One mole of an ideal gas undergoes a process in which `T = T_(0) + aV^(3)`, where `T_(0)` and `a` are positive constants and V is molar volume. The volume for which pressure with be minimum is

Options

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

Explanation

To find the volume at which the pressure of one mole of an ideal gas is minimized during the process described by \( T = T_0 + aV^3 \), we can follow these steps:

### Step 1: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
For one mole of gas (\( n = 1 \)), we can rewrite this as:
\[ PV = RT \]

### Step 2: Substitute the Expression for Temperature
We know that \( T = T_0 + aV^3 \). Substituting this into the ideal gas equation gives:
\[ PV = R(T_0 + aV^3) \]
This can be rearranged to:
\[ P = \frac{R(T_0 + aV^3)}{V} \]

### 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:
\[ \frac{dP}{dV} = \frac{d}{dV}\left(\frac{R(T_0 + aV^3)}{V}\right) \]
Using the quotient rule, we have:
\[ \frac{dP}{dV} = \frac{V \cdot R \cdot (3aV^2) - R(T_0 + aV^3)}{V^2} \]
This simplifies to:
\[ \frac{dP}{dV} = \frac{R(3aV^3 - T_0 - aV^3)}{V^2} \]
\[ = \frac{R(2aV^3 - T_0)}{V^2} \]

### Step 4: Set the Derivative Equal to Zero
Setting the derivative equal to zero for minimization:
\[ 2aV^3 - T_0 = 0 \]
This gives:
\[ 2aV^3 = T_0 \]

### Step 5: Solve for Volume
Rearranging the equation to find \( V \):
\[ V^3 = \frac{T_0}{2a} \]
Taking the cube root, we find:
\[ V = \left(\frac{T_0}{2a}\right)^{1/3} \]

### Conclusion
Thus, the volume at which the pressure is minimized is:
\[ V = \left(\frac{T_0}{2a}\right)^{1/3} \]