How much does a cloud weigh?

To answer this question, we have to get our hands dirty and do a little bit of math. Lets take an example of where a fair weather cumulus cloud (such as in the picture to the left) is about 1 km by 1 km by 1 km in dimension and located approximately 2 km above the ground. The mass of a differential volume element (such as our "cube" cloud), is simply the density times the volume of that element.

So, to figure this out, we need to first calculate the density of dry air and compare that to the density of the air in our "cube" cloud, which is made up of a small percent of water vapor along with a lot of air molecules. Water has an atomic weight of 18 while Nitrogen (N₂) has an atomic weight of 28 and Oxygen (0₂) has an atomic weight of 32. Therefore we can already hypothesis that when water molecules displace Nitrogen or Oxygen molecules, the "weight" of the air mixture will decrease. Therefore, a cloud should "weigh" less than the air it is displacing.

The ideal gas law states that:

DENSITY = P / TR (pressure(P) divided by temperature(T) multiplied by a constant(R))

From the average values found in our atmosphere, we can use a temperature at about 2 km altitude of 275.15K (+2.15C) and a pressure at 2 km of 79.495KPa (795 mb). [Note: don't worry so much about the units given, just follow the general idea of the discussion.] For dry air, the constant(R) equals 287 J/K*kg. This gives us a density of 1.007 kg/m³ for dry air. For pure water vapor, we would use a constant(R) of 461 J/K*kg which ends up giving us a density for pure water vapor of 0.627 kg/m³. Because the cloud is made up of a small amount of water vapor and a large amount of air, we would need to calculate the partial pressure of the water vapor, which in this case comes out to around 7 mb, or 0.9% of the total pressure (795 mb).

So, [99.1 X (dry air density) + 0.9 X (moist air density)]/100 is approximately the density of the cloud itself. Calculating this gives a density in the cloud of 1.003 kg/m³, compared to 1.007 kg/m³ in the dry air surrounding the cloud. This shows that a cloud is less dense than the air around it, which is why it floats!

Now that density is taken care of, we only need to calculate the volume of the cloud. To make the example as easy as possible, we are assuming a 1 km by 1 km by 1 km cloud which gives us a volume of 1 cubic km (km³).

Finally, we can now calculate the mass of the cloud. Remember, the mass is density times the volume. So, we multiply 1 km³ by the density which is 1.003 kg/m³ and then multiply that by a factor of 1000 to get the units to be in kilograms(kg).

If you get out your calculators and do the math, you will see that our "cube" cloud weighs 1,003,000,000 kg or approximately 2,211,200,000 lbs. Thats almost 2.2 BILLION pounds!!!

However, remember that air also has mass. By doing the same calculation, but this time using the density of dry air, we come out with a mass of 1,007,000,000 kg or approximately 2,220,000,000 lbs. So, dry air indeed weighs more than moist air, which is why clouds can exist where they do, seeming to 'float' across the skies.

In conclusion, a "typical" fair weather cumulus cloud (such as seen in the picture) "weighs" about over 2 billion pounds, or about 9 million pounds less than dry air of equal volume. The main point to get from this discussion is not that clouds weigh a lot, since in reality they cannot be weighed, but rather, that moist air is less dense than dry air. This is an important factor during the summer thunderstorm season when afternoon thunderstorms develop rapidly after a cloudless morning.