One mole of an ideal gas undergoes a process `P = P_(0) [1 + ((2 V_(0))/(V))^(2)]^(-1)`, where `P_(0) V_(0)` are constants. Change in temperature of the gas when volume is changed from `V = V_(0) to V = 2 V_(0)` is:

Options

A. `(4)/(5) (P_(0) V_(0))/(nR)`
B. `(3)/(4) (P_(0) V_(0))/(nR)`
C. `(2)/(3) (P_(0) V_(0))/(nR)`
D. `(9)/(7) (P_(0) V_(0))/(nR)`

Explanation

To find the change in temperature of the gas when the volume changes from \( V = V_0 \) to \( V = 2V_0 \), we can follow these steps:

### Step 1: Determine Initial Pressure and Temperature
Given the equation for pressure:
\[ P = P_0 \left(1 + \frac{2V_0}{V}\right)^{-1} \]
First, calculate the initial pressure when \( V = V_0 \):
\[ P_{\text{initial}} = P_0 \left(1 + \frac{2V_0}{V_0}\right)^{-1} = P_0 \left(1 + 2\right)^{-1} = P_0 \left(3\right)^{-1} = \frac{P_0}{3} \]

Now, using the ideal gas law \( PV = nRT \), where \( n = 1 \) mole, we can find the initial temperature:
\[ T_{\text{initial}} = \frac{P_{\text{initial}} V_0}{nR} = \frac{\left(\frac{P_0}{3}\right) V_0}{1 \cdot R} = \frac{P_0 V_0}{3R} \]

### Step 2: Determine Final Pressure and Temperature
Next, calculate the pressure when \( V = 2V_0 \):
\[ P_{\text{final}} = P_0 \left(1 + \frac{2V_0}{2V_0}\right)^{-1} = P_0 \left(1 + 1\right)^{-1} = P_0 \left(2\right)^{-1} = \frac{P_0}{2} \]

Now, find the final temperature:
\[ T_{\text{final}} = \frac{P_{\text{final}} (2V_0)}{nR} = \frac{\left(\frac{P_0}{2}\right) (2V_0)}{1 \cdot R} = \frac{P_0 V_0}{R} \]

### Step 3: Calculate Change in Temperature
Now, we can find the change in temperature:
\[ \Delta T = T_{\text{final}} - T_{\text{initial}} = \frac{P_0 V_0}{R} - \frac{P_0 V_0}{3R} \]

To combine these, find a common denominator:
\[ \Delta T = \frac{P_0 V_0}{R} - \frac{P_0 V_0}{3R} = \frac{3P_0 V_0}{3R} - \frac{P_0 V_0}{3R} = \frac{2P_0 V_0}{3R} \]

### Final Result
Thus, the change in temperature of the gas is:
\[ \Delta T = \frac{2P_0 V_0}{3R} \]