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A cubical box of side 2 ~m contains helium gas. It was observed that in a time of 1 second, an atom travelling with the root-mean-square speed parallel to one of the edges of the cube, made 250 hits with one of the walls, without any collision with other atoms. The average kinetic energy of the helium gas is Take R= 25 3 ~J / mol - K and kB =1.38 10⁻²³ JK ⁻¹

Options

  1. A3.31 10⁻²¹ ~J
  2. B82.8 10⁻²¹ ~J
  3. C1 10⁻²¹ ~J
  4. D82.8 10⁻¹⁹ ~J

Correct answer

A. 3.31 10⁻²¹ ~J

Step-by-step solution

The side length of the cubical box is L = 2 m . An atom travelling with root-mean-square speed v_ rms parallel to one of the edges makes 250 hits in 1 second. The distance covered by the atom between two consecutive hits on the same wall is 2L . The time taken for one round trip is t = 2L v_ rms . The number of hits per second is n = 1 t = v_ rms 2L . Given n = 250 s ⁻¹ and L = 2 m , we have 250 = v_ rms 2 2 = v_ rms 4 . v_ rms = 250 4 = 1000 m/s . The average kinetic energy of a single helium atom is K_ avg = 1 2

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