Why is the air less dense higher up you go?

[scilearner]

Hello everyone

 

I have a quick question why is the air less dense higher up you go. I can understand air expands and the volume is bigger hence less dence but that only happens if it is warm, higher you go it is cooler so how does air expand. Thank you!!

[CaptainPanic]

It's easy:

 

The air that's on top is pushing down (gravity is pulling it down). So, all the air in the upper atmosphere pushes down, and compacts the air below it.

[scilearner]

It's easy:

 

The air that's on top is pushing down (gravity is pulling it down). So, all the air in the upper atmosphere pushes down, and compacts the air below it.

 

Thanks for the reply Sorry my science knowledge is not very good and I couldn't really understand. So if it compacts wouldn't it be more dense higher up.

[CaptainPanic]

See it like this:

 

All the way at the edge of space, there are still a few molecules, but really few. However, these do have a little weight, and gravity pulls them down a bit.

That means they push on the next couple of molecules below them. The molecules below them are therefore a little more compact. Same as when you sit on a ball, you will compress the air inside because you're so heavy. - the effect of air is less obvious, because air is very thin - but we have a few hundred kilometers of air above us, pushing down!

 

So, those molecules a bit further down, together with the ones all the way at the edge of space, are attracted, and together push down on the ones below them.

 

The molecules that they all push on are therefore compacted even a bit more. And those all push down on the ones below them, and therefore compact those even more.

 

And therefore, the lower you go, the more molecules are pushing down on that particular layer of air.

[scilearner]

See it like this:

 

All the way at the edge of space, there are still a few molecules, but really few. However, these do have a little weight, and gravity pulls them down a bit.

That means they push on the next couple of molecules below them. The molecules below them are therefore a little more compact. Same as when you sit on a ball, you will compress the air inside because you're so heavy. - the effect of air is less obvious, because air is very thin - but we have a few hundred kilometers of air above us, pushing down!

 

So, those molecules a bit further down, together with the ones all the way at the edge of space, are attracted, and together push down on the ones below them.

 

The molecules that they all push on are therefore compacted even a bit more. And those all push down on the ones below them, and therefore compact those even more.

 

And therefore, the lower you go, the more molecules are pushing down on that particular layer of air.

 

Thanks once again I can understand how this would create higher pressure lower down but how is this related to density. Warm air is more dense than cold air so how is cold air at the top be less dense.

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Merged post follows:

Consecutive posts merged

I'll make my question in form of a picture

 

[CaptainPanic]

Thanks once again I can understand how this would create higher pressure lower down but how is this related to density. Warm air is more dense than cold air so how is cold air at the top be less dense.

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Merged post follows:

Consecutive posts merged

I'll make my question in form of a picture

 

 

You're right about the hot and cold. Cold air IS more dense. Hot air IS less dense. Those density differences are in addition to the things I just described... the picture gets more complicated as we write more posts

 

And to make things more complicated: air heats up and cools down all the time... and that, my dear Watson, is how wind is created. Wind and pressure differences (low pressure, high pressure zones) are caused mostly by heating / cooling (although water has a big hand in this: if it evaporates, the air cools, and if it condenses, it heats).

It's quite complicated, which is why weather forecasts are reliable for only a few days ahead.

[Sisyphus]

I can understand how this would create higher pressure lower down but how is this related to density.

 

It is directly related. I think reading about the ideal gas law, which roughly desribes the relationship between temperature, pressure, and volume of gases, will answer most of your questions. As pressure increases, volume decreases, and vice versa. And remember, density is just mass per unit volume. Put the same mass in a smaller volume and you've got higher density. Give the link a look, and come back with any questions you have about it.

[scilearner]

Thanks for the reply Captain Panic even though I got confused

 

It is directly related. I think reading about the ideal gas law, which roughly desribes the relationship between temperature, pressure, and volume of gases, will answer most of your questions. As pressure increases, volume decreases, and vice versa. And remember, density is just mass per unit volume. Put the same mass in a smaller volume and you've got higher density. Give the link a look, and come back with any questions you have about it.

