One mole of an ideal gas undergoes a process `p=(p_(0))/(1+((V_(0))/(V))^(2))`. Here, `p_(0)` and `V_(0)` are constants. Change in temperature of the gas when volume is changed from `V=V_(0)` to `V=2V_(0)` is

Options

A. `-(2p_(0)V_(0))/(5R)`
B. `(11p_(0)V_(0))/(10R)`
C. `-(5p_(0)V_(0))/(4R)`
D. `p_(0)V_(0)`

Answer

To solve the problem of finding the change in temperature of one mole of an ideal gas undergoing the specified process, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
We start with the initial volume \( V_1 = V_0 \). The pressure at this volume can be calculated using the given equation:
\[ p_1 = \frac{p_0}{1 + \left(\frac{V_0}{V_0}\right)^2} = \frac{p_0}{1 + 1} = \frac{p_0}{2} \]

2. **Calculate Initial Temperature**:
Using the ideal gas law \( PV = nRT \), where \( n = 1 \) mole, we can find the initial temperature \( T_1 \):
\[ T_1 = \frac{p_1 V_1}{nR} = \frac{\left(\frac{p_0}{2}\right) V_0}{R} = \frac{p_0 V_0}{2R} \]

3. **Identify Final Conditions**:
Now, we change the volume to \( V_2 = 2V_0 \). We need to find the pressure at this new volume:
\[ p_2 = \frac{p_0}{1 + \left(\frac{V_0}{2V_0}\right)^2} = \frac{p_0}{1 + \left(\frac{1}{2}\right)^2} = \frac{p_0}{1 + \frac{1}{4}} = \frac{p_0}{\frac{5}{4}} = \frac{4p_0}{5} \]

4. **Calculate Final Temperature**:
Now, we can find the final temperature \( T_2 \) using the ideal gas law:
\[ T_2 = \frac{p_2 V_2}{nR} = \frac{\left(\frac{4p_0}{5}\right)(2V_0)}{R} = \frac{8p_0 V_0}{5R} \]

5. **Calculate Change in Temperature**:
The change in temperature \( \Delta T \) is given by:
\[ \Delta T = T_2 - T_1 = \left(\frac{8p_0 V_0}{5R}\right) - \left(\frac{p_0 V_0}{2R}\right) \]
To subtract these two fractions, we need a common denominator:
\[ \Delta T = \frac{8p_0 V_0}{5R} - \frac{2.5p_0 V_0}{5R} = \frac{(8 - 2.5)p_0 V_0}{5R} = \frac{5.5p_0 V_0}{5R} = \frac{11p_0 V_0}{10R} \]

### Final Answer:
The change in temperature of the gas when the volume is changed from \( V_0 \) to \( 2V_0 \) is:
\[ \Delta T = \frac{11p_0 V_0}{10R} \]