The internal energy of a gas is given by U = 3 pV. 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. `p_(0)V_(0)`
B. `2p_(0)V_(0)`
C. `3p_(0)V_(0)`
D. `4p_(0)V_(0)`

Answer

To solve the problem, we need to calculate the heat absorbed by the gas during its expansion from \( V_0 \) to \( 2V_0 \) against a constant pressure \( p_0 \).

### Step-by-Step Solution:

1. **Identify the given parameters:**
- Internal energy formula: \( U = 3pV \)
- Initial volume: \( V_i = V_0 \)
- Final volume: \( V_f = 2V_0 \)
- Constant pressure: \( p = p_0 \)

2. **Calculate the change in internal energy (\( \Delta U \)):**
- The change in internal energy is given by:
\[ \Delta U = U_f - U_i = 3pV_f - 3pV_i \]
- Substituting the values:
\[ \Delta U = 3p_0(2V_0) - 3p_0(V_0) = 3p_0(2V_0 - V_0) = 3p_0V_0 \]

3. **Calculate the work done (\( W \)):**
- The work done by the gas during expansion at constant pressure is given by:
\[ W = p \Delta V = p_0(V_f - V_i) \]
- Substituting the values:
\[ W = p_0(2V_0 - V_0) = p_0V_0 \]

4. **Apply the first law of thermodynamics:**
- The first law of thermodynamics states:
\[ Q = \Delta U + W \]
- Substituting the values we calculated:
\[ Q = 3p_0V_0 + p_0V_0 = 4p_0V_0 \]

5. **Final Result:**
- The heat absorbed by the gas during the process is:
\[ Q = 4p_0V_0 \]

### Conclusion:
The heat absorbed by the gas in the process is \( 4p_0V_0 \).