# Is increase in atmospheric $\mathrm{CO}_2$ concentration affecting Earth's atmospheric pressure?

It's well known that the $\mathrm{CO}_2$ levels around the world are increasing, leading to an increase in overall surface temperature, etc.

But I'm wondering, are these gases (or the effects from these gases) increasing the global atmospheric pressure to any great degree, or is the atmospheric composition changing without any major change to air pressure levels, compared to centuries or millennia ago?

The air in the atmosphere is a fluid. The pressure exerted by a fluid on a horizontal surface can be calculated using the formula.

$$P = \rho\,\mathrm{g}\, h$$

Where:

• $P$ is pressure
• $\rho$ is the density of the fluid
• $\mathrm{g}$ is the acceleration due to gravity
• $h$ is the height of the fluid

The acceleration due to gravity is a constant and the height of atmosphere can be considered constant. That leaves only the density of the air in the atmosphere as the only variable that will affect pressure on the surface of the Earth, in this situation.

The amount of carbon dioxide in the atmosphere is around 400 parts per million — a very small proportion.

Air density varies according to temperature and pressure but is within the range of $1.146$ to $1.422\ \mathrm{kg/m}^3$. By comparison, water has a density of $1000\ \mathrm{kg/m}^3$.

If the quantity of carbon dioxide were to increase, even tenfold, it would have a negligible affect on the density of the air and thus a negligible affect on the atmospheric pressure on the surface of the Earth. Its effect on temperature is another matter.

### Edit

Answer to the additional question: Can you expound on what effect the increase in temperature will have on pressure?

The relationship between pressure, temperature and volume for gases is given by the universal gas law which states,

$$PV = m\, \mathrm{R}\, T$$

Where:

• $P$ is pressure
• $V$ is volume
• $m$ is mass
• $\mathrm{R}$ is the universal gas constant
• $T$ is temperature

Rearranging the equation,

$$P = \frac{m\mathrm{R}T}{V}$$

When considering the affect of temperature on atmospheric pressure, $\mathrm{R}$ is already a constant and the mass and volume of the atmosphere can be considered to be constant because any changes due to increased $\ce{CO_2}$ will be exceedingly small and can be regarded as being negligible.

Consequently $m\mathrm{R}/V$ is a constant $\mathrm{k}$ and the equation becomes,

$$P = \mathrm{k}T$$

The Kelvin temperature scale, used in the equation, is an extrapolation of the Celsius temperature scale, thus a rise in temperature of $1^\circ\mathrm{C}$ ($1\ \mathrm{K}$) will increase atmospheric pressure by $1\ \mathrm{pascal}$.

The average atmospheric pressure is $101\,325\ \mathrm{Pa}$. Assuming a very bad case scenario of a rise of $10^\circ\mathrm{C}$, the pressure would increase $10\ \mathrm{Pa}$, or $0.0099\%$, which is insignificant.

Weather reports report atmospheric pressure in units of hectopascals ($\mathrm{hPa}$). A pressure of $101\,325\ \mathrm{Pa}$ is $1013.25\ \mathrm{hPa}$. A rise of $10\ \mathrm{Pa}$ is the same as $0.1\ \mathrm{hPa}$. Atmospheric pressures vary between $890\ \mathrm{hPa}$ and $1090\ \mathrm{hPa}$.

• Can you expound on what effect the increase in temperature will have on pressure? – farrenthorpe Aug 30 '15 at 1:31
• -1 This answer in incorrect. First relationship holds for incompressible fluids - air is not incompressible. You may want to substantially revise the first part of the answer. Second, your ideal gas law calculation is not correct - a change in 1 K air temperature results in a change in surface pressure of approximately 3 hPa, which is 300 times larger than what your calculation suggests. And this doesn't even touch on the issues of CO2 concentration affecting heat and moisture budgets affecting global circulation affecting atmospheric pressure. – milancurcic Aug 30 '15 at 17:20

Fred is right, any change is virtually nothing, certainly not noticeable to anything but very finely tuned instruments.

But lets look at the Chemical Process.

$\ce{CH4}$ (mined from the ground) + $\ce{2O2 -> CO2 + 2 H2O}$. The $\ce{H2O}$ while it enters the air as water vapor doesn't stay water vapor for any significant time so the Atmospheric scale tells us $\ce{2 O2}$ to start (64 atomic weight), $\ce{CO2}$ out (44 atomic weight), so burning methane lightens the atmosphere.

