<p>"Look like" is not quite right, since gaseous CO2 is transparent to visible light.  To get a sense of the volume, we can do some simple math, which unless I've made a mistake somewhere gives an area 21 km/13 miles on a side, stretching from sea level to space.</p>

<p>Showing my work:</p>

<p>x = 5 Gt =  5000000000 * 2000 lbs/ton = 10000000000000 lbs</p>

<p>Sea level pressure = 14.7 lbs/in^2, so 680272108843 in^2</p>

<p>144 in^2 per ft^2, so an area of 4724111866 ft^2, or a square 68732 ft on a side</p>

<p>Convert to miles = 13 miles on a side
Convert to km = 20.9 km</p>

CC BY-SA 4.0
2018-05-11T05:25:35.370Z
OK, my bad.  One day I'll get used to all those prefixes...  Post will be edited to reflect Million rather than 1000 Million.  However, I think it is fair to say we're looking at a Very Big 100Km-high (Karman Line) Box.
gorblimey
CC BY-SA 4.0
2018-05-11T05:42:44.197Z
This should not be the correct answer. It is very wrong. Major problem: 5 million is 5000000, not 5000000000. Another problem: A pound mass and a pound force are different units. It happens to be a wash in this case, but equating the two (pounds mass and pounds force) is a sign of sloppiness. Rather minor problem: The article talked about tonnes (1000 kg) rather than tons (2000 lb).
David Hammen
CC BY-SA 4.0
2018-05-11T06:05:14.057Z
@David Hammen - Apart from my complete ineptitude with Mega-Giga-, jamesqf's arithmetic looks OK to me.  My own check using his logic but my corrected number shows a 222.14m square, which sanity-checks with removing excess zeroes.  So the Very Big Box meta-answer is still good enough for me.
gorblimey
CC BY-SA 4.0
2018-05-11T06:15:08.637Z
About "proper" units.  In Oz we have been saddled with the Pascal as the unit of pressure,  While this is technically correct, it is in fact rather useless for the laity.  We note that in most of Europe (and France is the Home of SI) pressure is routinely expressed as Kg/cm^2.  This avoids the possiblity of introducing headaches like momentum and acceleration.
gorblimey
CC BY-SA 4.0
2018-05-11T06:37:57.027Z
Using this approach (but in SI units, and with 5Mt rather than 5Gt), I reckon it's a block of the atmosphere resting on a square of sides around 700m. Done in a very quick 'n dirty fashion, though, so might well be wrong. Also, kinda wrong in principle, since the mass of that volume of CO2 would be significantly more than the mass of that volume of air.
Semidiurnal Simon
CC BY-SA 4.0
2018-05-11T06:52:36.287Z
We-e-e-ll, my own quick and dirty was (222.14^2x10x5)^-3=1,702.5m cube, but I did not adjust for pressure...  And yes, none of my work has accounted for mass.  OTOH, the solution I needed is very much along the lines of "Humongous!"
gorblimey
CC BY-SA 4.0
2018-05-11T12:34:24.717Z
@SemidiurnalSimon is correct, both in his comment above and in his answer. A 696x696 meter patch reaching from the ground to the top of the atmosphere would contain 5 million tonnes of carbon dioxide.
David Hammen
CC BY-SA 4.0
2018-05-11T17:55:58.280Z
@Semidiurnal Simon: Right about the mass correction.  I did think of that, but with the original amount of Gtonnes (and I think there's a bit of a correction there too :-() I figured the "really big box up to space" would be close enough :-)  Now for extra credit: suppose we freeze all that CO2 into dry ice: how big a pile do we have?
jamesqf
CC BY-SA 4.0
2018-05-11T19:18:23.580Z
@jamesqf I reckon about 7.5e6 m^3, or a cube of ~196m per side (though a pile of crushed material would be somewhat bigger!).
The amount of energy extracted in the refrigeration process is left as an exercise for the reader :-P
Semidiurnal Simon
CC BY-SA 4.0
2018-05-11T04:46:33.993Z
2018-05-11T04:46:33.993Z
jamesqf
<p>The question is hard to give a definitive answer to, as it depends on the pressure at which the gas is kept in the tank. But to try to give an intuitive sense of how much gas this is, let's work out two things: Firstly, how big the box would be at atmospheric pressure; and secondly, how many standard gas bottles it would take.</p>

<h2>At atmospheric pressure</h2>

<p>The mass of one <a href="https://en.wikipedia.org/wiki/Mole_(unit)" rel="noreferrer">mole</a> of CO<sub>2</sub> is 44 grams. So 5 MT of CO<sub>2</sub> is $\frac{5\times 10^{12}}{44} = 114\times 10^9$ moles of gas.</p>

<p>The <a href="https://en.wikipedia.org/wiki/Ideal_gas_law" rel="noreferrer">ideal gas law</a> tells us that</p>

