During an isothermal expansion, a confined ideal gas does `-150 J` of work aginst its surroundings. This implies that

Options

A. `150 J` of heat has been removed from the gas
B. `300 J` of heat has been added to the gas
C. no heat is transferred because the process is isothermal
D. `150 J` of heat been added to the gas

Explanation

To solve the problem, we need to analyze the situation of an ideal gas undergoing an isothermal expansion while doing work against its surroundings.

### Step-by-Step Solution:

1. **Understanding the Process**:
- The process is isothermal, which means the temperature of the gas remains constant throughout the expansion.

2. **Work Done by the Gas**:
- The gas does `-150 J` of work against its surroundings. The negative sign indicates that the system (gas) is doing work on the surroundings.

3. **First Law of Thermodynamics**:
- According to the first law of thermodynamics, the change in internal energy (ΔU) of the system is given by:
\[ \Delta U = Q - W \]
- Where:
- \( Q \) is the heat added to the system.
- \( W \) is the work done by the system (in this case, it is negative because the gas is doing work).

4. **Internal Energy Change in Isothermal Process**:
- For an ideal gas undergoing an isothermal process, the change in internal energy (ΔU) is zero:
\[ \Delta U = 0 \]
- Therefore, we can simplify the first law equation to:
\[ 0 = Q - W \]
- Rearranging gives us:
\[ Q = W \]

5. **Calculating Heat Transfer**:
- Since the work done \( W = -150 J \), we substitute this into the equation:
\[ Q = -(-150 J) = 150 J \]
- This indicates that the gas absorbs 150 J of heat from its surroundings.

6. **Conclusion**:
- During the isothermal expansion, since the gas does work against the surroundings, it must absorb heat to maintain a constant temperature. Therefore, **150 J of heat is added to the gas**.

### Final Answer:
The implication of the gas doing `-150 J` of work against its surroundings during isothermal expansion is that **150 J of heat is added to the gas**.