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 need to find the heat absorbed by the gas during its expansion from volume \( V_0 \) to \( 2V_0 \) against a constant pressure \( P_0 \).

### Step-by-Step Solution:

1. **Identify the Change in Internal Energy (\( \Delta U \))**:
The internal energy of the gas is given by the equation:
\[ U = 5 + 2PV \]
Since the pressure is constant at \( P_0 \), we can express the change in internal energy as:
\[ \Delta U = U_{final} - U_{initial} \]
where \( U_{initial} = 5 + 2P_0V_0 \) and \( U_{final} = 5 + 2P_0(2V_0) \).

Calculating these:
\[ U_{initial} = 5 + 2P_0V_0 \]
\[ U_{final} = 5 + 4P_0V_0 \]
Therefore, the change in internal energy is:
\[ \Delta U = (5 + 4P_0V_0) - (5 + 2P_0V_0) = 2P_0V_0 \]

2. **Calculate the Work Done (\( W \))**:
The work done by the gas during expansion at constant pressure is given by:
\[ W = P \Delta V \]
Here, \( \Delta V = V_{final} - V_{initial} = 2V_0 - V_0 = V_0 \).
Thus, the work done is:
\[ W = P_0 \times V_0 \]

3. **Apply the First Law of Thermodynamics**:
The first law of thermodynamics states:
\[ Q = \Delta U + W \]
where \( Q \) is the heat absorbed by the gas. Substituting the values we calculated:
\[ Q = \Delta U + W = 2P_0V_0 + P_0V_0 \]
Combining these gives:
\[ Q = 3P_0V_0 \]

### Final Answer:
The heat absorbed by the gas in the process is:
\[ Q = 3P_0V_0 \]