How many times is the average velocity of an air molecule greater than the average velocity

How many times is the average velocity of an air molecule greater than the average velocity of a grain of dust suspended in the air? the mass of a dust particle is 5 ng.

To calculate the ratio of the average velocities of an air molecule and a grain of dust, we use the formula: k = Vm / Vp = √ (3 * R * T / MV) / √ (3 * k * T / mp) = √ (R / MV) / (k / mp)) = √ (R * mp) / (k * MV)) = √ (Na * mp) / MV).

Data: Na – Avogadro’s number (Na = 6.02 * 10 ^ 23 mol-1); mp is the mass of a grain of dust (mp = 5 ng = 5 * 10 ^ -12 kg); Мв – molar mass of air (Мв = 29 * 10 ^ -3 kg / mol).

Let’s calculate: k = √ (Na * mp) / MV) = √ (6.02 * 10 ^ 23 * 5 * 10 ^ -12) / (29 * 10 ^ -3)) = 10.19 * 10 ^ 6 R.

Answer: An air molecule has an average velocity of 10.19 * 10 ^ 6 times more.




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