The root mean square speed of molecules of nitrogen gas at 27° C is approximately : (Given mass of a nitrogen molecule =4.6 × 10⁻²⁶ kg and take Boltzmann constant k_B=1.4 × 10⁻²³ JK⁻¹ )

Solution

Physics sir · 4 steps

Use vᵣₘₛ = √(3k_BT/m) with the mass of one molecule, and change 27°C to kelvin first.

1. Pick the formula. You are given the mass of one molecule and the Boltzmann constant, so use the per-molecule form of the rms speed. Temperature must be in kelvin: T = 27 + 273 = 300 K.

   vᵣₘₛ = √(3k_BT/m)

2. Put in numbers. The numerator is 3 × 1.4 × 10⁻²³ × 300 = 1.26 × 10⁻²⁰ J. Dividing by the mass gives (1.26 × 10⁻²⁰)/(4.6 × 10⁻²⁶) = 2.74 × 10⁵ m² s⁻².
3. Take the root. vᵣₘₛ = √(2.74 × 10⁵) ≈ 523 m s⁻¹.
4. So. The rms speed of nitrogen molecules at 300 K is about 523 m/s, which is option (d).

The idea to keep. If you are given the mass of one molecule, pair it with k_B. If you are given molar mass, pair it with R. Always put temperature in kelvin.

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