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

Options

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

Answer

To solve the problem, we need to analyze the situation of an ideal gas undergoing an isothermal expansion while doing work against its surroundings. Here's a step-by-step breakdown:

### Step 1: Understand the Process
In an isothermal process, the temperature of the system remains constant. For an ideal gas, this means that any heat added to the system will be used to do work, and there will be no change in internal energy.

**Hint:** Remember that in an isothermal process, the internal energy change (\( \Delta U \)) for an ideal gas is zero.

### Step 2: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta Q = \Delta U + W \]
Where:
- \( \Delta Q \) is the heat added to the system,
- \( \Delta U \) is the change in internal energy,
- \( W \) is the work done by the system.

Since the process is isothermal for an ideal gas, we have:
\[ \Delta U = 0 \]
Thus, the equation simplifies to:
\[ \Delta Q = W \]

**Hint:** Identify that for an isothermal process involving an ideal gas, the change in internal energy is zero.

### Step 3: Analyze the Work Done
The problem states that the gas does \(-150 J\) of work against its surroundings. The negative sign indicates that work is done by the system (the gas) on the surroundings.

**Hint:** Recall that work done by the system is considered negative in thermodynamics when the system expands.

### Step 4: Relate Work to Heat Transfer
From the simplified first law equation:
\[ \Delta Q = W \]
Since \( W = -150 J \), we have:
\[ \Delta Q = -150 J \]
This means that the system has released 150 J of heat to the surroundings.

**Hint:** Pay attention to the sign of work done; it indicates the direction of energy transfer.

### Step 5: Conclusion
Since the heat transfer \( \Delta Q \) is negative, we conclude that 150 J of heat has been removed from the gas.

**Final Answer:** The correct implication is that **150 J of heat has been removed from the gas.**