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 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 the Initial Temperature \( T_1 \)

Using the ideal gas law, we have:

\[ T_1 = \frac{p_1 V_1}{nR} \]

Where:
- \( p_1 = \frac{p_0}{1 + \left(\frac{V_0}{V_0}\right)^2} = \frac{p_0}{1 + 1} = \frac{p_0}{2} \)
- \( V_1 = V_0 \)
- \( n = 1 \) (one mole of gas)
- \( R \) is the ideal gas constant.

Substituting these values:

\[ T_1 = \frac{\left(\frac{p_0}{2}\right) V_0}{1 \cdot R} = \frac{p_0 V_0}{2R} \]

### Step 2: Determine the Final Temperature \( T_2 \)

Now we calculate the final temperature \( T_2 \) when the volume is \( V = 2V_0 \):

\[ T_2 = \frac{p_2 V_2}{nR} \]

Where:
- \( p_2 = \frac{p_0}{1 + \left(\frac{V_0}{2V_0}\right)^2} = \frac{p_0}{1 + \frac{1}{4}} = \frac{p_0}{\frac{5}{4}} = \frac{4p_0}{5} \)
- \( V_2 = 2V_0 \)

Substituting these values:

\[ T_2 = \frac{\left(\frac{4p_0}{5}\right)(2V_0)}{1 \cdot R} = \frac{8p_0 V_0}{5R} \]

### Step 3: Calculate the Change in Temperature \( \Delta T \)

Now, we find the change in temperature:

\[ \Delta T = T_2 - T_1 \]

Substituting the expressions for \( T_2 \) and \( T_1 \):

\[ \Delta T = \frac{8p_0 V_0}{5R} - \frac{p_0 V_0}{2R} \]

To combine these fractions, we need a common denominator. The common denominator is \( 10R \):

\[ \Delta T = \frac{16p_0 V_0}{10R} - \frac{5p_0 V_0}{10R} = \frac{(16 - 5)p_0 V_0}{10R} = \frac{11p_0 V_0}{10R} \]

### Final Answer

Thus, the change in temperature of the gas is:

\[ \Delta T = \frac{11p_0 V_0}{10R} \]