JEE Main 2020 · 5 September · morning · Single correct · +4 / −1

Number of molecules in a volume of 4 cm³ of a perfect monoatomic gas at some temperature T and at a pressure of 2 cm of mercury is close to?
(Given, mean kinetic energy of a molecule
(at T) is 4 × 10 –14 erg, g = 980 cm/s², density of
mercury = 13.6 g/cm³)

  1. (a) 5.8 × 10¹⁸
  2. (b) 4.0 × 10¹⁶
  3. (c) 5.8 × 10¹⁶
  4. (d) 4.0 × 10¹⁸
Answer and solution

Answer (d) 4.0 × 10¹⁸

Solution

Physics sir · 4 steps

Work fully in CGS: the mean kinetic energy gives kT, and then N = PV/kT.

  1. Pressure in CGS. A mercury column of height h exerts P = hρ g. So P = 2 × 13.6 × 980 = 26656 dyn cm⁻².
  2. Find kT. For a monoatomic gas the mean kinetic energy of a molecule is 3/2kT. So kT = 2/3 × 4 × 10⁻¹⁴ = 2.67 × 10⁻¹⁴ erg.

    PV = NkT ⇒ N = PV/kT

  3. Put in numbers. PV = 26656 × 4 = 1.066 × 10⁵ erg. Then N = (1.066 × 10⁵)/(2.67 × 10⁻¹⁴) = 4.0 × 10¹⁸.
  4. So. The gas holds about 4.0 × 10¹⁸ molecules, which is option (d).

The idea to keep. Mean kinetic energy per molecule is 3/2kT, so it hands you kT directly. Then N = PV/kT, as long as P, V and energy all sit in one unit system.

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

Study Kinetic Theory of Gases

Asked about 130 times in JEE Main since 2020. This question is from 2020.