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 specific process where the product of internal energy (U) and density (ρ) is constant (Uρ = constant).

### Step-by-Step Solution:

1. **Understand the Given Relation**:
We start with the relation given in the problem:
\[ U \cdot \rho = \text{constant} \]
This implies that:
\[ U \propto \frac{1}{\rho} \]
Let's denote this as Equation (1).

2. **Relate 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 universal gas constant. This means:
\[ U \propto T \]
Therefore, from Equation (1), we can conclude:
\[ T \propto \frac{1}{\rho} \]
Let's denote this as Equation (2).

3. **Relate Density to Volume**:
Density (ρ) is defined as:
\[ \rho = \frac{m}{V} \]
where \( m \) is the mass and \( V \) is the volume. Thus, we can say:
\[ \rho \propto \frac{1}{V} \implies \frac{1}{\rho} \propto V \]
Let's denote this as Equation (3).

4. **Combine Relationships**:
From Equations (2) and (3), we can conclude:
\[ T \propto V \]
This indicates that the temperature increases with volume.

5. **Identify the Type of Process**:
Since temperature is proportional to volume and not changing in a way that would indicate a change in pressure, we can conclude that the process occurs at constant pressure (p = constant).

6. **Use the First Law of Thermodynamics**:
The first law of thermodynamics states:
\[ Q = \Delta U + W \]
Rearranging gives:
\[ W = Q - \Delta U \]

7. **Express Work Done**:
For a constant pressure process, the work done (W) can be expressed as:
\[ W = P \Delta V \]
The change in internal energy (ΔU) for an ideal gas can be expressed as:
\[ \Delta U = n C_V \Delta T \]
where \( C_V \) for a monoatomic gas is \( \frac{3}{2} R \).

8. **Express Heat Added (Q)**:
For a constant pressure process, the heat added can be expressed as:
\[ Q = n C_P \Delta T \]
where \( C_P \) for a monoatomic gas is \( \frac{5}{2} R \).

9. **Calculate the Ratio**:
Now, substituting these into the expression for the ratio of change in internal energy to work done:
\[ \frac{\Delta U}{W} = \frac{n C_V \Delta T}{n C_P \Delta T - n C_V \Delta T} \]
Simplifying gives:
\[ \frac{\Delta U}{W} = \frac{C_V}{C_P - C_V} \]
From Mayer's relation, we know:
\[ C_P - C_V = R \]
Thus, substituting \( C_V = \frac{3}{2} R \):
\[ \frac{\Delta U}{W} = \frac{\frac{3}{2} R}{R} = \frac{3}{2} \]

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