One mole of a gas changed from its initial state (15L,2 atm) to final state (4L,10 atm) reversibly. If this change can be represented by a straight line in P - V curve maximum temperature (approximate), the gas attained is `x xx10^(2) K.` Then find the value of x.

Explanation

To solve the problem step by step, we will follow the process outlined in the video transcript, ensuring we derive the maximum temperature the gas attained during the reversible change from its initial state to its final state.

### Step 1: Understand the Initial and Final States
We have:
- Initial state: \( V_1 = 15 \, \text{L} \), \( P_1 = 2 \, \text{atm} \)
- Final state: \( V_2 = 4 \, \text{L} \), \( P_2 = 10 \, \text{atm} \)

### Step 2: Plot the Points on a P-V Diagram
On a P-V diagram, we plot the initial state at (15 L, 2 atm) and the final state at (4 L, 10 atm). The line connecting these two points will represent the reversible process.

### Step 3: Find the Equation of the Line
Using the two points, we can find the equation of the line in the form \( P - P_1 = \frac{P_2 - P_1}{V_2 - V_1}(V - V_1) \).

Substituting the values:
- \( P_1 = 2 \)
- \( P_2 = 10 \)
- \( V_1 = 15 \)
- \( V_2 = 4 \)

The slope \( m \) of the line is:
\[ m = \frac{10 - 2}{4 - 15} = \frac{8}{-11} = -\frac{8}{11} \]

Now, we can write the equation of the line:
\[ P - 2 = -\frac{8}{11}(V - 15) \]
Rearranging gives:
\[ P = -\frac{8}{11}V + \frac{120}{11} + 2 \]
\[ P = -\frac{8}{11}V + \frac{142}{11} \]

### Step 4: Relate Pressure, Volume, and Temperature
Using the ideal gas law:
\[ PV = nRT \]
For one mole of gas (\( n = 1 \)):
\[ T = \frac{PV}{R} \]
Substituting for \( P \):
\[ T = \frac{V}{R} \left(-\frac{8}{11}V + \frac{142}{11}\right) \]
\[ T = \frac{1}{R} \left(-\frac{8}{11}V^2 + \frac{142}{11}V\right) \]

### Step 5: Find the Maximum Temperature
To find the maximum temperature, we differentiate \( T \) with respect to \( V \) and set the derivative to zero:
\[ \frac{dT}{dV} = \frac{1}{R} \left(-\frac{16}{11}V + \frac{142}{11}\right) = 0 \]
Setting the equation to zero gives:
\[ -\frac{16}{11}V + \frac{142}{11} = 0 \]
Solving for \( V \):
\[ V = \frac{142}{16} = 8.875 \, \text{L} \]

### Step 6: Calculate the Maximum Temperature
Substituting \( V = 8.875 \) L back into the temperature equation:
\[ T = \frac{1}{R} \left(-\frac{8}{11}(8.875)^2 + \frac{142}{11}(8.875)\right) \]
Using \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \):
\[ T = \frac{1}{0.0821} \left(-\frac{8}{11}(79.515625) + \frac{142}{11}(8.875)\right) \]
Calculating the values:
\[ T = \frac{1}{0.0821} \left(-\frac{636.125}{11} + \frac{1260.25}{11}\right) \]
\[ T = \frac{1}{0.0821} \left(\frac{624.125}{11}\right) \]
\[ T \approx \frac{56.83}{0.0821} \approx 693.07 \, \text{K} \]

### Step 7: Express in the Required Form
The temperature can be expressed as:
\[ T \approx 6.93 \times 10^2 \, \text{K} \]
Thus, \( x \approx 7 \).

### Final Answer
The value of \( x \) is **7**.