A gas is found to be obeyed the law `p^2V = constant`. The initial temperature and volume are `T_0 and V_0`. If the gas expands to a volume `3 V_0`, then the final temperature becomes.

Options

A. `sqrt (3) T_0`
B. `sqrt (2) T_0`.
C. `(T_0)/(sqrt (3)`.
D. `(T_(0))/(sqrt(2))`.

Explanation

To solve the problem step by step, we will use the relationship given in the question and the ideal gas law.

### Step 1: Understand the given relationship
The gas obeys the law \( p^2 V = \text{constant} \). This means that the product of the square of the pressure and the volume remains constant during the process.

### Step 2: Relate pressure to temperature and volume
From the ideal gas law, we know that:
\[ pV = nRT \]
where \( n \) is the number of moles of gas, \( R \) is the universal gas constant, and \( T \) is the temperature. Rearranging this gives:
\[ p = \frac{nRT}{V} \]

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

### Step 4: Establish the relationship between temperature and volume
Since \( \frac{n^2 R^2 T^2}{V} = \text{constant} \), we can express this as:
\[ T^2 \propto V \]
This implies that:
\[ T \propto \sqrt{V} \]

### Step 5: Apply the initial conditions
Let the initial temperature and volume be \( T_0 \) and \( V_0 \), respectively. When the gas expands to a new volume \( V' = 3V_0 \), we can relate the new temperature \( T' \) to the initial temperature:
\[ T' \propto \sqrt{V'} \]
Substituting \( V' = 3V_0 \):
\[ T' \propto \sqrt{3V_0} \]

### Step 6: Relate the new temperature to the initial temperature
Since we know that \( T_0 \propto \sqrt{V_0} \), we can express the new temperature as:
\[ T' = k \sqrt{3V_0} \]
where \( k \) is a proportionality constant that relates \( T_0 \) and \( V_0 \):
\[ T_0 = k \sqrt{V_0} \]

### Step 7: Find the final temperature
Now substituting \( k \sqrt{V_0} \) into the equation for \( T' \):
\[ T' = k \sqrt{3} \sqrt{V_0} = \sqrt{3} T_0 \]

### Final Answer
Thus, the final temperature when the gas expands to a volume of \( 3V_0 \) is:
\[ T' = \sqrt{3} T_0 \]