What is the average kinetic energy of a molecule of an ideal gas at temperatures of 27°c? what is the total random kinetic energy of the molecule in 1 mole of this gas .what is the RMS of speed of oxygen molecule .​

Respuesta :

(a) The average kinetic energy of a molecule of an ideal gas is 6.21 x 10⁻²¹ J.

(b) The total random kinetic energy of the molecule in 1 mole of this gas is 3,741.3 J.

(c) The RMS speed of the oxygen molecule is 215.25 m/s.

Average kinetic energy of ideal gas

The average kinetic energy of a molecule of an ideal gas is calculated as follows;

[tex]K .E = \frac{3}{2} K T[/tex]

Where;

  • K is Boltzmann constant = 1.38066 x 10⁻²³ J/K
  • T is temperature in Kelvin, = 27°c  + 273 = 300 K

[tex]K.E = \frac{3}{2} \times (1.38 066 \times 10^{-23}) \times 300\\\\K.E = 6.21 \times 10^{-21} \ J[/tex]

Total random kinetic energy

The total random kinetic energy of the molecule in 1 mole of this gas is calculated as follows;

[tex]K.E = \frac{3}{2} nRT\\\\K.E = \frac{3}{2} (1) (8.314)(300)\\\\K.E = 3,741.3 \ J[/tex]

RMS of speed of oxygen molecule

The RMS speed of the oxygen molecule is calculated as follows;

[tex]K.E = \frac{1}{2} mV_{rms}^2\\\\V_{rms}^2 = \frac{2K.E}{m} \\\\V_{rms} = \sqrt{\frac{2 K.E}{m} }[/tex]

one mole of oxygen gas = 32 g = 0.032 kg

[tex]V_{rms} = \sqrt{\frac{2 \times 741.3 }{0.032} }\\\\V_{rms} = 215.25 \ m/s[/tex]

Learn more about average kinetic energy of gas molecules here: https://brainly.com/question/952575

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