What would be the work done on`//` by the system if the internal energy of the system falls by the 100 joules even when 200 joules of heat is supplied to it ?

Answer

To solve the problem, we will use the first law of thermodynamics, which is expressed by the equation:

\[ \Delta U = Q + W \]

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

### Step-by-Step Solution:

1. **Identify the given values**:
- The change in internal energy (\(\Delta U\)) is given as -100 joules (since it falls).
- The heat supplied to the system (Q) is +200 joules.

2. **Substitute the values into the first law of thermodynamics**:
\[ \Delta U = Q + W \]
Plugging in the known values:
\[ -100 \, \text{J} = 200 \, \text{J} + W \]

3. **Rearrange the equation to solve for W**:
\[ W = \Delta U - Q \]
Substituting the values:
\[ W = -100 \, \text{J} - 200 \, \text{J} \]

4. **Calculate W**:
\[ W = -300 \, \text{J} \]

5. **Interpret the result**:
The negative sign indicates that the work is done on the system.

### Final Answer:
The work done on the system is -300 joules.

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