The first law of thermodynamics for a closed system is dU = dq + dw, where dw = `dw_(pv)+dw_("non-pv")`. The most common type of `w_("non-pv")` is electrical work. As per IUPAC convention work done on the system is positive.
A system generates 50 J electrical energy, has 150 J of pressure-volume work done on it by the surroundings while releasing 300 J of heat energy. What is the change in the internal energy of the sytem?

Options

A. `-500`
B. `-100`
C. `-300`
D. `-200`

Explanation

To solve the problem, we will use the first law of thermodynamics, which states:

\[ dU = dq + dw \]

Where:
- \( dU \) is the change in internal energy.
- \( dq \) is the heat added to the system.
- \( dw \) is the work done on the system.

In this case, we can break down the work done into two components:
- \( dw_{(pv)} \): Pressure-volume work done on the system.
- \( dw_{(non-pv)} \): Non-pressure-volume work, which in this case is electrical work.

### Given Data:
1. Electrical energy generated (non-pv work) = 50 J (This is work done by the system, so it will be negative)
2. Pressure-volume work done on the system = 150 J (This is work done on the system, so it will be positive)
3. Heat energy released = 300 J (Since heat is released, it will be negative)

### Step-by-Step Solution:

1. **Identify the heat transfer (dq)**:
\[ dq = -300 \, \text{J} \quad (\text{heat is released}) \]

2. **Identify the work done (dw)**:
- Pressure-volume work done on the system:
\[ dw_{(pv)} = +150 \, \text{J} \quad (\text{work done on the system is positive}) \]
- Electrical work done by the system:
\[ dw_{(non-pv)} = -50 \, \text{J} \quad (\text{work done by the system is negative}) \]

3. **Calculate total work done (dw)**:
\[ dw = dw_{(pv)} + dw_{(non-pv)} = 150 \, \text{J} - 50 \, \text{J} = 100 \, \text{J} \]

4. **Apply the first law of thermodynamics**:
\[ dU = dq + dw \]
Substituting the values:
\[ dU = -300 \, \text{J} + 100 \, \text{J} = -200 \, \text{J} \]

5. **Final Result**:
The change in internal energy of the system is:
\[ dU = -200 \, \text{J} \]

### Summary:
The change in internal energy of the system is \(-200 \, \text{J}\).

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