The internal energy of a gas is given by `U=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

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between internal energy, work done, and heat absorbed.
According to the first law of thermodynamics, the relationship is given by:
\[ \Delta Q = \Delta U + \Delta W \]
where:
- \(\Delta Q\) is the heat absorbed by the gas,
- \(\Delta U\) is the change in internal energy,
- \(\Delta W\) is the work done by the gas.

### Step 2: Calculate the change in internal energy (\(\Delta U\)).
The internal energy \(U\) is given by:
\[ U = 2pV \]
The gas expands from \(V_0\) to \(2V_0\) against a constant pressure \(p_0\).

First, we find the initial and final internal energy:
- Initial internal energy when \(V = V_0\):
\[ U_i = 2p_0 V_0 \]
- Final internal energy when \(V = 2V_0\):
\[ U_f = 2p_0 (2V_0) = 4p_0 V_0 \]

Now, we can calculate the change in internal energy:
\[ \Delta U = U_f - U_i = 4p_0 V_0 - 2p_0 V_0 = 2p_0 V_0 \]

### Step 3: Calculate the work done (\(\Delta W\)).
Since the process is isobaric (constant pressure), the work done by the gas can be calculated as:
\[ \Delta W = p \Delta V \]
where \(\Delta V = V_f - V_i = 2V_0 - V_0 = V_0\).

Thus, the work done is:
\[ \Delta W = p_0 \cdot V_0 \]

### Step 4: Substitute \(\Delta U\) and \(\Delta W\) into the first law equation.
Now we can substitute our values into the first law of thermodynamics:
\[ \Delta Q = \Delta U + \Delta W = (2p_0 V_0) + (p_0 V_0) = 3p_0 V_0 \]

### Final Answer
The heat absorbed by the gas in the process is:
\[ \Delta Q = 3p_0 V_0 \]

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