- #1

Hak

- 709

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- Homework Statement
- Physics puzzle: Visibility on the road is 100 m. Assuming that the diameter of a fog droplet is 1 micron, estimate the concentration of fog droplets in the air.

- Relevant Equations
- /

To estimate the concentration of fog droplets in the air, I thought of the mass of water in a given volume of air: I searched this value on the net, and I found that it is typically around ##0.05 g/m^3##.

The mass of a single fog droplet can be calculated from its diameter and density. Assuming that the fog droplet is spherical and has a diameter of ##1## micron (##10^{-6} m##), its volume is

$$V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi(0.5 \times 10^{-6})^3 \approx 5.24 \times 10^{-19}m^3$$

The density of water is about ##1000 kg/m^3##, so the mass of a single fog droplet is

$$m=\rho V=1000 \times 5.24 \times 10^{-19}m^3 \approx 5.24 \times 10^{-16} Kg$$

Now, we can estimate the concentration of fog droplets in the air by dividing the mass of water in a given volume of air by the mass of a single droplet. This gives us

$$n= \frac{0.05 \times 10^3}{5.24 \times10^{-16}} \approx 9.54 \times 10^{16} droplets/m^3$$

(Edit)~~This would mean that there are about 95 trillion fog droplets in every cubic meter of air...~~

This process does not seem convincing to me. In my opinion there is something deeper that I cannot decipher. Where am I going wrong? Could you give me some hints?

The mass of a single fog droplet can be calculated from its diameter and density. Assuming that the fog droplet is spherical and has a diameter of ##1## micron (##10^{-6} m##), its volume is

$$V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi(0.5 \times 10^{-6})^3 \approx 5.24 \times 10^{-19}m^3$$

The density of water is about ##1000 kg/m^3##, so the mass of a single fog droplet is

$$m=\rho V=1000 \times 5.24 \times 10^{-19}m^3 \approx 5.24 \times 10^{-16} Kg$$

Now, we can estimate the concentration of fog droplets in the air by dividing the mass of water in a given volume of air by the mass of a single droplet. This gives us

$$n= \frac{0.05 \times 10^3}{5.24 \times10^{-16}} \approx 9.54 \times 10^{16} droplets/m^3$$

(Edit)

This process does not seem convincing to me. In my opinion there is something deeper that I cannot decipher. Where am I going wrong? Could you give me some hints?