The internal energy of a gas is given by `U= 5 + 2PV`. It expands from `V_(0)` to `2V_(0)` against a constant pressure `P_(0)`. The heat absorbed by the gas in the process is :-

Options

A. `-3P_(0)V_(0)`
B. `3P_(0)V_(0)`
C. `2P_(0)V_(0)`
D. `P_(0)V_(0)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the Work Done (W)
The work done by the gas during expansion against a constant pressure is given by the formula:
\[ W = P \Delta V \]
where \( \Delta V = V_f - V_i \). In this case, the initial volume \( V_i = V_0 \) and the final volume \( V_f = 2V_0 \). Therefore:
\[ \Delta V = 2V_0 - V_0 = V_0 \]
Substituting this into the work formula with \( P = P_0 \):
\[ W = P_0 \cdot V_0 \]

### Step 2: Calculate the Change in Internal Energy (ΔU)
The change in internal energy is given by:
\[ \Delta U = U_f - U_i \]
We know that the internal energy \( U \) is given by:
\[ U = 5 + 2PV \]
Now, we will calculate \( U_f \) and \( U_i \):
- For the final state (at volume \( V_f = 2V_0 \)):
\[ U_f = 5 + 2P_0(2V_0) = 5 + 4P_0V_0 \]
- For the initial state (at volume \( V_i = V_0 \)):
\[ U_i = 5 + 2P_0(V_0) = 5 + 2P_0V_0 \]
Now, substituting these into the change in internal energy:
\[ \Delta U = (5 + 4P_0V_0) - (5 + 2P_0V_0) = 4P_0V_0 - 2P_0V_0 = 2P_0V_0 \]

### Step 3: Apply the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ Q = \Delta U + W \]
Substituting the values we found for \( \Delta U \) and \( W \):
\[ Q = 2P_0V_0 + P_0V_0 = 3P_0V_0 \]

### Final Answer
Thus, the heat absorbed by the gas in the process is:
\[ \boxed{3P_0V_0} \]