One mole of an ideal gas undergoes a process
`P=P_(0)-alphaV^(2)`
where `alpha` and `P_(0)` are positive constant and V is the volume of one mole of gas
Q. When temperature is maximum, volume is

Options

A. `sqrt((P_(0))/(3alpha))`
B. `sqrt((P_(0))/(alpha))`
C. `sqrt((P_(0))/(2alpha))`
D. `sqrt(P_(0)alpha)`

Answer

To solve the problem, we need to find the volume \( V \) when the temperature \( T \) of the gas is at its maximum during the given process \( P = P_0 - \alpha V^2 \).

### Step-by-Step Solution:

1. **Understand the given equation**:
The pressure \( P \) is given by the equation:
\[ P = P_0 - \alpha V^2 \]
Here, \( P_0 \) and \( \alpha \) are constants, and \( V \) is the volume of one mole of the gas.

2. **Relate pressure, volume, and temperature**:
For one mole of an ideal gas, the ideal gas law states:
\[ PV = nRT \]
Since \( n = 1 \) (one mole), we can write:
\[ PV = RT \quad \Rightarrow \quad T = \frac{PV}{R} \]

3. **Substitute for \( P \)**:
Substitute the expression for \( P \) from step 1 into the equation for \( T \):
\[ T = \frac{(P_0 - \alpha V^2)V}{R} \]
Simplifying this gives:
\[ T = \frac{P_0 V - \alpha V^3}{R} \]

4. **Find the condition for maximum temperature**:
To find the volume at which the temperature is maximum, we need to take the derivative of \( T \) with respect to \( V \) and set it to zero:
\[ \frac{dT}{dV} = 0 \]

5. **Differentiate \( T \)**:
Differentiate \( T \):
\[ \frac{dT}{dV} = \frac{1}{R} \left( P_0 - 3\alpha V^2 \right) \]
Setting this equal to zero for maximum temperature:
\[ P_0 - 3\alpha V^2 = 0 \]

6. **Solve for \( V \)**:
Rearranging gives:
\[ 3\alpha V^2 = P_0 \quad \Rightarrow \quad V^2 = \frac{P_0}{3\alpha} \]
Taking the square root:
\[ V = \sqrt{\frac{P_0}{3\alpha}} \]

### Final Answer:
The volume \( V \) when the temperature is maximum is:
\[ V = \sqrt{\frac{P_0}{3\alpha}} \]