In a given process on an ideal gas, dW = 0 and `dQ lt 0`. Then for the gas

Options

A. the temperature will decrease
B. the volume will increases
C. the pressure will remain constant
D. the temperature will increase

Explanation

To solve the problem, we need to analyze the given conditions for an ideal gas and apply the first law of thermodynamics.

### Step-by-Step Solution:

1. **Understanding the Given Conditions:**
- We are given that \( dW = 0 \) (work done is zero).
- We also have \( dQ < 0 \) (heat added to the system is negative, meaning heat is removed).

2. **Applying the First Law of Thermodynamics:**
- The first law of thermodynamics states:
\[ dQ = dU + dW \]
- Since \( dW = 0 \), we can simplify this to:
\[ dQ = dU \]

3. **Analyzing the Sign of \( dU \):**
- Given that \( dQ < 0 \), it follows that:
\[ dU < 0 \]
- This means that the internal energy of the gas is decreasing.

4. **Relating Internal Energy to Temperature:**
- For an ideal gas, the change in internal energy is related to the change in temperature by the equation:
\[ dU = nC_v dT \]
- Here, \( n \) is the number of moles and \( C_v \) is the specific heat at constant volume, which is a positive constant.

5. **Determining the Change in Temperature:**
- Since \( dU < 0 \) and \( nC_v \) is positive, it follows that:
\[ dT < 0 \]
- This indicates that the temperature of the gas is decreasing.

6. **Conclusion:**
- Based on the analysis, since the internal energy is decreasing, the temperature of the gas must also decrease. Therefore, the correct option is:
- **Option A: The temperature will decrease.**