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

Options

A. (a) `2p_0V_0`
B. (b) `4p_0V_0`
C. (c) `3p_0V_0`
D. (d) `p_0V_0`

Explanation

To solve the problem, we will use the first law of thermodynamics, which states:

\[ Q = \Delta U + W \]

where \(Q\) is the heat absorbed by the gas, \(\Delta U\) is the change in internal energy, and \(W\) is the work done by the gas.

### Step 1: Calculate the Work Done (W)
The gas expands against a constant pressure \(p_0\) from an initial volume \(V_0\) to a final volume \(2V_0\). The work done by the gas during this expansion can be calculated using the formula:

\[ W = p_0 \times (V_{final} - V_{initial}) \]

Substituting the values:

\[ W = p_0 \times (2V_0 - V_0) = p_0 \times V_0 \]

### Step 2: Calculate the Change in Internal Energy (\(\Delta U\))
The internal energy \(U\) of the gas is given by:

\[ U = 2pV \]

We need to find the change in internal energy as the volume changes from \(V_0\) to \(2V_0\):

\[ \Delta U = U_{final} - U_{initial} \]

Calculating \(U_{final}\) and \(U_{initial}\):

\[ U_{final} = 2p_0 \times (2V_0) = 4p_0V_0 \]
\[ U_{initial} = 2p_0 \times V_0 = 2p_0V_0 \]

Now, substituting these into the change in internal energy:

\[ \Delta U = 4p_0V_0 - 2p_0V_0 = 2p_0V_0 \]

### Step 3: Substitute into the First Law of Thermodynamics
Now we can substitute the values of \(\Delta U\) and \(W\) into the first law equation:

\[ Q = \Delta U + W \]
\[ Q = 2p_0V_0 + p_0V_0 \]

Combining the terms:

\[ Q = 3p_0V_0 \]

### Final Answer
The heat absorbed by the gas in the process is:

\[ Q = 3p_0V_0 \]

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