Question

What is the average translational kinetic energy of molecules in an ideal gas at ${37^0}C$.$?$

Answer 1

First we have to know what the average translational kinetic energy of an ideal gas. Then we need to convert the temperature from celsius to kelvin by adding $273$. Using the kinetic energy of gas formula $E = \dfrac{3}{2}{k_B}T$ find the average translational kinetic energy of an ideal gas at ${37^0}C$.

Complete answer:
The translational kinetic energy of an object (rigid body) of a given mass is the work required to accelerate it from rest to a given velocity. The word translational refers to motion along a linear path, from one point to another.
The difference between rotational and translational kinetic energy is that translational is straight line motion while rotational is not. The rotational motion of the tire means it has rotational kinetic energy while the movement of the bike along the path means the tire also has translational kinetic energy.
A pure substance in the gaseous state contains more energy than in the liquid state, which in turn contains more energy than in the solid state. Particles has the highest kinetic energy when they are in the gaseous state. Kinetic energy is related to heat (also called thermal energy).
Kinetic Energy of Gas Formula: In an ideal gas, there are no attractive forces between the gas molecules, and there is no rotation or vibration within the molecules. The kinetic energy of the translational motion of an ideal gas depends on its temperature. The formula for the kinetic energy of a gas defines the average kinetic energy per molecule. The kinetic energy is measured in Joules ($J$), and the temperature is measured in Kelvin ($K$).
Average kinetic energy per molecule$= \dfrac{3}{2}$(Boltzmann’s constant) $\times$ (Temperature)
$E = \dfrac{3}{2}{k_B}T$
Where $E =$ average kinetic energy per molecule of gas ($J$)
${k_B} =$Boltzmann's constant ($1.38 \times {10^{ - 23}}J/K$)
$T =$ temperature ($K$)
Given $T = {37^o}C$ then $T = (37 + 273) = 310K$
then $E = \dfrac{3}{2} \times 1.38 \times {10^{ - 23}} \times 310 = 641.7 \times {10^{ - 23}}J$.

Note:
Note that Boltzmann’s constant is the physical constant relating energy at the particle level with temperature observed at the bulk level. It is the gas constant $R$ divided by Avogadro’s number ${N_A}$. Noble gas is any of the elements of group 18 of the periodic table, being monatomic and inert.