Temperature - Effect on atmospheric pressure.

[ExcessRed]

What is correlation between temperature and atmospheric pressure?

For example, why does Titan have 1.6 times the atmospheric pressure of Earth despite having pretty much the same atmospheric 'weight', for lack of better term. (Total atmospheric mass times gravity.)

EDIT: Is this the right forum?

 

[qraal]

What is correlation between temperature and atmospheric pressure?

For example, why does Titan have 1.6 times the atmospheric pressure of Earth despite having pretty much the same atmospheric 'weight', for lack of better term. (Total atmospheric mass times gravity.)

EDIT: Is this the right forum?

Not sure where you got the mass and gravity product being the same between Earth and Titan from, but it follows from the fact that surface pressure is due to the weight of the air, and weight is mass times gravity.

The correlation between temperature and pressure is that the two related by the density - Titan's atmosphere at the surface is several times denser than Earth's. As the gas is roughly three times colder and 1.5 times higher pressure, the density is 4.5 times Earth's.

 

[ExcessRed]

Titan's atmosphere is about 1.19 times as massive as Earth's overall. [Coustenis, Athéna and Taylor, F. W. (2008). Titan: Exploring an Earthlike World. World Scientific. p. 130.] Since only relative change matters in this context, we can say titan has 1.19 "earth atmospheric masses" - or EAMs - for simplicity instead of converting to a particular unit.

Gravity of titan is 1.35 m/s^2 and titan has about 8.3×10^7 km^2 surface area.

So Titan has an atmospheric pressure of (1.19*1.35) / 8.3×10^7 = 1.94×10^-8 EAM m/s^2 / km^2

Earth, on the other hand has an EAM of 1 (Duh!), gravity of 9.8 m/s^2, and surface area of 5.1×10^8 meaning an atmospheric pressure of (1*9.8)/5.1×10^8 = 1.92×10^-8 EAM m/s^2 / km^2

This implies that the surface pressure of Titan and Earth is roughly the same (1% increase), so obviously there is more to it than just the weight of the atmosphere divided by surface area since the figure is supposed to be more like a 60% increase.

I was told that discrepancy was due to temperature, but if that's not the case... please tell me what cause is, because I'm clearly missing something!

 

[qraal]

That total mass figure for Titan is incorrect. You might want to check against another source.

 

[ExcessRed]

The figure for atmospheric mass was correct.

I found the answer: temperature increase the density of the atmosphere. As density increases, the atmosphere compresses and sits closer to the surface of the planet. Gravity now has a greater effect on the mass of the atmosphere, since the mass is closer to the planet and the atmospheric weight increases.

Mass times gravity divided by surface area only holds true if gravity constant regardless of distance.

EDIT: I'm reading more now to confirm, but I believe to properly calculate the pressure you have to use the 'http://en.wikipedia.org/wiki/Scale_height" ' of the atmosphere - for which temperature is a factor.

 

[qraal]

The figure for atmospheric mass was correct.

I found the answer: temperature increase the density of the atmosphere. As density increases, the atmosphere compresses and sits closer to the surface of the planet. Gravity now has a greater effect on the mass of the atmosphere, since the mass is closer to the planet and the atmospheric weight increases.

Mass times gravity divided by surface area only holds true if gravity constant regardless of distance.

EDIT: I'm reading more now to confirm, but I believe to properly calculate the pressure you have to use the 'http://en.wikipedia.org/wiki/Scale_height" ' of the atmosphere - for which temperature is a factor.

Hi ExcessRed

Possibly. One thing to keep in mind is that the mass of an atmosphere can't be measured directly, but must be computed according to various assumptions.

Another point is that the volume of the column of atmosphere which presses down on a given area of surface increases as the radial distance increases, exactly compensating for the decrease in the force of gravity.

To see why, consider a square metre of surface area bounded by radii from the centre of a planet. As we move outwards, the area bounded by those radii increases. At twice the planet's radius, the area is now 4 square metres. Thus the volume contained within a unit height times the area enclosed has increased four-fold. But that larger volume's weight (enclosed mass by gravity) has also declined due to the inverse square law of gravity, thus decreasing four-fold.

To work out the scale-height is complicated by gravity's decline with radial distance, but you can probably figure it out for yourself.

 

[ExcessRed]

I have figured out how to calculate the scale height - it's a pretty simple formula - but I don't know exactly how to use this value to get a more accurate estimation of air pressure at sea level.

Any suggestions?

 

[qraal]

I have figured out how to calculate the scale height - it's a pretty simple formula - but I don't know exactly how to use this value to get a more accurate estimation of air pressure at sea level.

Any suggestions?

Sea-Level pressure isn't related to the scale height. It's an input variable, not one you can compute without other data (for example pressure at a given altitude and the behaviour of the atmosphere's temperature with altitude.) Scale height is related directly to the ratio of the kinetic energy of the air at a given location, and the gravitational potential needed to match that energy. Pressure doesn't feature in the equation. But, of course, the rate at which pressure declines with altitude is directly related to the scale-height via the Barometric Formula. Look it up.

 

[ExcessRed]

I have looked it up. Assuming scale height is not related to air pressure at sea level, then what is a different way to estimate the affect of temperature on air pressure at sea level?

 

[Drakkith]

Look here: http://en.wikipedia.org/wiki/Atmospheric_pressure

There is an equation for atmospheric pressure that includes temp. It might require some modifications to the values and such, but maybe it will work.

 

[ExcessRed]

That's the same formula with scale height integrated rather than calculating it separately.

That equation uses one height to determine the pressure at another height by calculating the change based on scale height. Basically it requires you to have one pressure before you can solve for the other. I have no pressure - that's what I want to estimate. Woe.

If there is some fancy maths I can use to circumvent this problem ... advise would be great!

 

[Drakkith]

Ahh, ok. My mistake.

 

[ExcessRed]

Ahh, ok. My mistake.

Meh - no worries. This question has become more math than science at this point! I should probably post a simplified query with nothing but the equation in the math forum. There's a math forum right? :P

 

[qraal]

I have looked it up. Assuming scale height is not related to air pressure at sea level, then what is a different way to estimate the affect of temperature on air pressure at sea level?

If the temperature rises in a small volume then for a short time pressure increases - but it immediately tries to reach a new equilibrium by expanding. That's why the atmosphere is in ceaseless motion over differently heated parts of the surface - convection lifts parcels of hot air and vapour into the sky to cool off, and cold air and rain drops down again. Then there's different winds carrying energy back and forth between sea and land. The net result, over the whole surface, is the heat added by the Sun is equal to the heat radiated by the atmosphere. And, when the inputs and outputs are added up, the equatorial regions gain more, while the polar regions lose more, thus a net flow of heat from the equator to the poles.