Practice portal › Kinetic Theory of Gases › Speed, Temperature and Molar Mass

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 J K⁻¹)

Asked in JEE Main 11th April 2nd Shift 2023 · Speed with temperature and molar mass

Answer: (2) 523 m/s

Step-by-step solution

Per molecule, vᵣₘₛ=√(3k_BT)/m.

T=27+273=300 K.

3k_BT=3×1.4×10⁻²³×300=1.26×10⁻²⁰.

(3k_BT)/m=(1.26×10⁻²⁰)/(4.6×10⁻²⁶)=2.74×10⁵.

vᵣₘₛ=√2.74×10⁵≈523 m/s.

Why the other options are wrong

More Speed, Temperature and Molar Mass questionsAll Speed, Temperature and Molar Mass questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer