Find the minimum attainable pressure of an ideal gas in the process `T = T_(0) + alpha V^(2)`, Where `T_(0)` and `alpha` are positive constant and `V` is the volume of one mole of gas. Draw the approximate `T -V` plot of this process.

Options

A. `2Rsqrt(alphaT_(0))`
B. `3Rsqrt(alphaT_(0))`
C. 3R
D. `3Rsqrt((alphaT_(0))/(2))`

Answer

To find the minimum attainable pressure of an ideal gas in the process defined by the equation \( T = T_0 + \alpha V^2 \), where \( T_0 \) and \( \alpha \) are positive constants, we can follow these steps:

### Step 1: Relate Pressure, Volume, and Temperature
For one mole of an ideal gas, the ideal gas law states:
\[ PV = RT \]
Where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( R \) is the universal gas constant,
- \( T \) is the temperature.

### Step 2: Substitute for Volume
From the ideal gas law, we can express volume \( V \) in terms of pressure \( P \) and temperature \( T \):
\[ V = \frac{RT}{P} \]

### Step 3: Substitute Volume into the Temperature Equation
We substitute \( V \) into the given temperature equation:
\[ T = T_0 + \alpha \left(\frac{RT}{P}\right)^2 \]
This simplifies to:
\[ T = T_0 + \alpha \frac{R^2 T^2}{P^2} \]

### Step 4: Rearranging the Equation
Rearranging this equation gives:
\[ P^2 = \frac{\alpha R^2 T^2}{T - T_0} \]

### Step 5: Express Pressure in Terms of Temperature
Taking the square root of both sides, we find:
\[ P = \sqrt{\frac{\alpha R^2 T^2}{T - T_0}} \]

### Step 6: Differentiate Pressure with Respect to Temperature
To find the minimum pressure, we differentiate \( P \) with respect to \( T \) and set the derivative equal to zero:
\[ \frac{dP}{dT} = \frac{1}{2} \left(\frac{\alpha R^2 T^2}{T - T_0}\right)^{-1/2} \cdot \left( \frac{d}{dT}\left(\frac{\alpha R^2 T^2}{T - T_0}\right) \right) = 0 \]

### Step 7: Solve the Derivative
Using the quotient rule, we differentiate:
\[ \frac{d}{dT}\left(\frac{\alpha R^2 T^2}{T - T_0}\right) = \frac{(T - T_0)(2\alpha R^2 T) - \alpha R^2 T^2}{(T - T_0)^2} \]
Setting this derivative equal to zero gives:
\[ (T - T_0)(2\alpha R^2 T) - \alpha R^2 T^2 = 0 \]
This simplifies to:
\[ 2T - T_0 = 0 \implies T = 2T_0 \]

### Step 8: Substitute Back to Find Minimum Pressure
Now substituting \( T = 2T_0 \) back into the expression for \( P \):
\[ P_{min} = \sqrt{\frac{\alpha R^2 (2T_0)^2}{2T_0 - T_0}} = \sqrt{4\alpha R^2 T_0} = 2R\sqrt{\alpha T_0} \]

### Conclusion
Thus, the minimum attainable pressure of the ideal gas is:
\[ P_{min} = 2R\sqrt{\alpha T_0} \]