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

Options

A. `q=W=500 J , DeltaE=0`
B. `q=DeltaE=500 J, W=0`
C. `q=-W=500 J, DeltaE=0`
D. `DeltaE=0,q=W=-500 J`

Explanation

To solve the problem step by step, we will analyze the situation using the principles of thermodynamics.

### Step 1: Understand the Given Information
We have 1 mole of gas heated at constant volume. The initial temperature (T1) is 298 K, and the final temperature (T2) is 308 K. The heat supplied to the gas (Q) is 500 J.

### Step 2: Identify the Conditions
Since the gas is heated at constant volume, we know that:
- The change in volume (ΔV) is 0.
- Therefore, the work done (W) on or by the gas is also 0, because work done is given by the equation:
\[ W = -P \Delta V \]
Since ΔV = 0, it follows that W = 0.

### Step 3: Apply the First Law of Thermodynamics
According to the first law of thermodynamics, the change in internal energy (ΔU) of the system can be expressed as:
\[ \Delta U = Q + W \]
Substituting the known values:
- Q = 500 J (heat supplied)
- W = 0 J (work done)

So, we can write:
\[ \Delta U = 500 J + 0 J = 500 J \]

### Step 4: Conclusion
From our calculations, we find that the change in internal energy (ΔU) is equal to the heat supplied (Q), which is 500 J. Therefore, the correct statement regarding the system is that the change in internal energy is equal to the heat supplied.

### Answer
The correct statement is that the change in internal energy (ΔU) is 500 J.

---