Pressure and volume of a gas changes from `(p_0V_0)` to `(p_0/4, 2V_0)` in a process `pV^2=` constant. Find work done by the gas in the given process.

Explanation

To find the work done by the gas during the process where the pressure and volume change from \( (p_0, V_0) \) to \( \left(\frac{p_0}{4}, 2V_0\right) \) under the condition that \( pV^2 = \text{constant} \), we can follow these steps:

### Step 1: Establish the constant
Given that the process follows the relation \( pV^2 = k \) (where \( k \) is a constant), we can find the value of \( k \) using the initial state of the gas.

\[ k = p_0 V_0^2 \]

### Step 2: Express pressure in terms of volume
From the equation \( pV^2 = k \), we can express pressure \( p \) in terms of volume \( V \):

\[ p = \frac{k}{V^2} \]

Substituting the value of \( k \):

\[ p = \frac{p_0 V_0^2}{V^2} \]

### Step 3: Set up the work done integral
The work done \( W \) by the gas during a volume change can be calculated using the integral:

\[ W = \int_{V_0}^{2V_0} p \, dV \]

Substituting the expression for \( p \):

\[ W = \int_{V_0}^{2V_0} \frac{p_0 V_0^2}{V^2} \, dV \]

### Step 4: Solve the integral
Now, we can solve the integral:

\[ W = p_0 V_0^2 \int_{V_0}^{2V_0} \frac{1}{V^2} \, dV \]

The integral of \( \frac{1}{V^2} \) is:

\[ \int \frac{1}{V^2} \, dV = -\frac{1}{V} \]

Thus, we evaluate the definite integral:

\[ W = p_0 V_0^2 \left[-\frac{1}{V}\right]_{V_0}^{2V_0} \]

Calculating the limits:

\[ W = p_0 V_0^2 \left(-\frac{1}{2V_0} + \frac{1}{V_0}\right) \]

This simplifies to:

\[ W = p_0 V_0^2 \left(\frac{1}{V_0} - \frac{1}{2V_0}\right) = p_0 V_0^2 \left(\frac{1}{2V_0}\right) \]

### Step 5: Final simplification
Now, simplifying further:

\[ W = p_0 V_0 \cdot \frac{1}{2} = \frac{p_0 V_0}{2} \]

### Conclusion
Thus, the work done by the gas during the process is:

\[ \boxed{\frac{p_0 V_0}{2}} \]