An ideal gas is at a temperature of 320 K. What is the average translational kinetic energy of one of its molecules

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An ideal gas is at a temperature of 320 K. What is the average translational kinetic energy of one of its molecules

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Sigridomena 4 years 2021-08-07T06:00:38+00:00 1 Answers 45 views 0

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  1. Farah
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    2021-08-07T06:02:26+00:00
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    Answer:

    6.624 x 10^-21 J

    Explanation:

    The temperature of the ideal gas = 320 K

    The average translational energy of an ideal gas is gotten as

    K_{ave} = \frac{3}{2}K_{b}T

    where

    K_{ave}  is the average translational energy of the molecules

    K_{b} = Boltzmann constant = 1.38 × 10^-23 m^2 kg s^-2 K^-1

    T is the temperature of the gas = 320 K

    substituting value, we have

    K_{ave} = \frac{3}{2} * 1.38*10^{-23} * 320 = 6.624 x 10^-21 J

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