<p>$PV = nRT$</p>

<p>where $P$ is pressure, $V$ is volume, $n$ is the number of moles of the gas, $R$ is a constant, and $T$ is the temperature in Kelvin. If we assume a pressure of 100,000 Pa (roughly atmospheric pressure) and a temperature of 300 K (roughly room temperature), then</p>

<p>$V = \frac{nRT}{P} = \frac{114 \times 10^9 \times 8.314 \times 300}{10^5} = 2.8\times 10^9$ m<sup>3</sup>. </p>

<p>That volume works out to <strong>a cube with sides of 1.4 km</strong>.</p>

<h2>In gas bottles</h2>

<p>A standard large CO<sub>2</sub> bottle of the type that is familiar in the UK looks like the one on the right of this picture:</p>

<p><a href="https://i.sstatic.net/PbnJ5.jpg" rel="noreferrer"><img src="https://i.sstatic.net/PbnJ5.jpg" alt="enter image description here"></a></p>

<p>and <a href="https://www.boconline.co.uk/internet.lg.lg.gbr/en/images/Carbon-Dioxide-factsheet410_54560.pdf?v=2.0" rel="noreferrer">according to BOC</a> contains 34 kg of gas, at a pressure of up to 50 bar. So that's 29.4 cylinders per tonne or, to contain 5 million tonnes of CO<sub>2</sub>, <strong>147 million of these cylinders.</strong></p>

CC BY-SA 4.0
2018-05-11T06:29:45.200Z
2018-05-11T06:29:45.200Z
Semidiurnal Simon
<p>Reading a news article <a href="https://www.watoday.com.au/national/western-australia/probe-after-gorgon-releases-millions-of-tonnes-of-greenhouse-gas-20180510-p4zeis.html" rel="nofollow noreferrer">Probe after Gorgon releases millions of tonnes of greenhouse gas</a>, I am confused.  CO2 is a gas, right?  Gases are not very dense, right?  So what does 5MT of the stuff look like?  How big a tank do we need to hold it?</p>

CC BY-SA 4.0
2018-05-11T05:38:40.130Z
Please deselect the answer you have chosen. It is very wrong.
David Hammen
CC BY-SA 4.0
2018-05-11T05:42:57.143Z
Ummm... How so?
gorblimey
CC BY-SA 4.0
2018-05-11T06:40:48.867Z
In general, it's polite to wait at least a few hours (or days) before accepting an answer, in case a much better one comes along :-)
Semidiurnal Simon
CC BY-SA 4.0
2018-05-11T06:54:13.800Z
I think there may well be several best answers here. :))
gorblimey
CC BY-SA 4.0
2018-05-12T19:22:13.553Z
@David Hammen: Read the comments to that answer.  It is not (very) wrong GIVEN THE ORIGINAL INPUT.  If you know how to write answers that automatically adjust themselves when the question is changed, please feel free to tell us lesser minds how you do it :-)
jamesqf
CC BY-SA 4.0
2018-05-12T23:46:32.407Z
@jamesqf It's bad practice to substantially change a question once it has answers; but in this case, as it happened, would you consider editing your answer to be correct for the current question rather than the old one? While I agree that it isn't your fault, the current answer is definitely wrong.
Semidiurnal Simon
<climate-change><geochemistry><environmental-protection>
What does 5,000,000 tonnes of CO2 look like?
gorblimey
# What does 5,000,000 tonnes of CO2 look like?

(Asked by: username_0 on 2018-05-11)

Reading a news article [Probe after Gorgon releases millions of tonnes of greenhouse gas](https://www.watoday.com.au/national/western-australia/probe-after-gorgon-releases-millions-of-tonnes-of-greenhouse-gas-20180510-p4zeis.html), I am confused. CO2 is a gas, right? Gases are not very dense, right? So what does 5MT of the stuff look like? How big a tank do we need to hold it?

**Tags:** climate-change, geochemistry, environmental-protection

> **Comments:**
> **username_1**: Please deselect the answer you have chosen. It is very wrong.  
> **username_0**: Ummm... How so?  
> **username_2**: In general, it's polite to wait at least a few hours (or days) before accepting an answer, in case a much better one comes along :-)  
> **username_0**: I think there may well be several best answers here. :))  
> **username_3**: @username_1: Read the comments to that answer.  It is not (very) wrong GIVEN THE ORIGINAL INPUT.  If you know how to write answers that automatically adjust themselves when the question is changed, please feel free to tell us lesser minds how you do it :-)  
> **username_2**: @username_3 It's bad practice to substantially change a question once it has answers; but in this case, as it happened, would you consider editing your answer to be correct for the current question rather than the old one? While I agree that it isn't your fault, the current answer is definitely wrong.  

---

## 2 answers

### Answer by username_3 on 2018-05-11. Score: -1 | Accepted answer

"Look like" is not quite right, since gaseous CO2 is transparent to visible light. To get a sense of the volume, we can do some simple math, which unless I've made a mistake somewhere gives an area 21 km/13 miles on a side, stretching from sea level to space.

