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What is the average translational kinetic energy of a molecule of an ideal gas at temperature of ${{27}^{0}}C$.

Last updated date: 17th Jun 2024
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Hint: Molecules of gas strike the walls of the container and each other randomly. The pressure exerted by gas on the walls is due to their collision. The force exerted is equal and opposite. We can calculate the average force using impulse and also calculate the average velocity of molecules then we can calculate pressure using force and use it to calculate kinetic energy by ideal gas law.

Formulae used:
$I=F\Delta t$

Complete step-by-step solution:
Let us assume a cube shaped box of length $L$filled with molecules of an ideal gas. When a molecule collides with the wall of the box, it rebounds with the same kinetic energy and when two molecules of an ideal gas collide, collisions are elastic this means kinetic energy is conserved.

The change in momentum when the gas molecules collide with the wall is-
  & \Delta p=m{{v}_{x}}-(-m{{v}_{x}}) \\
 & \Rightarrow \Delta p=2m{{v}_{x}} \\

We know that impulse is the product of force and time taken to exert the force, it is given by-
$I=F\Delta t$
Here, $I$ is the impulse
$F$ is the force exerted
$\Delta t$ is time taken

Impulse is also calculated as change in momentum-
$F\Delta t=\Delta p$
$\Rightarrow F\Delta t=2m{{v}_{x}}$ - (1)

Since the molecules after colliding with a wall having the same kinetic energy move across the container and then come back after collision with another wall, so the total distance it covers is $2L$.
We know that-
Here, $v$ is the velocity
$d$ is distance covered
$t$ is time taken

Therefore, substituting values of the ideal gas molecule in the above equation, we get,
  & {{v}_{x}}=\dfrac{2L}{\Delta t} \\
 & \therefore \Delta t=\dfrac{2L}{{{v}_{x}}} \\

Substituting in eq (1), we get,
  & F\Delta t=2m{{v}_{x}} \\
 & \Rightarrow F\times \dfrac{2L}{{{v}_{x}}}=2m{{v}_{x}} \\
 & \therefore F=\dfrac{mv_{x}^{2}}{L} \\

This is the average force exerted by the molecule on the wall. The total force exerted on the wall will be the sum of forces exerted by the molecules on the wall, therefore,

Multiplying and dividing by $N$, number of molecules on RHS, we get,
$\Rightarrow F=\dfrac{Nm}{L}v{{_{x}^{2}}_{rms}}$ - (2)

The force on the wall can be expressed in terms of average velocity with all components, ${{v}_{x}},\,{{v}_{y}},\,{{v}_{z}}$.
  & v_{rms}^{2}=v{{_{x}^{2}}_{rms}}+v{{_{y}^{2}}_{rms}}+v{{_{z}^{2}}_{rms}}=3v{{_{x}^{2}}_{rms}} \\
 & \Rightarrow \dfrac{v_{rms}^{2}}{3}=v{{_{x}^{2}}_{rms}} \\

Substituting in eq (2), we get,
  & F=\dfrac{Nm}{L}v{{_{x}^{2}}_{rms}} \\
 & \Rightarrow F=\dfrac{Nm}{3L}v_{rms}^{2} \\
We know that,
$P$ is pressure
$F$ is force
$A$ is area

Therefore, we substitute given values in the above equation to get,
  & P=\dfrac{\dfrac{Nm}{3L}v_{rms}^{2}}{{{L}^{2}}} \\
 & \Rightarrow P=\dfrac{Nmv_{rms}^{2}}{3{{L}^{3}}} \\
$V={{L}^{3}}$ therefore,
  & \Rightarrow P=\dfrac{Nmv_{rms}^{2}}{3{{L}^{3}}} \\
 & \Rightarrow PV=\dfrac{Nmv_{rms}^{2}}{3} \\
 & \Rightarrow PV=\dfrac{2N}{3}\left( \dfrac{1}{2}mv_{rms}^{2} \right) \\
 & \therefore PV=\dfrac{2N}{3}K{{E}_{av}} \\
Here, $K{{E}_{av}}$ is the average kinetic energy of a molecule

From the ideal gas law, we have,
$PV=NkT$ Substituting in above equation, we have,
$k$ is the Boltzmann’s constant
$T$ is the temperature
  & NkT=\dfrac{2N}{3}K{{E}_{av}} \\
 & \Rightarrow kT=\dfrac{2}{3}K{{E}_{av}} \\
 & \therefore K{{E}_{av}}=\dfrac{3}{2}kT \\

Therefore the average kinetic energy of an ideal gas molecule is $\dfrac{3}{2}kT$.

The molecules have components of velocity in three directions, x-axis, y-axis and z-axis. The movement of molecules of ideal gas inside the container is completely random. For a monatomic gas, as there is negligible attraction between molecules, the kinetic energy is the total internal energy or the gas.