If E is the energy density in an ideal gas, then the pressure of the ideal gas is

Options

A. `P = (2)/(3) E`
B. `P = (3)/(2) E`
C. `P = (5)/(2) E`
D. `P = (2)/(5) E`

Answer

To find the pressure \( P \) of an ideal gas in terms of the energy density \( E \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand Energy Density**:
Energy density \( E \) is defined as the kinetic energy per unit volume of the gas.

2. **Relate Pressure and Kinetic Energy**:
The pressure \( P \) of an ideal gas can be related to the average kinetic energy of its particles. For an ideal gas, the average kinetic energy per molecule is given by:
\[ KE = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant and \( T \) is the temperature.

3. **Express Kinetic Energy in Terms of Density**:
The total kinetic energy of the gas can be expressed as:
\[ KE_{total} = \frac{3}{2} N k T \]
where \( N \) is the number of molecules. The energy density \( E \) can then be expressed as:
\[ E = \frac{KE_{total}}{V} = \frac{3}{2} \frac{N k T}{V} \]

4. **Relate Number of Molecules to Density**:
The number of molecules \( N \) can be related to the density \( \rho \) of the gas:
\[ N = \frac{\rho V}{m} \]
where \( m \) is the mass of one molecule.

5. **Substitute into Energy Density**:
Substituting \( N \) into the expression for energy density:
\[ E = \frac{3}{2} \frac{\rho V k T}{m V} = \frac{3}{2} \frac{\rho k T}{m} \]

6. **Relate Pressure to Energy Density**:
From the ideal gas law, we know that:
\[ P = \frac{1}{3} \rho v_{rms}^2 \]
where \( v_{rms} \) is the root mean square speed of the gas molecules.

7. **Express \( v_{rms} \) in Terms of Energy Density**:
The root mean square speed can also be related to energy density:
\[ E = \frac{1}{2} \rho v_{rms}^2 \]
Rearranging gives:
\[ v_{rms}^2 = \frac{2E}{\rho} \]

8. **Substituting Back to Find Pressure**:
Substitute \( v_{rms}^2 \) back into the pressure equation:
\[ P = \frac{1}{3} \rho \left(\frac{2E}{\rho}\right) = \frac{2E}{3} \]

9. **Final Result**:
Thus, the pressure \( P \) of the ideal gas in terms of the energy density \( E \) is:
\[ P = \frac{2E}{3} \]

### Conclusion:
The pressure of the ideal gas is given by:
\[ \boxed{P = \frac{2E}{3}} \]