The internal energy (U), pressure (P) and volume (V) of an ideal gas are related as `U=3PV+4`. The gas is :-

Options

A. Diatomic only
B. Polyatomic only
C. Either monoatomic or diatomic
D. Either monoatomic or diatomic

Answer

To solve the problem, we need to analyze the relationship given between internal energy (U), pressure (P), and volume (V) of an ideal gas, which is expressed as:

\[ U = 3PV + 4 \]

### Step-by-Step Solution:

1. **Understanding Internal Energy**:
The internal energy \( U \) of an ideal gas is related to its degrees of freedom. The total internal energy can be expressed as:
\[ U = n \cdot C_v \cdot T \]
where \( n \) is the number of moles, \( C_v \) is the molar heat capacity at constant volume, and \( T \) is the temperature.

2. **Relating Pressure and Volume**:
From the ideal gas equation, we know:
\[ PV = nRT \]
where \( R \) is the universal gas constant.

3. **Substituting \( nRT \)**:
We can substitute \( nRT \) in the expression for internal energy:
\[ U = 3PV + 4 \]
Replacing \( nRT \) with \( PV \):
\[ U = 3PV + 4 = nC_vT \]

4. **Expressing \( C_v \)**:
Now, we can express \( C_v \) in terms of \( P \) and \( V \):
\[ U = nC_vT \implies C_v = \frac{U}{nT} = \frac{3PV + 4}{nT} \]

5. **Finding Degrees of Freedom**:
The degrees of freedom \( f \) of a gas is related to its internal energy. For an ideal gas:
\[ U = \frac{f}{2} nRT \]
Therefore, we can equate:
\[ \frac{f}{2} nRT = 3PV + 4 \]

6. **Using \( PV = nRT \)**:
Substitute \( nRT \) with \( PV \):
\[ \frac{f}{2} PV = 3PV + 4 \]

7. **Simplifying the Equation**:
Dividing both sides by \( PV \) (assuming \( PV \neq 0 \)):
\[ \frac{f}{2} = 3 + \frac{4}{PV} \]

8. **Finding \( f \)**:
Rearranging gives:
\[ f = 6 + \frac{8}{PV} \]
Since \( PV \) is positive, \( \frac{8}{PV} \) is also positive, thus:
\[ f > 6 \]

9. **Identifying the Type of Gas**:
- For a monoatomic gas, \( f = 3 \).
- For a diatomic gas, \( f = 5 \).
- For a polyatomic gas, \( f \) is generally greater than 6.

Since we have found that \( f > 6 \), the gas must be polyatomic.

### Conclusion:
The gas is a **polyatomic gas**.