Showing my work:

x = 5 Gt = 5000000000 \* 2000 lbs/ton = 10000000000000 lbs

Sea level pressure = 14.7 lbs/in^2, so 680272108843 in^2

144 in^2 per ft^2, so an area of 4724111866 ft^2, or a square 68732 ft on a side

Convert to miles = 13 miles on a side
Convert to km = 20.9 km

> **Comments:**
> **username_0**: OK, my bad.  One day I'll get used to all those prefixes...  Post will be edited to reflect Million rather than 1000 Million.  However, I think it is fair to say we're looking at a Very Big 100Km-high (Karman Line) Box.  
> **username_1**: This should not be the correct answer. It is very wrong. Major problem: 5 million is 5000000, not 5000000000. Another problem: A pound mass and a pound force are different units. It happens to be a wash in this case, but equating the two (pounds mass and pounds force) is a sign of sloppiness. Rather minor problem: The article talked about tonnes (1000 kg) rather than tons (2000 lb).  
> **username_0**: @username_1 - Apart from my complete ineptitude with Mega-Giga-, username_3's arithmetic looks OK to me.  My own check using his logic but my corrected number shows a 222.14m square, which sanity-checks with removing excess zeroes.  So the Very Big Box meta-answer is still good enough for me.  
> **username_0**: About "proper" units.  In Oz we have been saddled with the Pascal as the unit of pressure,  While this is technically correct, it is in fact rather useless for the laity.  We note that in most of Europe (and France is the Home of SI) pressure is routinely expressed as Kg/cm^2.  This avoids the possiblity of introducing headaches like momentum and acceleration.  
> **username_2**: Using this approach (but in SI units, and with 5Mt rather than 5Gt), I reckon it's a block of the atmosphere resting on a square of sides around 700m. Done in a very quick 'n dirty fashion, though, so might well be wrong. Also, kinda wrong in principle, since the mass of that volume of CO2 would be significantly more than the mass of that volume of air.  
> **username_0**: We-e-e-ll, my own quick and dirty was (222.14^2x10x5)^-3=1,702.5m cube, but I did not adjust for pressure...  And yes, none of my work has accounted for mass.  OTOH, the solution I needed is very much along the lines of "Humongous!"  
> **username_1**: @username_2 is correct, both in his comment above and in his answer. A 696x696 meter patch reaching from the ground to the top of the atmosphere would contain 5 million tonnes of carbon dioxide.  
> **username_3**: @username_2: Right about the mass correction.  I did think of that, but with the original amount of Gtonnes (and I think there's a bit of a correction there too :-() I figured the "really big box up to space" would be close enough :-)  Now for extra credit: suppose we freeze all that CO2 into dry ice: how big a pile do we have?  
> **username_2**: @username_3 I reckon about 7.5e6 m^3, or a cube of ~196m per side (though a pile of crushed material would be somewhat bigger!). The amount of energy extracted in the refrigeration process is left as an exercise for the reader :-P  

---

### Answer by username_2 on 2018-05-11. Score: 6

The question is hard to give a definitive answer to, as it depends on the pressure at which the gas is kept in the tank. But to try to give an intuitive sense of how much gas this is, let's work out two things: Firstly, how big the box would be at atmospheric pressure; and secondly, how many standard gas bottles it would take.

## At atmospheric pressure

The mass of one [mole](https://en.wikipedia.org/wiki/Mole_(unit)) of CO2 is 44 grams. So 5 MT of CO2 is $\frac{5\times 10^{12}}{44} = 114\times 10^9$ moles of gas.

The [ideal gas law](https://en.wikipedia.org/wiki/Ideal_gas_law) tells us that

$PV = nRT$

where $P$ is pressure, $V$ is volume, $n$ is the number of moles of the gas, $R$ is a constant, and $T$ is the temperature in Kelvin. If we assume a pressure of 100,000 Pa (roughly atmospheric pressure) and a temperature of 300 K (roughly room temperature), then

$V = \frac{nRT}{P} = \frac{114 \times 10^9 \times 8.314 \times 300}{10^5} = 2.8\times 10^9$ m3.

That volume works out to **a cube with sides of 1.4 km**.

## In gas bottles

A standard large CO2 bottle of the type that is familiar in the UK looks like the one on the right of this picture:

[![enter image description here](https://i.sstatic.net/PbnJ5.jpg)](https://i.sstatic.net/PbnJ5.jpg)

and [according to BOC](https://www.boconline.co.uk/internet.lg.lg.gbr/en/images/Carbon-Dioxide-factsheet410_54560.pdf?v=2.0) contains 34 kg of gas, at a pressure of up to 50 bar. So that's 29.4 cylinders per tonne or, to contain 5 million tonnes of CO2, **147 million of these cylinders.**


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