The pressure (P) and temperature (T) relationship of an ideal gas obeys the equation `PT^(2)=` constant. The volume expansion coefficient of the gas will be:

Options

A. `3T^(2)`
B. `(3)/(T^(2))`
C. `(3)/(T^(3))`
D. `(3)/(T)`

Answer

To find the volume expansion coefficient (β) of an ideal gas given the relationship \( PT^2 = \text{constant} \), we can follow these steps:

### Step-by-Step Solution:

1. **Start with the given relationship**:
\[ PT^2 = C \]
where \( C \) is a constant.

2. **Use the Ideal Gas Law**:
The ideal gas law states:
\[ PV = nRT \]
Rearranging this gives:
\[ P = \frac{nRT}{V} \]

3. **Substitute \( P \) into the first equation**:
Substitute \( P \) from the ideal gas law into the relationship \( PT^2 = C \):
\[ \left(\frac{nRT}{V}\right)T^2 = C \]
This simplifies to:
\[ \frac{nRT^3}{V} = C \]

4. **Rearranging the equation**:
Rearranging gives:
\[ T^3 = \frac{CV}{nR} \]
Let's denote \( \frac{C}{nR} \) as a new constant \( C' \):
\[ T^3 = C'V \]

5. **Differentiate with respect to time**:
Differentiate both sides with respect to \( T \):
\[ 3T^2 \frac{dT}{dt} = C' \frac{dV}{dt} \]

6. **Express \( \frac{dV}{dt} \)**:
Rearranging the differentiation gives:
\[ \frac{dV}{dt} = \frac{3T^2}{C'} \frac{dT}{dt} \]

7. **Define the volume expansion coefficient**:
The volume expansion coefficient \( \beta \) is defined as:
\[ \beta = \frac{1}{V} \frac{dV}{dT} \]
From the previous equation, we can express \( \frac{dV}{dT} \):
\[ \frac{dV}{dT} = \frac{3T^2}{C'} \]

8. **Substituting into the volume expansion coefficient**:
Substitute \( \frac{dV}{dT} \) into the equation for \( \beta \):
\[ \beta = \frac{1}{V} \cdot \frac{3T^2}{C'} \]

9. **Simplifying \( \beta \)**:
Since \( C' \) is a constant, we can express \( \beta \) in terms of \( T \):
\[ \beta = \frac{3}{T} \]

### Final Answer:
Thus, the volume expansion coefficient of the gas is:
\[ \beta = \frac{3}{T} \]