Let Q and W denote the amount of heat given to an ideal gas the work done by it in an isothermal process.

Options

A. `Q = 0`
B. `W = 0 `
C. `Q != W`
D. `Q = W `

Answer

To solve the problem, we need to analyze the relationship between heat (Q), work done (W), and internal energy change (ΔU) for an ideal gas undergoing an isothermal process.

### Step-by-Step Solution:

1. **Understanding the Isothermal Process**:
- In an isothermal process, the temperature of the system remains constant. Therefore, the change in temperature (ΔT) is zero.
- Since temperature is constant, the internal energy of an ideal gas, which depends only on temperature, also remains constant.

2. **Change in Internal Energy**:
- The change in internal energy (ΔU) for an ideal gas can be expressed as:
\[ \Delta U = \frac{f}{2} n R \Delta T \]
where \( f \) is the degrees of freedom, \( n \) is the number of moles, and \( R \) is the universal gas constant.
- Since ΔT = 0 for an isothermal process, we have:
\[ \Delta U = 0 \]

3. **Applying the First Law of Thermodynamics**:
- The First Law of Thermodynamics states:
\[ Q = \Delta U + W \]
- Substituting ΔU = 0 into the equation gives:
\[ Q = 0 + W \]
or simply,
\[ Q = W \]

4. **Conclusion**:
- From the above analysis, we conclude that for an ideal gas undergoing an isothermal process, the heat added to the gas (Q) is equal to the work done by the gas (W).
- Therefore, the correct answer is:
\[ Q = W \]

### Final Answer:
The correct option is D: \( Q = W \).