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)`

Answer

To solve the problem, we need to find the volume for which the pressure of one mole of an ideal gas is at a minimum, given the relationship between temperature and volume as \( T = T_0 + aV^3 \).

### Step-by-Step Solution:

1. **Start with the Ideal Gas Law**:
The ideal gas law is given by:
\[ PV = nRT \]
For one mole of gas (\( n = 1 \)), this simplifies to:
\[ PV = RT \]

2. **Substitute the Expression for Temperature**:
We know from the problem that:
\[ T = T_0 + aV^3 \]
Substituting this into the ideal gas equation gives:
\[ PV = R(T_0 + aV^3) \]

3. **Rearranging the Equation**:
Rearranging the equation, we have:
\[ P = \frac{R(T_0 + aV^3)}{V} \]
This can be expressed as:
\[ P = \frac{RT_0}{V} + RaV^2 \]

4. **Finding the Condition for Minimum Pressure**:
To find the volume at which pressure is minimized, we need to take the derivative of \( P \) with respect to \( V \) and set it to zero:
\[ \frac{dP}{dV} = -\frac{RT_0}{V^2} + 2RaV = 0 \]

5. **Setting the Derivative to Zero**:
Setting the derivative equal to zero gives:
\[ -\frac{RT_0}{V^2} + 2RaV = 0 \]
Rearranging this, we get:
\[ 2RaV = \frac{RT_0}{V^2} \]

6. **Multiplying through by \( V^2 \)**:
Multiplying both sides by \( V^2 \) results in:
\[ 2RaV^3 = RT_0 \]

7. **Solving for \( V^3 \)**:
Dividing both sides by \( 2aR \):
\[ V^3 = \frac{T_0}{2a} \]

8. **Taking the Cube Root**:
Taking the cube root gives us:
\[ V = \left(\frac{T_0}{2a}\right)^{1/3} \]

### Final Answer:
The volume for which pressure will be minimum is:
\[ V = \left(\frac{T_0}{2a}\right)^{1/3} \]