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

Options

A. (a) `3/2`
B. (b) `2/3`
C. (c) `1/3`
D. (d) `3/5`

Explanation

To solve the problem, we need to find the ratio of the change in internal energy (ΔU) to the work done (W) by an ideal monoatomic gas undergoing a process where the product of internal energy (U) and density (ρ) is constant.

### Step-by-Step Solution:

1. **Understand the relationship**: Given that \( U \cdot \rho = \text{constant} \), we can express this as:
\[ U = \frac{C}{\rho} \]
where \( C \) is a constant.

2. **Express internal energy in terms of density**: For an ideal monoatomic gas, the internal energy \( U \) is related to temperature \( T \) by:
\[ U = \frac{3}{2} nRT \]
where \( n \) is the number of moles and \( R \) is the universal gas constant.

3. **Relate temperature to density**: Since \( U \) is proportional to \( \frac{1}{\rho} \), we can also express temperature as:
\[ T \propto \frac{1}{\rho} \]
This implies that as the density increases, the temperature decreases.

4. **Using the ideal gas law**: The ideal gas law states:
\[ PV = nRT \]
Since \( \rho = \frac{m}{V} \) and mass \( m \) is constant, we can say:
\[ \rho \propto \frac{1}{V} \implies V \propto \frac{1}{\rho} \]

5. **Establish the relationship between T and V**: From the previous steps, we have:
\[ T \propto V \]

6. **Work done by the gas**: The work done by the gas at constant pressure can be expressed as:
\[ W = P \Delta V \]
Using the ideal gas law, we can relate pressure to temperature and volume:
\[ P = \frac{nRT}{V} \]

7. **First Law of Thermodynamics**: According to the first law,
\[ \Delta U = Q - W \]
For a process at constant pressure, the heat added \( Q \) can be expressed as:
\[ Q = nC_p \Delta T \]

8. **Expressing ΔU and W**:
- The change in internal energy for a monoatomic gas is:
\[ \Delta U = nC_v \Delta T \]
- The work done is:
\[ W = nC_p \Delta T - \Delta U \]

9. **Finding the ratio**: Now we can find the ratio of change in internal energy to work done:
\[ \frac{\Delta U}{W} = \frac{nC_v \Delta T}{nC_p \Delta T - nC_v \Delta T} \]
Simplifying gives:
\[ \frac{\Delta U}{W} = \frac{C_v}{C_p - C_v} \]
For a monoatomic ideal gas, we know:
\[ C_p - C_v = R \]
Thus,
\[ \frac{\Delta U}{W} = \frac{C_v}{R} \]

10. **Substituting the values**: For a monoatomic gas, \( C_v = \frac{3R}{2} \):
\[ \frac{\Delta U}{W} = \frac{\frac{3R}{2}}{R} = \frac{3}{2} \]

### Final Answer:
The ratio of change in internal energy to the work done by the gas is:
\[ \frac{\Delta U}{W} = \frac{3}{2} \]