One mole of ideal gas goes through process`P = (2V^2)/(1+V^2)`. Then change in temperature of gas when volume changes from `V = 1m^2` to `2m^2` is :

Options

A. `- 4/(5R)K`
B. `11/(5R)K`
C. `(-5)/(2R)K`
D. `2K`

Explanation

To solve the problem, we need to find the change in temperature of one mole of an ideal gas as it goes through a specific process defined by the equation \( P = \frac{2V^2}{1 + V^2} \), when the volume changes from \( V = 1 \, m^2 \) to \( V = 2 \, m^2 \).

### Step-by-Step Solution:

1. **Use the Ideal Gas Law**:
The ideal gas law is given by the equation:
\[ PV = nRT \]
For one mole of gas (\( n = 1 \)), this simplifies to:
\[ PV = RT \]
Rearranging gives:
\[ T = \frac{PV}{R} \]

2. **Substitute the Given Pressure Equation**:
We substitute the expression for pressure \( P = \frac{2V^2}{1 + V^2} \) into the ideal gas law:
\[ T = \frac{(\frac{2V^2}{1 + V^2})V}{R} \]
This simplifies to:
\[ T = \frac{2V^3}{R(1 + V^2)} \]

3. **Calculate Temperature at \( V = 1 \, m^2 \)**:
Substitute \( V = 1 \) into the temperature equation:
\[ T_1 = \frac{2(1)^3}{R(1 + (1)^2)} = \frac{2}{R(1 + 1)} = \frac{2}{2R} = \frac{1}{R} \]

4. **Calculate Temperature at \( V = 2 \, m^2 \)**:
Now substitute \( V = 2 \) into the temperature equation:
\[ T_2 = \frac{2(2)^3}{R(1 + (2)^2)} = \frac{2 \cdot 8}{R(1 + 4)} = \frac{16}{5R} \]

5. **Calculate Change in Temperature**:
The change in temperature \( \Delta T \) is given by:
\[ \Delta T = T_2 - T_1 = \frac{16}{5R} - \frac{1}{R} \]
To combine these fractions, find a common denominator:
\[ \Delta T = \frac{16}{5R} - \frac{5}{5R} = \frac{16 - 5}{5R} = \frac{11}{5R} \]

### Final Answer:
The change in temperature of the gas when the volume changes from \( 1 \, m^2 \) to \( 2 \, m^2 \) is:
\[ \Delta T = \frac{11}{5R} \, \text{K} \]