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kotykmax [81]
4 years ago
7

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

Physics
1 answer:
Anastasy [175]4 years ago
3 0

Answer:

<em>6.624 x 10^-21 J</em>

<em></em>

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 = <em>6.624 x 10^-21 J</em>

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Answer:

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Explanation:

Information provided

Angle between the normal to the slip plane with tensile axis, \alpha=60^{o}&#10;

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