Now, lets look at gasoline, which is a variety of hydrocarbons, but lets say Pentane is the most abundant one (not sure it is, but it might be) $\ce{C5H12}$, here we get the following: $$\ce{C5H12 + 8 O2 -> 5CO2 + 6H2O}$$. With this one we have 256 atomic weight to start and 220 atomic weight to finish.

As the carbon to hydrogen ratio goes up, the mass of the gas after the fuel is burned goes up, but on average, because burning fossil fuel takes oxygen and makes water as well as $\ce{CO2}$, mostly you're losing mass in the atmosphere and gaining water - but in negligible amounts.

There are other, larger factors, which, I don't understand all that well, but oceanic absorption for example. Oceans absorb atmosphere (less as they get warmer) so as the Earth warms the oceans will store less air and the atmosphere will gain in mass, a bit. Similarly, warmer air holds more water vapor, so, while burning $\ce{CO2}$ takes a little bit of mass out of the atmosphere, the warming will have the reverse effect. More water vapor will be in the air and less air will be in the water.

A 3rd effect, photosynthesis for example takes in $\ce{CO2}$ and puts out $\ce{O2}$, so as plants grow, the mass of the atmosphere is reduced, but overall, plants are simply not growing fast enough to match our output of $\ce{CO2}$. Not close. But the respiration of animals and plants on the planet does play a role, but as near as I can figure, the mathematics of this is pretty simple. Plants take in $\ce{CO2}$ and release $\ce{O2}$, animals, by eating plants, do the reverse, we breath in $\ce{O2}$ and exhale $\ce{CO2}$, so any net change due to animal life would probably register in total $\ce{CO2}$. (I think).

So there's your answer. The burning of the fuels makes the atmosphere a teeny tiny bit lighter but the warming of the atmosphere makes the atmosphere heavier and that's likely the trump ward. A warmer earth has a heavier atmosphere.

About 0.25 of the mass of the atmosphere at any given time is water vapor (Source), compared to about 400 PPM $\ce{CO2}$ or 0.04% by volume, but since $\ce{CO2}$ is about 50% heavier than the oxygen/nitrogen combination, about 0.06% by mass.

Because $\ce{CO2}$ isn't just added, but oxygen is taken away in greater mass when burning fossil fuels (44 to 64 with methane, 220 to 250 with pentane), it's tough to estimate but lets say about 15% of the mass of the increase in $\ce{CO2}$ is taken out of the atmosphere due to loss of $\ce{O2}$.

The increase in $\ce{CO2}$, from 280 PPM to 400 PPM, is 40%, so working out the math, 0.06% * 0.4* 0.15 = rough estimate 0.0036% of the mass of the atmosphere lost due to all our fossil fuel burning since the industrial revolution.

But, just 1 degree of warming can increase the atmospheric $\ce{H2O}$ by 6%-7%, so if we figure about .8 degrees C of warming this century and we run the math on atmospheric water vapor 0.25% * 0.06 * 0.8 = 0.012, or over 3 times as much mass added to the atmosphere from warming as is lost in fuel burning. And that doesn't take into account oceanic absorption, which as the oceans warm they hold less oxygen. Similarly, as Tundra thaws, gas is released from the Earth.

The net change (while quite small) is likely a heavier atmosphere (about 1/100th of 1% (so far, and more to come), but that's 1/10th of a millibar. A typical High to low pressure variation at sea level is about 30 millibars (Source), so we are talking about an unnoticeable change.

My math is rough but I think that's the correct answer to your question. You can't just weigh the $\ce{CO2}$ added, you have to also weigh the $\ce{O2}$ taken out and the bigger factors, atmospheric $\ce{H2O}$ and Oceanic absorption of atmospheric gas.

I have got a barometer in my hall (it's supposed to predict the weather). Sure does move about a lot. You know that of course. So the answer to your question is "YES" , in fact any change in the composition of a gas mixture is likely to change its pressure (other things being equal), but the effect will be small as the other 999,600 ppm of other stuff has a rather bigger effect BUT even if the effect was a 1 or 2% change I doubt you could confirm it by measurement as the normal variation in barometric pressure day to day and hour to hour will rather swamp it.