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 and \( V \) is the volume of one mole of gas, we can follow these steps:

### Step 1: Relate Volume to Pressure and Temperature
For one mole of an ideal gas, we use the ideal gas equation:
\[ PV = RT \]
From this, we can express the volume \( V \) in terms of pressure \( P \) and temperature \( T \):
\[ V = \frac{RT}{P} \]

### Step 2: Substitute Volume into the Given Temperature Equation
Now, substitute \( V \) into the temperature equation:
\[ T = t_0 + \alpha V^2 \]
Substituting for \( V \):
\[ T = t_0 + \alpha \left(\frac{RT}{P}\right)^2 \]
This simplifies to:
\[ T = t_0 + \alpha \frac{R^2 T^2}{P^2} \]

### Step 3: Rearranging the Equation
Rearranging the equation gives us:
\[ TP^2 = t_0 P^2 + \alpha R^2 T^2 \]
This can be rearranged to isolate \( P \):
\[ P^2 (T - t_0) = \alpha R^2 T^2 \]
Thus, we can express \( P \) as:
\[ P = \sqrt{\frac{\alpha R^2 T^2}{T - t_0}} \]

### Step 4: Differentiate to Find Minimum Pressure
To find the minimum pressure, we need to differentiate \( P \) with respect to \( T \) and set the derivative equal to zero:
\[ \frac{dP}{dT} = 0 \]
Using the quotient and chain rule, we differentiate:
\[ P = \sqrt{\frac{\alpha R^2 T^2}{T - t_0}} \]
Let \( u = \alpha R^2 T^2 \) and \( v = T - t_0 \), then:
\[ \frac{dP}{dT} = \frac{1}{2} \left(\frac{u}{v}\right)^{-1/2} \left(\frac{v \frac{du}{dT} - u \frac{dv}{dT}}{v^2}\right) \]
Where:
\[ \frac{du}{dT} = 2\alpha R^2 T \]
\[ \frac{dv}{dT} = 1 \]
Setting the derivative to zero gives:
\[ v(2\alpha R^2 T) - u = 0 \]
Substituting back gives:
\[ (T - t_0)(2\alpha R^2 T) - \alpha R^2 T^2 = 0 \]

### Step 5: Solve for Temperature \( T \)
This simplifies to:
\[ 2T - 2t_0 = T \implies T = 2t_0 \]

### Step 6: Substitute Back to Find Minimum Pressure
Now substitute \( T = 2t_0 \) back into the pressure equation:
\[ P = \sqrt{\frac{\alpha R^2 (2t_0)^2}{2t_0 - t_0}} = \sqrt{\frac{4\alpha R^2 t_0^2}{t_0}} = \sqrt{4\alpha R^2 t_0} = 2R\sqrt{\alpha t_0} \]

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

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