A monoatomic gas at a temperature T has pressure P and heat energy per unit volume E. then

Options

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

Explanation

To solve the problem, we need to establish the relationship between the pressure (P), heat energy per unit volume (E), and the temperature (T) of a monoatomic gas.

### Step-by-Step Solution:

1. **Understanding the Kinetic Energy of a Monoatomic Gas**:
The kinetic energy (KE) of a monoatomic gas can be expressed in terms of pressure and volume. The formula for the total kinetic energy of the gas is given by:
\[ KE = \frac{3}{2} PV \]
where \(P\) is the pressure and \(V\) is the volume of the gas.

2. **Kinetic Energy per Unit Volume**:
To find the kinetic energy per unit volume, we divide the total kinetic energy by the volume \(V\):
\[ KE \text{ per unit volume} = \frac{KE}{V} = \frac{3}{2} P \]

3. **Relating Kinetic Energy to Heat Energy**:
In thermodynamics, the kinetic energy of the gas molecules is related to the heat energy. For a monoatomic ideal gas, the heat energy per unit volume \(E\) is equal to the kinetic energy per unit volume:
\[ E = KE \text{ per unit volume} \]

4. **Substituting the Expression for Kinetic Energy**:
From the previous steps, we can substitute the expression for kinetic energy per unit volume into the equation for heat energy:
\[ E = \frac{3}{2} P \]

5. **Rearranging the Equation**:
To express pressure \(P\) in terms of heat energy \(E\), we rearrange the equation:
\[ P = \frac{2}{3} E \]

6. **Conclusion**:
Thus, we have established that the relationship between the heat energy per unit volume \(E\) and the pressure \(P\) of a monoatomic gas at temperature \(T\) is:
\[ E = \frac{3}{2} P \quad \text{or} \quad P = \frac{2}{3} E \]

### Final Answer:
The correct relationship is:
\[ P = \frac{2}{3} E \]
So, the correct option is B.