The temperature of a monatomic ideal gas remains constant during a process in which 4700 J of heat flows out of the gas. How much work (including the proper + or - sign) is done?

Options

A. zero J
B. `+ 94 00 J`
C. `+47 00 J`
D. `-47 00 J`

Answer

To solve the problem, we will use the First Law of Thermodynamics, which states:

\[ \Delta U = Q + W \]

Where:
- \(\Delta U\) is the change in internal energy,
- \(Q\) is the heat added to the system,
- \(W\) is the work done on the system.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Heat flowing out of the gas, \(Q = -4700 \, \text{J}\) (negative because heat is leaving the system).
- The process is isothermal (constant temperature) for a monatomic ideal gas.

2. **Determine the Change in Internal Energy**:
- For an ideal gas, the change in internal energy (\(\Delta U\)) is related to the temperature. Since the temperature is constant, the change in internal energy is:
\[ \Delta U = 0 \]

3. **Apply the First Law of Thermodynamics**:
- Substitute the known values into the First Law equation:
\[ \Delta U = Q + W \]
\[ 0 = -4700 \, \text{J} + W \]

4. **Solve for Work Done (W)**:
- Rearranging the equation gives:
\[ W = 4700 \, \text{J} \]

5. **Determine the Sign of Work**:
- Since the work is done by the gas on the surroundings (as heat is leaving), we take it as positive:
\[ W = +4700 \, \text{J} \]

### Final Answer:
The work done by the gas is \(W = +4700 \, \text{J}\).

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