`A` gas is found to obey the law `P^(2)V`= constant. The initial temperature and volume are `T_(0)` and `V_(0^(@))` If the gas expands to a volume `3V_(0)`, then

Options

A. final temperature becomes `sqrt(3)T_(0)`
B. internal energy of the gas increases
C. final temperature becomes `(T_(0))/(sqrt(3))`
D. internal energy of the gas decreases

Answer

To solve the problem, we need to analyze the relationship between pressure, volume, and temperature of the gas that obeys the law \( P^2 V = \text{constant} \). We will also determine the final temperature and the change in internal energy when the gas expands to a volume of \( 3V_0 \).

### Step-by-Step Solution:

1. **Understanding the Law**:
The gas obeys the law \( P^2 V = \text{constant} \). This implies that for any two states of the gas, we can write:
\[ P_1^2 V_1 = P_2^2 V_2 \]

2. **Using the Ideal Gas Law**:
We know from the ideal gas law that \( PV = nRT \) for one mole of gas. Rearranging gives:
\[ P = \frac{RT}{V} \]
Substituting this into the law \( P^2 V = \text{constant} \):
\[ \left(\frac{RT}{V}\right)^2 V = \text{constant} \]
Simplifying this gives:
\[ \frac{R^2 T^2}{V} = \text{constant} \]
Thus, we can express this as:
\[ T^2 V = \text{constant} \]

3. **Setting Up the Initial and Final Conditions**:
Let the initial conditions be \( T_1 = T_0 \) and \( V_1 = V_0 \). The final conditions will be \( V_2 = 3V_0 \) and we need to find \( T_2 \).

4. **Using the Constant Relationship**:
From the relationship \( T^2 V = \text{constant} \), we can write:
\[ T_1^2 V_1 = T_2^2 V_2 \]
Substituting the known values:
\[ T_0^2 V_0 = T_2^2 (3V_0) \]

5. **Solving for Final Temperature**:
Cancel \( V_0 \) from both sides:
\[ T_0^2 = 3 T_2^2 \]
Rearranging gives:
\[ T_2^2 = \frac{T_0^2}{3} \]
Taking the square root:
\[ T_2 = \sqrt{3} T_0 \]

6. **Internal Energy Change**:
The internal energy \( U \) of an ideal gas is directly proportional to its temperature. Therefore, if the temperature increases, the internal energy also increases. Since we found that \( T_2 = \sqrt{3} T_0 \), the internal energy increases.

### Final Answers:
- The final temperature \( T_2 = \sqrt{3} T_0 \).
- The internal energy of the gas increases.