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 `3 V_(0)`, its final temperature becomes

Options

A. `(T_(0))/(3)`
B. `(T_(0))/sqrt(3)`
C. `3 T_(0)`
D. `sqrt(3) T_(0)`

Explanation

To solve the problem, we need to analyze the relationship given in the question and apply the ideal gas law. Let's go through the steps systematically.

### Step 1: Understand the relationship given
The gas obeys the law \( P^2 V = \text{constant} \). This means that for any two states of the gas, we can write:
\[ P_1^2 V_1 = P_2^2 V_2 \]

### Step 2: Write the ideal gas law
The ideal gas law states:
\[ PV = nRT \]
From this, we can express pressure \( P \) in terms of temperature \( T \) and volume \( V \):
\[ P = \frac{nRT}{V} \]

### Step 3: Substitute \( P \) into the given relationship
Substituting \( P \) into the relationship \( P^2 V = \text{constant} \):
\[ \left(\frac{nRT}{V}\right)^2 V = \text{constant} \]
This simplifies to:
\[ \frac{n^2 R^2 T^2}{V} = \text{constant} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ T^2 = \text{constant} \cdot V \]
This indicates that \( T^2 \) is directly proportional to \( V \):
\[ T^2 \propto V \]

### Step 5: Relate initial and final states
Let’s denote the initial state with subscript 0:
- Initial temperature: \( T_0 \)
- Initial volume: \( V_0 \)

When the gas expands to a volume of \( 3V_0 \):
- Final volume: \( V_f = 3V_0 \)

Using the proportional relationship:
\[ \frac{T_f^2}{T_0^2} = \frac{V_f}{V_0} \]
Substituting the values:
\[ \frac{T_f^2}{T_0^2} = \frac{3V_0}{V_0} = 3 \]

### Step 6: Solve for final temperature \( T_f \)
Taking the square root of both sides:
\[ \frac{T_f}{T_0} = \sqrt{3} \]
Thus, the final temperature \( T_f \) is:
\[ T_f = T_0 \sqrt{3} \]

### Conclusion
The final temperature when the gas expands to a volume of \( 3V_0 \) is:
\[ T_f = T_0 \sqrt{3} \]