The temperature (T) of one mole of an ideal gas varies with its volume (V) as `T= - alpha V^(3) + beta V^(2)`, where `alpha and beta` are positive constants. The maximum pressure of gas during this process is

Options

A. `(alpha beta)/(2 R)`
B. `(beta^(2) R)/(4 alpha)`
C. `((alpha + beta) R)/(2 beta^(2))`
D. `(alpha^(2) R)/(2 beta)`

Answer

To find the maximum pressure of one mole of an ideal gas when the temperature varies with volume as given by \( T = -\alpha V^3 + \beta V^2 \), we will follow these steps:

### Step 1: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
Since we have one mole of gas (\( n = 1 \)), we can rewrite this as:
\[ P = \frac{RT}{V} \]
Substituting the expression for temperature \( T \):
\[ P = \frac{R(-\alpha V^3 + \beta V^2)}{V} \]
This simplifies to:
\[ P = R(-\alpha V^2 + \beta V) \]

### Step 2: Differentiate Pressure with Respect to Volume
To find the maximum pressure, we need to differentiate \( P \) with respect to \( V \):
\[ \frac{dP}{dV} = R(-2\alpha V + \beta) \]
Setting the derivative equal to zero to find critical points:
\[ -2\alpha V + \beta = 0 \]
From this, we can solve for \( V \):
\[ V = \frac{\beta}{2\alpha} \]

### Step 3: Determine the Nature of the Critical Point
To confirm that this critical point is a maximum, we need to take the second derivative:
\[ \frac{d^2P}{dV^2} = -2R\alpha \]
Since \( \alpha \) is positive, \( \frac{d^2P}{dV^2} < 0 \), indicating that \( P \) has a maximum at this value of \( V \).

### Step 4: Substitute Back to Find Maximum Pressure
Now we substitute \( V = \frac{\beta}{2\alpha} \) back into the expression for \( P \):
\[ P_{\text{max}} = R\left(-\alpha\left(\frac{\beta}{2\alpha}\right)^2 + \beta\left(\frac{\beta}{2\alpha}\right)\right) \]
Calculating this gives:
\[ P_{\text{max}} = R\left(-\alpha\frac{\beta^2}{4\alpha^2} + \frac{\beta^2}{2\alpha}\right) \]
\[ = R\left(-\frac{\beta^2}{4\alpha} + \frac{2\beta^2}{4\alpha}\right) \]
\[ = R\left(\frac{\beta^2}{4\alpha}\right) \]
Thus, the maximum pressure is:
\[ P_{\text{max}} = \frac{R\beta^2}{4\alpha} \]

### Final Answer
The maximum pressure of the gas during this process is:
\[ P_{\text{max}} = \frac{R\beta^2}{4\alpha} \]