One mole of an ideal gas passes through a process where pressure and volume obey the relation `P=P_0 [1-1/2 (V_0/V)^2]` Here `P_0` and `V_0` are constants. Calculate the change in the temperature of the gas if its volume changes from `V_0` to `2V_0`.

Options

A. `3/4 (P_0 V_0)/R`
B. `1/2 (P_0 V_0)/R`
C. `1/4 (P_0 V_0)/R`
D. `5/4 (P_0 V_0)/R`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given relation
We are given the pressure-volume relation for the gas:
\[ P = P_0 \left( 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right) \]
where \( P_0 \) and \( V_0 \) are constants.

### Step 2: Calculate initial and final pressures
1. **Initial Volume \( V_1 = V_0 \)**:
\[ P_1 = P_0 \left( 1 - \frac{1}{2} \left( \frac{V_0}{V_0} \right)^2 \right) = P_0 \left( 1 - \frac{1}{2} \cdot 1 \right) = P_0 \left( \frac{1}{2} \right) = \frac{P_0}{2} \]

2. **Final Volume \( V_2 = 2V_0 \)**:
\[ P_2 = P_0 \left( 1 - \frac{1}{2} \left( \frac{V_0}{2V_0} \right)^2 \right) = P_0 \left( 1 - \frac{1}{2} \cdot \frac{1}{4} \right) = P_0 \left( 1 - \frac{1}{8} \right) = P_0 \left( \frac{7}{8} \right) = \frac{7P_0}{8} \]

### Step 3: Use the ideal gas law to find temperature change
The ideal gas law states:
\[ PV = nRT \]
For one mole of gas (\( n = 1 \)):
1. **Initial Temperature \( T_1 \)**:
\[ T_1 = \frac{P_1 V_1}{R} = \frac{\left(\frac{P_0}{2}\right) V_0}{R} = \frac{P_0 V_0}{2R} \]

2. **Final Temperature \( T_2 \)**:
\[ T_2 = \frac{P_2 V_2}{R} = \frac{\left(\frac{7P_0}{8}\right) (2V_0)}{R} = \frac{14P_0 V_0}{8R} = \frac{7P_0 V_0}{4R} \]

### Step 4: Calculate the change in temperature
The change in temperature \( \Delta T \) is given by:
\[ \Delta T = T_2 - T_1 = \frac{7P_0 V_0}{4R} - \frac{P_0 V_0}{2R} \]
To combine these fractions:
\[ \Delta T = \frac{7P_0 V_0}{4R} - \frac{2P_0 V_0}{4R} = \frac{(7 - 2)P_0 V_0}{4R} = \frac{5P_0 V_0}{4R} \]

### Final Result
Thus, the change in temperature of the gas is:
\[ \Delta T = \frac{5P_0 V_0}{4R} \]