An ideal gas is taken through a process `PT^3` = constant . The coefficient of thermal expansion of the gas in the given process in n/T. Find n.

Answer

To solve the problem, we need to find the coefficient of thermal expansion of an ideal gas undergoing a process where \( P T^3 = \text{constant} \). We will denote the coefficient of thermal expansion as \( \gamma \).

### Step-by-Step Solution:

1. **Understand the Coefficient of Thermal Expansion**:
The coefficient of thermal expansion \( \gamma \) is defined as:
\[ \gamma = \frac{1}{V} \frac{dV}{dT} \]

2. **Relate Pressure and Volume**:
From the given process \( P T^3 = \text{constant} \), we can express pressure \( P \) in terms of volume \( V \) and temperature \( T \). Using the ideal gas law \( PV = nRT \), we can write:
\[ P = \frac{nRT}{V} \]
Therefore, substituting into the process equation gives:
\[ \frac{nRT}{V} T^3 = C \quad \text{(where C is a constant)} \]
Simplifying this, we get:
\[ \frac{nR T^4}{V} = C \]
Rearranging gives:
\[ V = \frac{nR T^4}{C} \]

3. **Differentiate Volume with Respect to Temperature**:
To find \( \frac{dV}{dT} \), we differentiate \( V \) with respect to \( T \):
\[ V = \frac{nR}{C} T^4 \]
Thus,
\[ \frac{dV}{dT} = \frac{nR}{C} \cdot 4T^3 \]

4. **Substituting into the Coefficient of Thermal Expansion**:
Now, substituting \( \frac{dV}{dT} \) back into the expression for \( \gamma \):
\[ \gamma = \frac{1}{V} \frac{dV}{dT} = \frac{1}{\frac{nR}{C} T^4} \cdot \left( \frac{nR}{C} \cdot 4T^3 \right) \]
Simplifying this gives:
\[ \gamma = \frac{4T^3}{T^4} = \frac{4}{T} \]

5. **Identifying \( n \)**:
We are given that \( \gamma = \frac{n}{T} \). From our calculation, we have:
\[ \gamma = \frac{4}{T} \]
Therefore, comparing both expressions:
\[ \frac{n}{T} = \frac{4}{T} \]
This leads us to conclude:
\[ n = 4 \]

### Final Answer:
\[ n = 4 \]