 

Thanks for the reply I know the ideal gas law so in this case let's say I want to know the density of gas molecules near the red area in the picture. Do I take the volume as the blue box or the yellow box. Thanks

[D H]

Forget about temperature for a bit. That just complicates things, and you don't understand the basic concepts yet. Ignore air movement. Just look at pressure versus altitude.

 

Your little blue cuboid is a good place to start. Assume the area of the bottom and top faces is A and that the height is Δz. (The Δ just means the height is fairly small. More on this later.) The volume of the cuboid is thus V=AΔz.

 

Assume the cuboid isn't moving vertically: It is in hydrostatic equilibrium. Google that phrase; it is key to understanding your original question. This means the vertical forces on the cube must sum to zero. The vertical forces on the cuboid are

• The pressure on the bottom face. This force is directed upward and is equal in magnitude to the product of the pressure at the bottom of the cuboid with the area of the bottom face. Denoting increasing altitude as positive:
  [math]F_{\text{bot}} = P_{\text{bot}}A[/math]
  
• The pressure on the top face. This force is directed downward and is equal in magnitude to the product of the pressure at the bottom of the cuboid with the area of the bottom face:
  [math]F_{\text{top}} = -P_{\text{top}}A[/math]
  
• The weight of the gas in the cuboid. This force This force is directed downward and is equal in magnitude to the product of the mass of the gas in the cuboid and the gravitational acceleration:
  [math]F_{\text{grav}} = -m_{\text{gas}}g[/math]

 

Adding these forces and setting the sum to zero yields

 

[math]F_{text{net}} = F_{\text{bot}} + F_{\text{top}} + F_{\text{grav}}

= (P_{\text{bot}}-P_{\text{top}})A - m_{\text{gas}}g = 0[/math]

 

The pressure at the top of the cuboid is equal to the pressure at the bottom plus some delta pressure:

 

[math]P_{\text{top}} = P_{\text{bot}}+\Delta P[/math]

 

There's nothing new here; this is just defining ΔP. Using this in the force balance equation,

 

[math]-\Delta P A - m_{\text{gas}}g = 0[/math]

 

or

 

[math]\Delta P A = - m_{\text{gas}}g[/math]

 

Qualitatively, this means the pressure at the top of the cuboid is a bit less than the pressure at the bottom of the cuboid.

 

The mass of the gas is the product of the volume of the cuboid times the average density in the cuboid. Denoting the average density as the density at the bottom plus some (unknown) delta density,

 

[math]m_{\text{gas}} = V(\rho + \Delta \rho) = A\Delta z (\rho + \Delta \rho)

= A\rho \Delta z + A\Delta \rho \Delta z[/math]

 

As the height of the cuboid becomes smaller and smaller, the difference between the density at the bottom of the cuboid and the average density will become small, and the product [math]\Delta \rho \Delta z[/math] will become very, very small. For sufficiently small cuboid heights, we can ignore this extra term:

 

[math]m_{\text{gas}} \approx A\rho \Delta z[/math]

 

The squiggly equal sign means "approximately equal to". This becomes closer and closer to just plain "equals" as the height of the cuboid is made ever smaller. With this, the force balance equation becomes

 

[math]\Delta P A \approx - A\rho g \Delta z[/math]

 

The area cancels out, leaving

 

[math]\Delta P \approx - \rho g \Delta z[/math]

 

or

 

[math]\frac{\Delta P}{\Delta z} \approx - \rho g[/math]

 

The approximation becomes exact as [math]\Delta z \to 0[/math]. We write this mathematically as a differential equation:

 

[math]\frac{dP}{dz} = - \rho g[/math]

 

Because the pressure depends on things other than altitude, it is best to write this exact differential equation as a partial differential equation:

 

[math]\frac{\partial P}{\partial z} = - \rho g[/math]