find the minimum attainable pressure of an ideal gs in the process `T = t_0 + prop V^2`, where `T_(0)n` and `alpha` are positive constants and (V) is the volume of one mole of gas.

Explanation

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: Write the Ideal Gas Equation
The ideal gas equation for one mole of gas is given by:
\[ PV = nRT \]
Since we have one mole of gas (\( n = 1 \)), we can simplify this to:
\[ PV = RT \]
From this, we can express pressure \( P \) as:
\[ P = \frac{RT}{V} \]

### Step 2: Substitute the Temperature Expression
We know from the problem statement that the temperature \( T \) can be expressed as:
\[ T = t_0 + \alpha V^2 \]
Substituting this expression for \( T \) into the equation for pressure, we get:
\[ P = \frac{R(t_0 + \alpha V^2)}{V} \]
This simplifies to:
\[ P = \frac{Rt_0}{V} + R\alpha V \]

### Step 3: Differentiate Pressure with Respect to Volume
To find the minimum pressure, we need to take the derivative of \( P \) with respect to \( V \) and set it to zero:
\[ \frac{dP}{dV} = -\frac{Rt_0}{V^2} + R\alpha \]
Setting the derivative equal to zero for minimization:
\[ -\frac{Rt_0}{V^2} + R\alpha = 0 \]

### Step 4: Solve for Volume \( V \)
Rearranging the equation, we have:
\[ R\alpha = \frac{Rt_0}{V^2} \]
This leads to:
\[ \alpha V^2 = t_0 \]
Thus, we can solve for \( V \):
\[ V^2 = \frac{t_0}{\alpha} \quad \Rightarrow \quad V = \sqrt{\frac{t_0}{\alpha}} \]

### Step 5: Substitute Back to Find Minimum Pressure
Now that we have the value of \( V \), we can substitute it back into the pressure equation:
\[ P_{\text{min}} = \frac{R(t_0 + \alpha V^2)}{V} \]
Substituting \( V = \sqrt{\frac{t_0}{\alpha}} \):
\[ P_{\text{min}} = \frac{R\left(t_0 + \alpha \cdot \frac{t_0}{\alpha}\right)}{\sqrt{\frac{t_0}{\alpha}}} \]
This simplifies to:
\[ P_{\text{min}} = \frac{R(2t_0)}{\sqrt{\frac{t_0}{\alpha}}} \]
\[ P_{\text{min}} = 2R\sqrt{\alpha t_0} \]

### Final Answer
Thus, the minimum attainable pressure of the ideal gas is:
\[ P_{\text{min}} = 2R\sqrt{\alpha t_0} \]

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