# The average translational kinetic energy of air molecules is 0.040 eV. Calculate the temperature...

## Question:

The average translational kinetic energy of air molecules is {eq}\rm 0.040\ eV {/eq}. Calculate the temperature of the air.

## Kinetic Energy:

Energy which is connected with the movement of an object is its kinetic energy. It is sometimes also called as translational kinetic energy since translational motion can be related to the movement of an object.

Given data

• Average translational kinetic energy of air molecule is {eq}0.040\;{\rm{eV}} {/eq}

Average translational kinetic energy of a gas molecule can be calculated using the formula,

{eq}{E_{avg}} = \dfrac{3}{2}{\rm{k}}T {/eq}

From this equation the temperature can be found out by,

{eq}T = \dfrac{2}{3} \times \dfrac{{{E_{avg}}}}{{\rm{k}}}......\left( {\rm{I}} \right) {/eq}

Since {eq}{\rm{1}}\;{\rm{eV = 1}}{\rm{.6}} \times {\rm{1}}{{\rm{0}}^{ - 19}}{\rm{J}} {/eq}

{eq}\begin{align*} 0.040\;{\rm{eV }} &= {\rm{ 0}}{\rm{.040}} \times {\rm{1}}{\rm{.6}} \times {\rm{1}}{{\rm{0}}^{ - 19}}{\rm{J}}\\ &= {\rm{6}}{\rm{.4}} \times {\rm{1}}{{\rm{0}}^{ - 21}}{\rm{J}} \end{align*} {/eq}

Boltzmann constant, k is given as,

{eq}{\rm{k}} = 1.38 \times {10^{ - 23}}{\rm{J }}{{\rm{K}}^{ - 1}} {/eq}

Substituting these values in equation (I),

{eq}\begin{align*} T &= \dfrac{2}{3} \times \dfrac{{6.40 \times {{10}^{ - 21}}}}{{1.38 \times {{10}^{ - 23}}}}\\ &= 309.17\;{\rm{K}}\\ &= 309.2\;{\rm{K}} \end{align*} {/eq}

Therefore, the temperature of the air is 309.2 K