1. Use the ideal gas relation between internal energy and temperature

For an ideal gas, $U = n C_V T$ and also $PV = nRT$ So, $U = \frac{C_V}{R} PV$

The given relation is $U = 3PV + 4$

For an ideal gas, internal energy depends only on temperature, so the physically relevant proportional form is $U \propto PV$ Hence the constant term $+4$ is irrelevant/unphysical for classification, and we compare the coefficient of $PV$: $\frac{C_V}{R} = 3$ So, $C_V = 3R$

2. Relate $C_V$ to degrees of freedom

For an ideal gas, $C_V = \frac{f}{2}R$ where $f$ is the number of degrees of freedom.

Thus, $\frac{f}{2}R = 3R$ $\frac{f}{2} = 3$ $f = 6$

3. Identify the type of gas

• Monoatomic gas: $f=3 \Rightarrow C_V = \frac{3}{2}R$
• Diatomic gas (ordinary temperature, vibrations neglected): $f=5 \Rightarrow C_V = \frac{5}{2}R$
• Polyatomic gas (non-linear): $f=6 \Rightarrow C_V = 3R$

Since we got $f=6$, the gas is polyatomic.

4. Check options

• A: either monoatomic or diatomic — false
• B: monoatomic only — false
• C: polyatomic only — true
• D: diatomic only — false

Therefore, the correct option is C.