An ideal gas `((C_(p))/(C_(v))=gamma)`has initial volume `V_(0)` is kept in a vessel. It undergoes a change and follows the following relation `P = kV^(2)` (where `P` is pressure, and `V` is volume) find the change in internal energy of the gas if its final pressure is `P_(0)`:

Explanation

To solve the problem, we will follow a systematic approach to find the change in internal energy of the ideal gas under the given conditions.

### Step-by-Step Solution:

1. **Understanding the Relationship Between \( C_p \) and \( C_v \)**:
We know that for an ideal gas, the ratio of specific heats is given by:
\[ \frac{C_p}{C_v} = \gamma \]
We also have the relation:
\[ C_p = C_v + R \]
From this, we can express \( C_v \) in terms of \( R \) and \( \gamma \):
\[ \gamma = \frac{C_v + R}{C_v} \implies C_v(\gamma - 1) = R \implies C_v = \frac{R}{\gamma - 1} \]

2. **Internal Energy Formula**:
The change in internal energy \( \Delta U \) for an ideal gas is given by:
\[ \Delta U = n C_v \Delta T \]
Substituting the expression for \( C_v \):
\[ \Delta U = n \left(\frac{R}{\gamma - 1}\right) \Delta T \]

3. **Using the Ideal Gas Law**:
The ideal gas law states:
\[ PV = nRT \]
Rearranging gives us:
\[ nR = \frac{PV}{T} \]
For a change, we can express this as:
\[ \Delta(PV) = nR \Delta T \]

4. **Relating Pressure and Volume**:
From the problem, we know that:
\[ P = kV^2 \]
Therefore, we can express the change in \( PV \) as:
\[ \Delta(PV) = P_f V_f - P_i V_i \]
where \( P_f \) and \( P_i \) are the final and initial pressures, and \( V_f \) and \( V_i \) are the final and initial volumes.

5. **Substituting Values**:
Given that the final pressure \( P_f = P_0 \) and the initial pressure \( P_i = kV_0^2 \), we can write:
\[ V_f = \sqrt{\frac{P_0}{k}} \]
Therefore, substituting these values into the equation for \( \Delta U \):
\[ \Delta U = \frac{\Delta(PV)}{\gamma - 1} = \frac{P_f V_f - P_i V_i}{\gamma - 1} \]

6. **Calculating \( \Delta U \)**:
Now substituting the values:
\[ \Delta U = \frac{P_0 \sqrt{\frac{P_0}{k}} - kV_0^2 V_0}{\gamma - 1} \]
Simplifying this gives:
\[ \Delta U = \frac{P_0^{3/2}/\sqrt{k} - kV_0^3}{\gamma - 1} \]

### Final Expression:
Thus, the change in internal energy of the gas is:
\[ \Delta U = \frac{P_0^{3/2}/\sqrt{k} - kV_0^3}{\gamma - 1} \]