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_(o))/3`
B. `(T_(o))/sqrt3`
C. `3 T_(o)`
D. None of these

Explanation

To solve the problem, we start with the given relationship for the gas:

1. **Understanding the Relationship**:
The gas obeys the law \( P^2 V = \text{constant} \). This means that as the volume changes, the pressure must change in such a way that the product of the square of the pressure and the volume remains constant.

2. **Using the Ideal Gas Law**:
From the ideal gas law, we know that:
\[ PV = nRT \]
Rearranging this gives us:
\[ P = \frac{nRT}{V} \]

3. **Substituting into the Given Law**:
Substitute \( P \) from the ideal gas law into the given relationship:
\[ P^2 V = \left(\frac{nRT}{V}\right)^2 V = \frac{n^2 R^2 T^2}{V^2} V = \frac{n^2 R^2 T^2}{V} \]
This implies:
\[ n^2 R^2 T^2 = \text{constant} \cdot V \]
Since \( n \) and \( R \) are constants, we can simplify this to:
\[ T^2 \propto V \]

4. **Setting Up the Proportionality**:
From the proportionality \( T^2 \propto V \), we can write:
\[ \frac{T_i^2}{T_f^2} = \frac{V_i}{V_f} \]
where \( T_i \) and \( V_i \) are the initial temperature and volume, and \( T_f \) and \( V_f \) are the final temperature and volume.

5. **Substituting Initial and Final Values**:
Given:
- Initial temperature \( T_i = T_0 \)
- Initial volume \( V_i = V_0 \)
- Final volume \( V_f = 3V_0 \)

Substitute these values into the equation:
\[ \frac{T_0^2}{T_f^2} = \frac{V_0}{3V_0} \]
This simplifies to:
\[ \frac{T_0^2}{T_f^2} = \frac{1}{3} \]

6. **Cross Multiplying and Solving for Final Temperature**:
Cross multiplying gives:
\[ T_0^2 = \frac{1}{3} T_f^2 \]
Rearranging this gives:
\[ T_f^2 = 3 T_0^2 \]

7. **Taking the Square Root**:
Taking the square root of both sides:
\[ T_f = \sqrt{3} T_0 \]

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