An ideal gas in a cylinder is slowly compressed to one third of its original volume. During this process, the temperature of the gas remains constant and the work done in compression is 75 J. How much heat flows into the gas ?

Answer

To solve the problem, we need to analyze the situation using the first law of thermodynamics and the properties of an isothermal process.

### Step-by-Step Solution:

1. **Identify the Process**:
The problem states that the gas is compressed to one third of its original volume while maintaining a constant temperature. This indicates that the process is isothermal.

2. **Understand the Work Done**:
The work done on the gas during compression is given as 75 J. Since work is done on the gas, we consider this as negative work in our calculations:
\[ W = -75 \, \text{J} \]

3. **Apply the First Law of Thermodynamics**:
The first law of thermodynamics is expressed as:
\[ \Delta Q = \Delta U + W \]
where:
- \(\Delta Q\) is the heat added to the system,
- \(\Delta U\) is the change in internal energy,
- \(W\) is the work done on the system.

4. **Calculate Change in Internal Energy**:
For an ideal gas undergoing an isothermal process, the change in internal energy (\(\Delta U\)) is zero because the internal energy of an ideal gas depends only on temperature, and the temperature remains constant:
\[ \Delta U = 0 \]

5. **Substitute Values into the First Law Equation**:
Now, substituting \(\Delta U = 0\) and \(W = -75 \, \text{J}\) into the first law equation:
\[ \Delta Q = 0 + (-75 \, \text{J}) = -75 \, \text{J} \]

6. **Interpret the Result**:
The negative sign indicates that heat is flowing out of the gas. Therefore, the amount of heat that flows into the gas is:
\[ \Delta Q = -75 \, \text{J} \]

### Final Answer:
The heat that flows into the gas is \(-75 \, \text{J}\), which means that 75 J of heat flows out of the gas.

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