The pressure of an ideal gas varies according to the law `P = P_(0) - AV^(2)`, where `P_(0)` and `A` are positive constants. Find the highest temperature that can be attained by the gas

Options

A. `(2P_(0))/(3R) (P_(0)/(3a))^(1//2)`
B. `(3P_(0))/(2R) (P_(0)/(3a))^(1//2)`
C. `(P_(0))/(R) (P_(0)/(3a))^(1//2)`
D. `(P_(0))/(3R) (P_(0)/(3a))^(1//2)`

Explanation

To find the highest temperature that can be attained by the gas given the pressure law \( P = P_0 - AV^2 \), we will use the ideal gas equation and perform some calculus. Here’s the step-by-step solution:

### Step 1: Write the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature.

### Step 2: Substitute the Pressure Equation
From the problem, we have:
\[ P = P_0 - AV^2 \]
Substituting this into the ideal gas law gives:
\[ (P_0 - AV^2)V = nRT \]

### Step 3: Rearrange to Find Temperature
Rearranging the equation for temperature \( T \):
\[ T = \frac{(P_0 - AV^2)V}{nR} \]
This can be rewritten as:
\[ T = \frac{P_0 V}{nR} - \frac{AV^3}{nR} \]

### Step 4: Differentiate Temperature with Respect to Volume
To find the maximum temperature, we need to differentiate \( T \) with respect to \( V \) and set the derivative equal to zero:
\[ \frac{dT}{dV} = \frac{P_0}{nR} - \frac{3AV^2}{nR} = 0 \]

### Step 5: Solve for Volume
Setting the derivative equal to zero:
\[ \frac{P_0}{nR} - \frac{3AV^2}{nR} = 0 \]
This simplifies to:
\[ P_0 = 3AV^2 \]
From this, we can solve for \( V \):
\[ V^2 = \frac{P_0}{3A} \quad \Rightarrow \quad V = \sqrt{\frac{P_0}{3A}} \]

### Step 6: Substitute Back to Find Maximum Temperature
Now, substitute \( V \) back into the temperature equation:
\[ T_{max} = \frac{P_0 \sqrt{\frac{P_0}{3A}}}{nR} - \frac{A\left(\sqrt{\frac{P_0}{3A}}\right)^3}{nR} \]
Calculating the second term:
\[ \left(\sqrt{\frac{P_0}{3A}}\right)^3 = \frac{P_0^{3/2}}{(3A)^{3/2}} = \frac{P_0^{3/2}}{3\sqrt{3}A^{3/2}} \]
Thus:
\[ T_{max} = \frac{P_0 \sqrt{\frac{P_0}{3A}}}{nR} - \frac{A \cdot \frac{P_0^{3/2}}{3\sqrt{3}A^{3/2}}}{nR} \]
This simplifies to:
\[ T_{max} = \frac{P_0^{3/2}}{nR \sqrt{3A}} - \frac{P_0^{3/2}}{3nR \sqrt{3A}} = \frac{2P_0^{3/2}}{3nR \sqrt{3A}} \]

### Final Result
Thus, the highest temperature that can be attained by the gas is:
\[ T_{max} = \frac{2P_0^{3/2}}{3nR \sqrt{3A}} \]