According to the first law of thermodynamics, `DeltaU=q+w`. In special cases the statement can be expressed in different ways. Which of the following is not a correct expression ?

Options

A. At constant temperature, `q=-w`
B. When no work is done , `DeltaU=q`
C. In gaseous system , `DeltaU=q+PDeltaV`
D. When work is done by the system : `DeltaU=q+w`

Explanation

To solve the question regarding the first law of thermodynamics, we need to analyze the given options based on the equation:

\[ \Delta U = Q + W \]

Where:
- \(\Delta U\) = Change in internal energy of the system
- \(Q\) = Heat absorbed or released
- \(W\) = Work done on or by the system

We will evaluate each option to determine which one is not a correct expression of the first law.

### Step-by-Step Solution:

**Step 1: Analyze Option 1**
- **Statement**: At constant temperature, \(Q = -W\).
- **Analysis**: In an isothermal process (constant temperature), the change in internal energy (\(\Delta U\)) is zero because internal energy is a function of temperature. Therefore, we have:
\[ 0 = Q + W \implies Q = -W \]
- **Conclusion**: This statement is correct.

**Step 2: Analyze Option 2**
- **Statement**: When no work is done, \(\Delta U = Q\).
- **Analysis**: If no work is done (\(W = 0\)), the equation simplifies to:
\[ \Delta U = Q + 0 \implies \Delta U = Q \]
- **Conclusion**: This statement is correct.

**Step 3: Analyze Option 3**
- **Statement**: In a gaseous system, \(\Delta U = Q + P\Delta V\).
- **Analysis**: The work done in a gaseous system can be expressed as \(W = P\Delta V\). Thus, substituting this into the first law gives:
\[ \Delta U = Q + P\Delta V \]
- **Conclusion**: This statement is also correct.

**Step 4: Analyze Option 4**
- **Statement**: When work is done by the system, \(\Delta U = Q + W\).
- **Analysis**: The confusion arises here. When work is done by the system, \(W\) is considered positive in the equation. However, if the system does work on the surroundings, the heat absorbed (\(Q\)) should be negative, leading to:
\[ \Delta U = Q - W \text{ (if work is done by the system)} \]
- **Conclusion**: This statement is incorrect.

### Final Answer:
The expression that is not correct is **Option 4**.

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