The root mean square speed of smoke particles of mass 5 × 10⁻¹⁷ kg in their Brownian motion in air at NTP is approximately. [Given k=1.38 × 10⁻²³ JK⁻¹]

Solution

Physics sir · 4 steps

A smoke particle is just a very heavy gas molecule, so use vᵣₘₛ = √(3kT/m) with T = 273 K.

1. Same rule applies. In Brownian motion the smoke particles are in thermal equilibrium with the air. Each one has average kinetic energy 3/2kT, the same as any air molecule. Here m is the mass of one particle, so use k, not R.

   1/2m vᵣₘₛ² = 3/2kT ⇒ vᵣₘₛ = √(3kT/m)

2. Put in numbers. NTP means T = 273 K. Then 3kT = 3 × 1.38 × 10⁻²³ × 273 = 1.13 × 10⁻²⁰ J. Dividing by m = 5 × 10⁻¹⁷ kg gives vᵣₘₛ² = 2.26 × 10⁻⁴ m² s⁻².
3. Take the root. vᵣₘₛ = √(2.26 × 10⁻⁴) = 1.5 × 10⁻² m s⁻¹. That is 15 mm s⁻¹.
4. So. The rms speed is about 15 mm s⁻¹, which is option (c).

The idea to keep. For a single particle, use vᵣₘₛ = √(3kT/m) with m in kg. Use R and molar mass only when you have moles. NTP means 273 K.

Answer from the NTA final answer key. Solution by PrepWiser.