Question

Pressure exerted by a perfect gas is equal to:
A.)mean kinetic energy per unit volume.
B.)half of mean kinetic energy per unit volume.
C.)two third of mean kinetic energy per unit volume.
D.)one third of mean kinetic energy per unit volume.

Answer 1

Hint: Recollect the kinetic theory equations for pressure and kinetic energy of a molecule.$P = \dfrac{1}{3}nm{\bar v^2}$ and$K{E_{avg}} = \dfrac{1}{2}m{\bar v^2}$. Also, we should be taking the root mean square average.

Complete step-by-step answer:
Let us consider an Ideal gas enclosed in a chamber of volume $V$, containing $N$ gas particles. We assume the chamber to contain only a single type of atom. So all the particles inside would have the same mass$m$. The particles inside the chamber would be travelling with different velocities and colliding with the walls of the container. This results in an outward force on each wall, which we call the pressure. Also, for further calculation purposes, let’s call the speed of ${i^{th}}$ particle${v_i}$.

We know the pressure inside an Ideal gas chamber is given by $P = \dfrac{1}{3}nm{\bar v^2}$ where $\bar v$ is the root mean square velocity, defined as
${\bar v^2} = \dfrac{{{v_1}^2 + {v_2}^2 + {v_3}^2 + \cdot \cdot \cdot {v_N}^2}}{N}$ - (1)

Let us substitute this root mean square expansion into equation (1).

$P = \dfrac{1}{3}nm{\text{ }}\left( {\dfrac{{{v_1}^2 + {v_2}^2 + {v_3}^2 + \cdot \cdot \cdot {v_N}^2}}{N}} \right)$ - (2)

Apart from this, we also know that the kinetic energy of a molecule is given as $\dfrac{1}{2}m{v^2}$. That means the total kinetic energy of the system can be said as the sum of all such individual kinetic energies. So

$K{E_{TOTAL}} = \dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}m{v_2}^2 + \dfrac{1}{2}m{v_3}^2 + \cdot \cdot \cdot = \dfrac{1}{2}m\left( {{v_1}^2 + {v_2}^2 + {v_3}^2 + \cdot \cdot \cdot {v_N}^2} \right)$ .
We see some similarities in equations and Multiplying and dividing equation with 2 gives

$P = \dfrac{2}{3}n\dfrac{1}{2}m{\text{ }}\left( {\dfrac{{{v_1}^2 + {v_2}^2 + {v_3}^2 + \cdot \cdot \cdot {v_N}^2}}{N}} \right)$
Substituting the value of $K{E_{TOTAL}}$from gives:

$P = \dfrac{2}{3}n\dfrac{{K{E_{TOTAL}}}}{N}$
Now, since $n$ is the number density or the number of molecules per unit volume, $n = \dfrac{N}{V}$ where $V$ is the volume of the container.
So we see that $P = \dfrac{2}{3}\dfrac{N}{V}\dfrac{{K{E_{TOTAL}}}}{N} = \dfrac{2}{3}\dfrac{{K{E_{TOTAL}}}}{V}$
Thus, pressure is two third the average Kinetic energy per unit volume, Option C is correct.

Additional Information:
Since the Total Kinetic energy of molecules is what we call the internal energy of gas, This gives a relation between the internal energy and Pressure.
Since we already know the relation connecting pressure with temperature, we can also find a relationship between Internal energy and temperature.

Note: The above result should not be misinterpreted as two-third of kinetic energy is with pressure. The result only says that these two quantities are numerically equal. This is a good example to prove that Dimensional equality does not imply physical equality.