An ideal monoatomic gas undergoes a process in which its internal energy U and density `rho` vary as `Urho`= constant. The ratio of change in internal energy and the work done by the gas is

Answer

To solve the problem, we need to find the ratio of the change in internal energy (ΔU) to the work done (W) by the gas during the specified process. The process is defined by the relationship \( U \cdot \rho = \text{constant} \).

### Step-by-Step Solution:

1. **Understanding the Given Relationship**:
We are given that \( U \cdot \rho = \text{constant} \). This implies that as the density \( \rho \) changes, the internal energy \( U \) changes in such a way that their product remains constant.

2. **Expressing Density**:
The density \( \rho \) can be expressed as:
\[ \rho = \frac{m}{V} \]
where \( m \) is the mass and \( V \) is the volume. Since the mass of the gas remains constant, we can say that \( \rho \) is inversely proportional to \( V \):
\[ \rho \propto \frac{1}{V} \]

3. **Relating Internal Energy to Temperature**:
For an ideal monoatomic gas, the internal energy \( U \) is directly proportional to the temperature \( T \):
\[ U = \frac{3}{2} nRT \]
where \( n \) is the number of moles and \( R \) is the gas constant.

4. **Connecting Temperature and Volume**:
Since \( U \cdot \rho = \text{constant} \), we can express this in terms of temperature and volume:
\[ U \cdot \rho = \frac{3}{2} nRT \cdot \frac{m}{V} = \text{constant} \]
This implies that \( T \) must vary with \( V \) such that the product remains constant.

5. **Using the First Law of Thermodynamics**:
The first law of thermodynamics states:
\[ \Delta U = Q - W \]
For an isobaric process (constant pressure), the work done \( W \) can be expressed as:
\[ W = P \Delta V \]
where \( P \) is the pressure.

6. **Finding the Ratio**:
From the first law, we can rearrange it to find \( W \):
\[ W = Q - \Delta U \]
If we consider the specific heat at constant volume \( C_V \) and at constant pressure \( C_P \), we know:
\[ C_P - C_V = R \]
For a monoatomic gas, \( C_V = \frac{3}{2} R \) and \( C_P = \frac{5}{2} R \).

7. **Calculating the Ratio**:
The ratio of the change in internal energy to the work done can be expressed as:
\[ \frac{\Delta U}{W} = \frac{n C_V \Delta T}{n C_P \Delta T} = \frac{C_V}{C_P} \]
Substituting the values:
\[ \frac{\Delta U}{W} = \frac{\frac{3}{2} R}{\frac{5}{2} R} = \frac{3}{5} \]

### Final Answer:
The ratio of the change in internal energy to the work done by the gas is:
\[ \frac{\Delta U}{W} = \frac{3}{5} \]