How can aerogel be lighter than air?

Question

Air has a density of about $\mathrm{1.3 kg/m^3}$.

From Carbon aerogels by Marcus A. Worsley and Theodore F. Baumann:

Though silica aerogels held the title of "world's lightest material" for a long time at $\sim \mathrm{ 1 mg/cm^3}$, recently, carbon-based aerogels have shattered that record with a density of less than $\mathrm{200 \mu g/cm^3}$.

So the above-named aerogels would have densities of $\sim \mathrm{1 kg/m^3}$ and $\mathrm{0.2 kg/m^3}$ respectively.

How can they be lighter than air if a part of them is a solid (silica or carbon) that is heavier than air?

Answer 1

While the summary you cited is a convenient and easy to understand phrase, it is a paraphrase of another cited paper:

• Sun H., Xu Z., Gao C., "Multifunctional, Ultra-Flyweight, Synergistically Assembled Carbon Aerogels", Adv. Mater. 25 (2013) 2554–2560.

The paper says:

The density was calculated by the weight of solid content without including the weight of entrapped air divided by the volume of aerogel (the density measured in a vacuum is identical to that in the air)

So indeed the other answers are correct: the air is not factored into the density, presumably so aerogels can be compared objectively (despite those at higher altitudes and lower humidity being measured less dense).

Answer 2

They're not lighter than air (or almost certainly are not). The mass of $1\,\mathrm{m^3}$ of such a material, in air, is $\rho_m f_m + \rho_a (1 - f_m)$, where $\rho_m$ is the density of the structure, $\rho_a$ is the density of air, and $f_m$ is the proportion of the bulk volume of the material which is structure rather than air. I think that $f_m \approx 0.002$ for a typical aerogel. The density they are quoting is then $\rho_m f_m$.

Answer 3

If one measured the density of a huge led zeppelin with a vacuum inside (like is done with aerogels; the until now lightest aerogel is aerographene, with a density of about 13% of air) this density could be lower than air if the led zeppelin is big enough. Can you imagine the led zeppelin floating when it's filled with air?

Answer 4

It sounds like they are measuring the weight in vacuum.

OK, try this. You weigh an open glass beaker. It not only contains air, it has about 15 pounds of air above it to the top of the atmosphere. Does the weight of the beaker include the weight of the air above it or in it? No. The air pressure is the same on all sides, and it doesn't count.

If you weigh a balloon that's full of air, you only count the air that's under pressure.

So an aerogel can be mostly empty space, and the weight of the aerogel itself is very low. But it isn't lighter than air unless the empty spaces are filled with vacuum.

Answer 5

It's lighter than the air it displaces.

However, it doesn't actually displace the air. The aerogel is almost entirely empty space, and air is free to occupy the voids in the aerogel. Thus the aerogel, in air, is less than twice the density of air alone.

If you tried to wrap the aerogel in plastic and suck all the air out, the considerable force of 100 kPa would surely collapse the aerogel. If the aerogel was strong enough to prevent that, then yes, you could build foam airships that way.

Answer 6

Nice question. For this problem is good to grasp Wigner–Seitz radius - it's radius of a sphere whose volume is equal to the mean volume per atom in a solid. Wigner radius relates to mass density in such way :

$$ r_{w}= \sqrt[\Large{3}~~~]{{\frac {3M}{4\pi Z\rho N_{A}}} } $$

where $M$ is molar mass, $Z$ is amount of free electrons per atom, $\rho$ is mass density, and $N_{A}$ is the Avogadro number. So when you will pack atoms/molecules in a scarce way,- Wigner radius will become big and so mass density of material will decrease. Using this methodology you can even make nano-structures composed from iron or other metals which density will be smaller than that of air. There's no any magic here.