To solve this problem, let's analyze the situation step by step using the first law of thermodynamics and the relationships between heat (\(q\)), work (\(w\)), and internal energy (\(\Delta U\)).
### Given Information:
- 1 mole of gas is heated at constant volume.
- Temperature increases from \(T_1 = 298 \, \text{K}\) to \(T_2 = 308 \, \text{K}\).
- Heat supplied to the gas, \(q = 500 \, \text{J}\).
### Key Concepts:
1. **Constant Volume Process:** In a constant volume process, the work done on or by the gas (\(w\)) is zero because work in a gas is given by \(w = P \Delta V\) and \(\Delta V = 0\) at constant volume.
Therefore, \(w = 0\).
2. **First Law of Thermodynamics:**
The first law of thermodynamics states that:
\[
\Delta U = q - w
\]
Since \(w = 0\):
\[
\Delta U = q - 0 = q
\]
Thus, in this case:
\[
\Delta U = q = 500 \, \text{J}
\]
### Conclusion:
Based on the analysis:
- The heat supplied to the gas (\(q\)) is 500 J.
- The work done (\(w\)) is 0 J.
- The change in internal energy (\(\Delta U\)) is equal to the heat supplied, which is also 500 J.
### Answer:
Therefore, the correct option is:
**B. \(q = \Delta U = 500 \, \text{J}\), \(w = 0\)**.
