What is the average kinetic energy of the molecules in a gas at 273K?

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What is the average kinetic energy of the molecules in a gas at 273K?

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Answer 1 · 2014-07-31T23:32:46

The average kinetic energy of the molecules is 5.66 × 10⁻²¹ J.

According to Kinetic Molecular Theory, the average kinetic energy of gas molecules is a function only of temperature.

The formula is

$K E = \frac{3}{2} k T$ $KE = 3/2kT$

where $T$ $T$ is the Kelvin temperature and $k$ $k$ is Boltzmann's constant.

$K E = \frac{3}{2} \times 1.381 \times 10 ⁻ ² ³ J\cdotK⁻¹ \times 273 K$ $KE = 3/2 × 1.381 × 10⁻²³ "J·K⁻¹" × 273"K"$ = 5.66 × 10⁻²¹ J

For a mole of molecules, the average kinetic energy is

$\frac{5.66 \times 10 ⁻ ² ¹ J}{1 molecule} \times \frac{6.022 \times 10 ² ³ molecules}{1 mol}$ $(5.66 × 10⁻²¹"J")/(1"molecule") × (6.022 × 10²³"molecules")/(1"mol")$ = 3.41 ×10³ J/mol = 3.41 kJ/mol

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What is the average kinetic energy of the molecules in a gas at 273K?

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