A cubical box of side 1 m contains helium gas (atomic weight 4) at a pressure of ${100 Nm}^{- 2}$ . During an observation time of 1 second, an atom travelling with root- mean-square speed parallel to one of the edges of the cube was found to make 500 hits with a particular wall, without any collision with other atoms. Take R = $\frac{25}{3}$ ${J mol}^{- 1} K^{- 1}$ and k = $1.38 \times 1 0^{- 23} {J K}^{- 1}$ . Evaluate average kinetic energy.

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$8 .143 \times 1 0^{- 20} J$

$6 .4 12 \times 1 0^{- 21} J$

$6 .4 12 \times 1 0^{- 20} J$

$3.312 \times 1 0^{- 21} J$

answer is A.

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Detailed Solution

Distance travelled in one second $= 500 \times 2 m = 1000 m . S o, v_{R M S} = 1000 m s^{- 1}$
But $v_{R M S} = \sqrt{\frac{3 R T}{M}} \Rightarrow T = 160 K$
Average kinetic energy per atom = $\frac{3}{2} k T = \frac{3}{2} \times 1.38 \times 1 0^{- 23} \times 160 = 3.312 \times 1 0^{- 21} J$