When 1 mol of gas is heated at constant volume temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct ?

Options

A. `q = Deltacup = -500 J, w = 0`
B. `q = Deltacup = + 500 J, w = 0`
C. `q = w - 500 J, Deltacup = 0`
D. `Delta cup = 0, q = w = -500 J`

Answer

To solve the problem, we will use the first law of thermodynamics and the information provided in the question.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Number of moles of gas (n) = 1 mol
- Initial temperature (T1) = 298 K
- Final temperature (T2) = 308 K
- Heat supplied (q) = 500 J
- Process: Constant volume

2. **Understand the First Law of Thermodynamics:**
The first law of thermodynamics states that:
\[ \Delta U = q + W \]
where:
- \(\Delta U\) = change in internal energy
- \(q\) = heat added to the system
- \(W\) = work done by the system

3. **Determine Work Done (W):**
Since the process is at constant volume, there is no change in volume (\(\Delta V = 0\)). Therefore, the work done (W) is:
\[ W = -P \Delta V = 0 \]

4. **Calculate Change in Internal Energy (\(\Delta U\)):**
Using the first law of thermodynamics, we can now substitute the values:
\[ \Delta U = q + W = 500 J + 0 = 500 J \]

5. **Conclusion:**
The change in internal energy (\(\Delta U\)) is 500 J. Since the work done is 0, the heat supplied is equal to the change in internal energy.

6. **Identify the Correct Statement:**
Based on the calculations, the correct statement is that the change in internal energy (\(\Delta U\)) is equal to the heat supplied (500 J) and the work done is zero.