Pressure of `1` mole ideal gas is given by
`P=P_(0)[1-(1)/(2)(V_(0)/(V))^(2)]` ,brgt If volume of gas change from `V` to `2 V`. Find change in temperature.

Options

A. `(1)/(2)(P_(O)V_(O))/(R)`
B. `(5)/(4)(P_(O)V_(O))/(R)`
C. `(3)/(4)(P_(O)V_(O))/(R)`
D. `(1)/(4)(P_(O)V_(O))/(R)`

Answer

To solve the problem of finding the change in temperature when the volume of an ideal gas changes from \( V \) to \( 2V \), we start with the given pressure equation:

\[ P = P_0 \left( 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right) \]

### Step 1: Relate Pressure to Temperature
Using the ideal gas law, we know that:

\[ PV = nRT \]

For 1 mole of gas (\( n = 1 \)), this simplifies to:

\[ PV = RT \]

### Step 2: Express Temperature in Terms of Pressure and Volume
From the ideal gas law, we can express temperature \( T \) as:

\[ T = \frac{PV}{R} \]

### Step 3: Substitute the Expression for Pressure
Now, substitute the expression for \( P \) from the given equation into the temperature equation:

\[ T = \frac{P_0 \left( 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right) V}{R} \]

### Step 4: Calculate Initial Temperature \( T_1 \)
For the initial volume \( V \):

\[ T_1 = \frac{P_0 \left( 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right) V}{R} \]

### Step 5: Calculate Final Temperature \( T_2 \)
Now, for the final volume \( 2V \):

\[ T_2 = \frac{P_0 \left( 1 - \frac{1}{2} \left( \frac{V_0}{2V} \right)^2 \right) (2V)}{R} \]

### Step 6: Simplify \( T_2 \)
Substituting \( 2V \) into the equation gives:

\[ T_2 = \frac{P_0 \left( 1 - \frac{1}{2} \left( \frac{V_0}{2V} \right)^2 \right) (2V)}{R} \]

Calculating the term inside the parentheses:

\[ \frac{1}{2} \left( \frac{V_0}{2V} \right)^2 = \frac{1}{2} \cdot \frac{V_0^2}{4V^2} = \frac{V_0^2}{8V^2} \]

Thus:

\[ T_2 = \frac{P_0 \left( 1 - \frac{V_0^2}{8V^2} \right) (2V)}{R} \]

### Step 7: Calculate Change in Temperature \( \Delta T \)
Now, we can find the change in temperature:

\[ \Delta T = T_2 - T_1 \]

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

\[ \Delta T = \left( \frac{P_0 \left( 1 - \frac{V_0^2}{8V^2} \right) (2V)}{R} \right) - \left( \frac{P_0 \left( 1 - \frac{V_0^2}{2V^2} \right) V}{R} \right) \]

### Step 8: Combine and Simplify
Finding a common denominator and simplifying will yield:

\[ \Delta T = \frac{P_0}{R} \left( \left( 2V \left( 1 - \frac{V_0^2}{8V^2} \right) - V \left( 1 - \frac{V_0^2}{2V^2} \right) \right) \right) \]

After simplifying this expression, we will arrive at:

\[ \Delta T = \frac{5P_0 V_0}{4R} \]

### Final Result
Thus, the change in temperature is:

\[ \Delta T = \frac{5P_0 V_0}{4R} \]