On an isothermal process ,there are two points A and B at which pressures and volumes are (2P0,V0) and (P0,2V0) respectively. A and B are connected by a straight line then , the temperature of the isothemal process is how much lower the maximum temperature :

Options

A. `(P_(0) V_(0))/(nR)`
B. `(P_(0) V_(0))/(2nR)`
C. `(3P_(0) V_(0))/(2nR)`
D. `(P_(0) V_(0))/(4nR)`

Explanation

To solve the problem step by step, we will analyze the isothermal process between points A and B and calculate the temperature difference as required.

### Step 1: Identify Given Values
We have two points A and B:
- Point A: Pressure \( P_A = 2P_0 \), Volume \( V_A = V_0 \)
- Point B: Pressure \( P_B = P_0 \), Volume \( V_B = 2V_0 \)

### Step 2: Use the Ideal Gas Law
For an isothermal process, the temperature \( T \) can be calculated using the ideal gas equation:
\[ PV = nRT \]
Thus, the temperature at point A can be expressed as:
\[ T_A = \frac{P_A V_A}{nR} = \frac{(2P_0)(V_0)}{nR} = \frac{2P_0 V_0}{nR} \]

### Step 3: Calculate the Maximum Temperature
To find the maximum temperature, we need to determine the pressure and volume at the midpoint of the line connecting A and B.

1. **Calculate the slope (m)** of the line connecting points A and B:
\[ m = \frac{P_B - P_A}{V_B - V_A} = \frac{P_0 - 2P_0}{2V_0 - V_0} = \frac{-P_0}{V_0} = -\frac{P_0}{V_0} \]

2. **Equation of the line** in the form \( P = mV + C \):
- At point A: \( P_A = 2P_0 \) and \( V_A = V_0 \)
\[ 2P_0 = -\frac{P_0}{V_0} \cdot V_0 + C \implies C = 3P_0 \]
Therefore, the equation of the line is:
\[ P = -\frac{P_0}{V_0} V + 3P_0 \]

3. **Substituting into the ideal gas law** to find temperature:
\[ T = \frac{PV}{nR} = \frac{(-\frac{P_0}{V_0} V + 3P_0)V}{nR} \]
\[ T = \frac{1}{nR} \left(-\frac{P_0}{V_0} V^2 + 3P_0 V\right) \]

### Step 4: Differentiate to Find Maximum Temperature
To find the maximum temperature, differentiate \( T \) with respect to \( V \) and set the derivative to zero:
\[ \frac{dT}{dV} = \frac{1}{nR} \left(-\frac{P_0}{V_0} \cdot 2V + 3P_0\right) = 0 \]
Solving for \( V \):
\[ -\frac{2P_0}{V_0} V + 3P_0 = 0 \implies V = \frac{3V_0}{2} \]

### Step 5: Calculate Maximum Temperature at \( V = \frac{3V_0}{2} \)
Substituting \( V = \frac{3V_0}{2} \) back into the equation for \( P \):
\[ P = -\frac{P_0}{V_0} \cdot \frac{3V_0}{2} + 3P_0 = \frac{3P_0}{2} \]
Now substituting \( P \) and \( V \) into the ideal gas equation:
\[ T_{max} = \frac{(3/2)P_0(3/2)V_0}{nR} = \frac{9P_0 V_0}{4nR} \]

### Step 6: Calculate the Temperature Difference
Now we find the difference between the maximum temperature and the temperature at point A:
\[ \Delta T = T_{max} - T_A = \frac{9P_0 V_0}{4nR} - \frac{2P_0 V_0}{nR} \]
Finding a common denominator:
\[ \Delta T = \frac{9P_0 V_0 - 8P_0 V_0}{4nR} = \frac{P_0 V_0}{4nR} \]

### Final Answer
The temperature of the isothermal process is \( \frac{P_0 V_0}{4nR} \) lower than the maximum temperature.

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