lost at night exceeds what is gained during daytime

Answer 4

The difference is that ideal gas pressure : $$ p=\frac {nkT}{V} \tag 1$$ is assumed to be homogeneous (same) across all gas container/column parts. So gravitational effects on gases are completely discarded in an ideal gas law.

While if you look at gas column pressure completely from the standpoint of gravity, then at the bottom, pressure (by definition) will be :

$$ p = \frac{mg}{A} \tag 2$$

where $m$ is gas column total mass and $A$ gas column cross-section area. This pressure does depend on gravity,- gravitational acceleration $g$ factor. So that same gas column at equator and in the poles will have different pressures due to different g values,- about $9.780~ m/s^2$ at the Equator vs $9.832~m/s^2$ at the poles.

Answer 5

You are correct, there is no difference, there is only one pressure in gas at rest.

The proposed explanation makes no sense.

The actual explanation is that in static atmosphere, the pressure indeed does not change with increase of temperature, because the pressure is determined by the weight of the air column, which is unlikely to change.

Assuming the ideal gas equation $$ p = nRT $$ applies (which it does in dry air), then the fact pressure $P$ does not change with increase of temperature implies the particle density $n$ decreases to counteract, so their product $nT$ remains constant.

In reality, atmosphere is not static, there are air streams which modify air pressure, so increase of temperature, if it causes convective motion, is likely to cause small changes of pressure as well.

Answer 6

The reason why you can't use $pV = nk_BT$ is that your gas isn't ideal. The definition of ideal gas requires the internal energy of the system to be equal to the kinetic energy of the particles in the gas. $$ U_\textrm{ideal} = \sum_{i=1}^N \frac{1}{2}mv^2_i $$ The atmosphere though is subject to gravity which changes the internal energy to $$ U_\textrm{atm} = \sum_{i=1}^N \left( \frac{1}{2}mv^2_i + mgy_i \right). $$ So, although the air particles are not interacting (in this model), their internal energy has an extra term. One interesting consequence is that on average particles closer to ground go faster than particles high in the sky. Another consequence is that there's a pressure gradient and $pV = nk_BT$ is no longer valid.