A cubical box with sides of length 0.368 m contains 1.980 moles of neon gas at a temperature of 298 K. What is the average rate (in atoms/s) at which neon atoms collide with one side of the container? The mass of a single neon atom is 3.35x10-26 kg.

Respuesta :

The average rate at which the neon atoms collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].

The given parameters;

  • length of the cubical box, L = 0.368
  • number of moles of the gas, n = 1.98
  • temperature, T = 298 K
  • mass of the gas, m = 3.35 x 10⁻²⁶ kg

The average kinetic energy of the gas molecules is calculated as follows;

[tex]K = \frac{3}{2} \frac{R}{N_a} T\\\\K = \frac{3 \times 8.314\times 298}{2\times 6.022 \times 10^{23}} \\\\K = 6.17\times 10^{-21} \ J/atoms[/tex]

The average speed of the gas molecules is calculated as follows;

[tex]K = \frac{1}{2}mv_{rms}^2\\\\v_{rms} = \sqrt{\frac{2K}{m} } \\\\v_{rms} = \sqrt{\frac{2\times 6.17 \times 10^{-21}}{3.35\times 10^{-26}} } \\\\v_{rms} = 607 \ m/s[/tex]

The time of collision of the gas molecules with the walls of the container is calculated as follows;

[tex]t = \frac{2d}{v} \\\\t = \frac{2\times 0.368}{607} \\\\t = 0.0012 \ s[/tex]

The average rate at which the gases collide with a single wall out of the 3 identical walls is calculated as follows;

[tex]rate =\frac{1}{3} \times \frac{n \times N_a}{t} \\\\rate = \frac{1.98 \times 6.02 \times 10^{23} \ atoms}{3 \times 0.0012 \ s} \\\\rate = 3.31 \times 10^{26} \ atoms/s[/tex]

Thus, the average rate at which the gases collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].

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