Prove that the pressure of an ideal gas is numerically equal to two third of the mean translational kinetic energy per unit volume of the gas.

Explanation

To prove that the pressure of an ideal gas is numerically equal to two-thirds of the mean translational kinetic energy per unit volume of the gas, we can follow these steps:

### Step 1: Define Mean Translational Kinetic Energy
The mean translational kinetic energy (KE) of a gas molecule can be expressed as:
\[ KE = \frac{1}{2} m v^2 \]
where \(m\) is the mass of a gas molecule and \(v\) is its velocity.

### Step 2: Calculate Kinetic Energy per Unit Volume
To find the kinetic energy per unit volume, we need to consider the number of molecules in a unit volume. The density (\(\rho\)) of the gas is given by:
\[ \rho = n \cdot m \]
where \(n\) is the number of molecules per unit volume.

The total kinetic energy per unit volume (E) can be expressed as:
\[ E = \text{(number of molecules per unit volume)} \times \text{(mean kinetic energy per molecule)} \]
Thus,
\[ E = n \cdot KE = n \cdot \frac{1}{2} m v^2 \]

### Step 3: Relate Number Density to Pressure
Using the ideal gas law, we know:
\[ PV = nRT \]
where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature.

We can also express the number of molecules per unit volume as:
\[ n = \frac{P}{kT} \]
where \(k\) is Boltzmann's constant.

### Step 4: Substitute for Kinetic Energy
Now, substituting \(n\) into the expression for kinetic energy per unit volume:
\[ E = n \cdot \frac{1}{2} m v^2 = \frac{P}{kT} \cdot \frac{1}{2} m v^2 \]

### Step 5: Relate Velocity to Pressure
From kinetic theory, we can relate the average velocity squared to pressure:
\[ \frac{1}{3} \rho v^2 = P \]
Thus,
\[ v^2 = \frac{3P}{\rho} \]

### Step 6: Substitute Back into Kinetic Energy Expression
Now substituting \(v^2\) back into the expression for \(E\):
\[ E = n \cdot \frac{1}{2} m \left(\frac{3P}{\rho}\right) \]
Replacing \(n\) with \(\frac{\rho}{m}\):
\[ E = \frac{\rho}{m} \cdot \frac{1}{2} m \cdot \frac{3P}{\rho} = \frac{3P}{2} \]

### Step 7: Relate Pressure to Kinetic Energy
Now, we can express the pressure in terms of kinetic energy per unit volume:
\[ P = \frac{2}{3} E \]

### Conclusion
Thus, we have proved that the pressure of an ideal gas is numerically equal to two-thirds of the mean translational kinetic energy per unit volume of the gas:
\[ P = \frac{2}{3} E \]