Question

What is the average translational kinetic energy of molecules in an ideal gas at $37°\mathrm{C}$?

Answer 1

Translational kinetic energy

• The energy necessary to accelerate a rigid body from rest to a certain velocity is known as the object's translational kinetic energy.
• Motion from one point to another that follows a straight line is referred to as translational motion.

Ideal gas translational kinetic energy

• A pure material has more energy in its gaseous state than in its liquid state, which has more energy than in its solid-state.
• When particles are in a gaseous form, their kinetic energy is at its maximum.
• There are no attraction forces between the gas molecules in an ideal gas, and the molecules themselves are neither rotating nor vibrating.
• An ideal gas's temperature affects the kinetic energy of its translational motion.

Formula of average translational kinetic energy

• The average translational kinetic energy is as follows:

$\begin{array}{rcl}E& =& \frac{3}{2}{K}_B\times T\end{array}$

• $$E=$$average kinetic energy per molecule of gas.
• $$k_B=$$Boltzmann's constant.
• $$T=$$Temperature

Convert degrees Celsius into Kelvin

• The conversion of degrees Celsius into Kelvin is done as follows:

$\begin{array}{rcl}T\left(\mathrm{K}\right)& =& T\left(°\mathrm{C}\right)+273\end{array}$

• Convert $37°\mathrm{C}$ into Kelvin as follows:

$\begin{array}{rcl}T\left(\mathrm{K}\right)& =& T\left(°\mathrm{C}\right)+273\\ & =& 37+273\\ & =& 310\mathrm{K}\end{array}$

Calculate average translational kinetic energy

• The average translational kinetic energy of molecules in an ideal gas is as follows:

$\begin{array}{rcl}E& =& \frac{3}{2}{K}_B\times T\\ & =& \frac{3}{2}\times 1.38\times {10}^{-23}\mathrm{J}\bcancel{{\mathrm{K}}^{-1}}\times 310\bcancel{\mathrm{K}}\\ & =& \frac{1283.4\times {10}^{-23}\mathrm{J}}{2}\\ & =& 641.7\times {10}^{-23}\mathrm{J}\end{array}$

• The average translational kinetic energy of molecules in an ideal gas is $641.7\times {10}^{-23}\mathrm{J}$.