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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