Find the maximum attainable temperature of ideal gas in each of the following process :
(a) `p = p_0 - alpha V^2` ,
(b) `p = p_0 e^(- beta v)`,
where `p_0, alpha` and `beta` are positive constants, and `V` is the volume of one mole of gas.

Explanation

To find the maximum attainable temperature of an ideal gas in the given processes, we will use the ideal gas law and calculus to determine the temperature as a function of volume and pressure.

### Step-by-Step Solution:

**(a)** For the process given by the equation \( p = p_0 - \alpha V^2 \):

1. **Use the Ideal Gas Law**: The ideal gas law states that \( PV = nRT \). For one mole of gas (\( n = 1 \)), this simplifies to:
\[ pV = RT \implies T = \frac{pV}{R} \]

2. **Substitute for Pressure**: Substitute \( p \) from the given equation into the ideal gas law:
\[ T = \frac{(p_0 - \alpha V^2)V}{R} \]

3. **Expand the Expression**: This gives:
\[ T = \frac{p_0 V - \alpha V^3}{R} \]

4. **Find the Maximum Temperature**: To find the maximum temperature, we take the derivative of \( T \) with respect to \( V \) and set it to zero:
\[ \frac{dT}{dV} = \frac{p_0 - 3\alpha V^2}{R} = 0 \]

5. **Solve for Volume**: Setting the derivative to zero gives:
\[ p_0 - 3\alpha V^2 = 0 \implies V^2 = \frac{p_0}{3\alpha} \implies V = \sqrt{\frac{p_0}{3\alpha}} \]

6. **Substitute Back to Find Maximum Temperature**: Substitute \( V \) back into the expression for \( T \):
\[ T_{max} = \frac{(p_0 - \alpha V^2)V}{R} = \frac{(p_0 - \alpha \cdot \frac{p_0}{3\alpha})\sqrt{\frac{p_0}{3\alpha}}}{R} \]
\[ = \frac{(p_0 - \frac{p_0}{3})\sqrt{\frac{p_0}{3\alpha}}}{R} = \frac{\frac{2p_0}{3}\sqrt{\frac{p_0}{3\alpha}}}{R} = \frac{2p_0^{3/2}}{3\sqrt{3\alpha}R} \]

**(b)** For the process given by the equation \( p = p_0 e^{-\beta V} \):

1. **Use the Ideal Gas Law**: Again, we start with:
\[ T = \frac{pV}{R} \]

2. **Substitute for Pressure**: Substitute \( p \):
\[ T = \frac{(p_0 e^{-\beta V})V}{R} \]

3. **Expand the Expression**: This gives:
\[ T = \frac{p_0 V e^{-\beta V}}{R} \]

4. **Find the Maximum Temperature**: To find the maximum temperature, we take the derivative of \( T \) with respect to \( V \) and set it to zero:
\[ \frac{dT}{dV} = \frac{p_0 e^{-\beta V} + p_0 V (-\beta e^{-\beta V})}{R} = 0 \]
\[ p_0 e^{-\beta V}(1 - \beta V) = 0 \]

5. **Solve for Volume**: Since \( p_0 e^{-\beta V} \neq 0 \), we have:
\[ 1 - \beta V = 0 \implies V = \frac{1}{\beta} \]

6. **Substitute Back to Find Maximum Temperature**: Substitute \( V \) back into the expression for \( T \):
\[ T_{max} = \frac{p_0 \cdot \frac{1}{\beta} \cdot e^{-1}}{R} = \frac{p_0 e^{-1}}{\beta R} \]

### Final Results:
- For process (a):
\[ T_{max} = \frac{2p_0^{3/2}}{3\sqrt{3\alpha}R} \]
- For process (b):
\[ T_{max} = \frac{p_0 e^{-1}}{\beta R} \]