On an isothermal process, there are two POINTs `A` and `B` at which pressures and volumes are `(2P_(0), V_(0))` and `(P_(0), 2V_(0))` respectively. If `A` and `B` are connected by a straight line, find the pressure at a POINT on this straight line at which temperature is maximum

Options

A. `(4 P_(0))/(3)`
B. `(5)/(4) P_(0)`
C. `(3)/(2) P_(0)`
D. `(7)/(5) P_(0)`

Explanation

To solve the problem step by step, we will follow the process outlined in the video transcript.

### Step 1: Understand the Points A and B
We have two points A and B on an isothermal process:
- 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: Equation of the Straight Line AB
The equation of the straight line connecting points A and B can be derived using the two-point form of the line equation. The general form is:
\[ \frac{y - y_A}{x - x_A} = \frac{y_B - y_A}{x_B - x_A} \]
Here, we replace \( y \) with pressure \( P \) and \( x \) with volume \( V \):
\[ \frac{P - 2P_0}{V - V_0} = \frac{P_0 - 2P_0}{2V_0 - V_0} \]
This simplifies to:
\[ \frac{P - 2P_0}{V - V_0} = \frac{-P_0}{V_0} \]

### Step 3: Rearranging the Equation
Cross-multiplying gives:
\[ P - 2P_0 = -\frac{P_0}{V_0}(V - V_0) \]
Rearranging this leads to:
\[ P = 2P_0 - \frac{P_0}{V_0}(V - V_0) \]

### Step 4: Expressing PV
Multiplying both sides by \( V \):
\[ PV = V(2P_0) - \frac{P_0}{V_0}(V^2 - VV_0) \]
This can be expressed as:
\[ PV = 2P_0V - \frac{P_0}{V_0}V^2 + \frac{P_0}{V_0}VV_0 \]

### Step 5: Relating to Temperature
Using the ideal gas law \( PV = nRT \), we can equate:
\[ nRT = 2P_0V - \frac{P_0}{V_0}V^2 + P_0V \]
Rearranging gives:
\[ nRT = 2P_0V - \frac{P_0}{V_0}V^2 + P_0V \]

### Step 6: Finding Maximum Temperature
To find the maximum temperature, we differentiate \( T \) with respect to \( V \) and set the derivative to zero:
\[ \frac{dT}{dV} = 0 \]
This leads to:
\[ 2P_0 - \frac{P_0}{V_0}(2V - V_0) = 0 \]

### Step 7: Solving for V
Solving the equation:
\[ 2P_0 = \frac{P_0}{V_0}(2V - V_0) \]
Cancelling \( P_0 \) and rearranging gives:
\[ 2 = \frac{2V}{V_0} - 1 \]
Thus,
\[ 3 = \frac{2V}{V_0} \]
This implies:
\[ V = \frac{3}{2}V_0 \]

### Step 8: Finding Pressure at Maximum Temperature
Substituting \( V = \frac{3}{2}V_0 \) back into the equation for pressure:
\[ P = 2P_0 - \frac{P_0}{V_0}\left(\frac{3}{2}V_0 - V_0\right) \]
This simplifies to:
\[ P = 2P_0 - \frac{P_0}{V_0}\left(\frac{1}{2}V_0\right) = 2P_0 - \frac{1}{2}P_0 = \frac{3}{2}P_0 \]

### Final Answer
The pressure at the point on the straight line where the temperature is maximum is:
\[ \boxed{\frac{3}{2}P_0} \]