JEE Main 2021 · 27 July · morning · Single correct · +4 / −1
The number of molecules in one litre of an ideal gas at 300 K and 2 atmospheric pressure with mean kinetic energy 2 × 10⁻⁹ J per molecules is :
- (a) 0.75 × 10¹¹
- (b) 3 × 10¹¹
- (c) 1.5 × 10¹¹
- (d) 6 × 10¹¹
Answer and solution
Answer (c) 1.5 × 10¹¹
Solution
Physics sir · 4 steps
Write pressure in terms of mean kinetic energy, then solve for N; you never need the temperature here.
- Pressure from KE. Kinetic theory gives P = 1/3N/V m v̅²̅. Since the mean kinetic energy per molecule is E = 1/2 m v̅²̅, we have m v̅²̅ = 2E, so P = 2/3N/VE.
N = 3PV/2E
- Put in numbers. P = 2 atm ≈ 2 × 10⁵ Pa and V = 1 L = 10⁻³ m³, so PV ≈ 200 J. Then N = (3 × 200)/(2 × 2 × 10⁻⁹) = 600/(4 × 10⁻⁹) = 1.5 × 10¹¹.
- Why not use T. The energy given per molecule already fixes the pressure relation. Using N = PV/kT instead gives about 4.9 × 10²², which is not an option, so the question intends you to use E.
- So. The number of molecules is 1.5 × 10¹¹, option (c).
The idea to keep. Remember PV = 2/3NE, where E is the mean KE per molecule. When a question gives E directly, use this form and keep track of the factor 2/3.
Answer from the NTA final answer key. Solution by PrepWiser.