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 at which the pressure of one mole of an ideal gas is minimized, given the relationship between temperature and volume as \( T = T_0 + aV^3 \).

### Step-by-Step Solution:

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

2. **Substitute the Expression for Temperature**:
We substitute the expression for temperature \( T = T_0 + aV^3 \) into the ideal gas law:
\[ PV = R(T_0 + aV^3) \]

3. **Rearrange to Find Pressure**:
Rearranging gives us:
\[ P = \frac{R(T_0 + aV^3)}{V} \]
This can be simplified to:
\[ P = \frac{RT_0}{V} + aRV^2 \]

4. **Differentiate Pressure with Respect to Volume**:
To find the volume at which pressure is minimized, we differentiate \( P \) with respect to \( V \):
\[ \frac{dP}{dV} = -\frac{RT_0}{V^2} + 2aRV \]

5. **Set the Derivative Equal to Zero**:
To find the minimum pressure, we set the derivative equal to zero:
\[ -\frac{RT_0}{V^2} + 2aRV = 0 \]

6. **Solve for Volume**:
Rearranging the equation gives:
\[ 2aRV = \frac{RT_0}{V^2} \]
Multiplying both sides by \( V^2 \):
\[ 2aRV^3 = RT_0 \]
Dividing both sides by \( 2aR \):
\[ V^3 = \frac{T_0}{2a} \]
Taking the cube root gives:
\[ V = \left(\frac{T_0}{2a}\right)^{1/3} \]

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