An ideal gas is expanded such that `PT^(2)=a` constant. The coefficient of volume expansion of the gas is

Options

A. `(1)/(T)`
B. `(2)/(T)`
C. `(3)/(T)`
D. `(4)/(T)`

Explanation

To find the coefficient of volume expansion of an ideal gas that expands such that \( P T^2 = a \) (a constant), we can follow these steps:

### Step 1: Start with the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature

### Step 2: Express Pressure in Terms of Volume and Temperature
From the ideal gas law, we can express pressure \( P \) as:
\[ P = \frac{nRT}{V} \]

### Step 3: Substitute Pressure into the Given Condition
We know that \( P T^2 = a \). Substituting the expression for \( P \) from Step 2 into this equation gives:
\[ \frac{nRT}{V} T^2 = a \]
This simplifies to:
\[ \frac{nRT^3}{V} = a \]

### Step 4: Rearranging the Equation
Rearranging the equation, we find:
\[ T^3 = \frac{aV}{nR} \]
This shows that \( T^3 \) is directly proportional to \( V \):
\[ T^3 \propto V \]

### Step 5: Differentiate the Relationship
To find the coefficient of volume expansion, we differentiate both sides with respect to time \( t \):
\[ \frac{d(T^3)}{dt} = \frac{d}{dt}\left(\frac{aV}{nR}\right) \]
Using the chain rule on the left side:
\[ 3T^2 \frac{dT}{dt} = \frac{a}{nR} \frac{dV}{dt} \]

### Step 6: Rearranging the Differentiated Equation
Rearranging gives:
\[ \frac{dV}{dt} = \frac{3nR}{a} T^2 \frac{dT}{dt} \]

### Step 7: Express the Coefficient of Volume Expansion
The coefficient of volume expansion \( \beta \) is defined as:
\[ \beta = \frac{1}{V} \frac{dV}{dT} \]
From our previous equation, we can express \( \frac{dV}{dT} \):
\[ \frac{dV}{dT} = \frac{3nR}{a} T^2 \]

### Step 8: Substitute into the Coefficient of Volume Expansion Formula
Now substituting back into the formula for \( \beta \):
\[ \beta = \frac{1}{V} \cdot \frac{3nR}{a} T^2 \]

### Step 9: Relate Volume to Temperature
Since \( T^3 \propto V \), we can express \( V \) in terms of \( T \):
\[ V = kT^3 \quad \text{(for some constant \( k \))} \]
Substituting this back into the equation for \( \beta \):
\[ \beta = \frac{1}{kT^3} \cdot \frac{3nR}{a} T^2 \]
This simplifies to:
\[ \beta = \frac{3nR}{a k T} \]

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