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 find the change in temperature of the ideal gas as it goes through the specified process, we will follow these steps:

### Step 1: Calculate the initial pressure at \( V = 1 \, m^2 \)

The given relationship for pressure is:
\[ P = \frac{2V^2}{1 + V^2} \]
Substituting \( V = 1 \):
\[ P_1 = \frac{2(1)^2}{1 + (1)^2} = \frac{2 \cdot 1}{1 + 1} = \frac{2}{2} = 1 \, \text{atm} \]

### Step 2: Calculate the initial temperature \( T_i \)

Using the ideal gas law:
\[ PV = nRT \]
Where:
- \( n = 1 \) mole
- \( R \) is the gas constant

Substituting the values:
\[ T_i = \frac{P_1 V_1}{nR} = \frac{1 \cdot 1}{1 \cdot R} = \frac{1}{R} \, \text{K} \]

### Step 3: Calculate the pressure at \( V = 2 \, m^2 \)

Now substituting \( V = 2 \):
\[ P_2 = \frac{2(2)^2}{1 + (2)^2} = \frac{2 \cdot 4}{1 + 4} = \frac{8}{5} \, \text{atm} \]

### Step 4: Calculate the final temperature \( T_f \)

Using the ideal gas law again:
\[ T_f = \frac{P_2 V_2}{nR} = \frac{\left(\frac{8}{5}\right) \cdot 2}{1 \cdot R} = \frac{16}{5R} \, \text{K} \]

### Step 5: Calculate the change in temperature \( \Delta T \)

The change in temperature is given by:
\[ \Delta T = T_f - T_i = \frac{16}{5R} - \frac{1}{R} \]
Finding a common denominator (which is \( 5R \)):
\[ \Delta T = \frac{16}{5R} - \frac{5}{5R} = \frac{16 - 5}{5R} = \frac{11}{5R} \, \text{K} \]

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

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