In a given process on an ideal gas, `dW=0` and `dQ` is negative, then for the gas:

Options

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

Explanation

To solve the problem, we will apply the first law of thermodynamics and analyze the implications of the given conditions for the ideal gas.

### Step-by-Step Solution:

1. **Understand the Given Conditions**:
- We have \( dW = 0 \) (work done is zero).
- We have \( dQ < 0 \) (heat added to the system is negative, meaning heat is lost).

2. **Apply 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. **Substitute the Given Values**:
- We know \( dQ < 0 \), therefore:
\[ dU < 0 \]
- This indicates that the change in internal energy is negative.

4. **Relate Internal Energy to Temperature**:
- For an ideal gas, the internal energy \( U \) is related to temperature \( T \) by the equation:
\[ dU = nC_v dT \]
- Here, \( n \) is the number of moles and \( C_v \) is the specific heat at constant volume.
- Since \( dU < 0 \) and \( nC_v > 0 \) (as both \( n \) and \( C_v \) are positive), it follows that:
\[ dT < 0 \]

5. **Conclusion**:
- A negative change in temperature (\( dT < 0 \)) implies that the temperature of the gas decreases.

### Final Answer:
The temperature of the gas decreases.