A system undergoes a process which absorbed `0.5 KJ` of heat and undergoing an expansion againt external pressure of `1 atm`, during the process change in internal energy is `300 J`. Then predict the change in volume (lit)

Options

A. 1
B. 2
C. 3
D. 4

Explanation

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

\[ \Delta U = q + w \]

Where:
- \(\Delta U\) = change in internal energy
- \(q\) = heat absorbed by the system
- \(w\) = work done by the system

### Step 1: Identify the given values
- Heat absorbed, \(q = 0.5 \, \text{kJ} = 500 \, \text{J}\) (since \(1 \, \text{kJ} = 1000 \, \text{J}\))
- Change in internal energy, \(\Delta U = 300 \, \text{J}\)
- External pressure, \(P_{\text{external}} = 1 \, \text{atm}\)

### Step 2: Convert external pressure to appropriate units
To use the work formula, we need to convert the pressure from atm to Joules per liter:
- \(1 \, \text{atm} = 101.325 \, \text{J/L}\)

### Step 3: Calculate work done by the system
The work done by the system during expansion against external pressure is given by:

\[ w = -P_{\text{external}} \Delta V \]

Substituting the pressure in Joules per liter:

\[ w = -101.325 \, \text{J/L} \cdot \Delta V \]

### Step 4: Substitute values into the first law of thermodynamics
Now we can substitute the values into the first law equation:

\[ \Delta U = q + w \]

Substituting the known values:

\[ 300 \, \text{J} = 500 \, \text{J} - 101.325 \, \text{J/L} \cdot \Delta V \]

### Step 5: Rearranging the equation to solve for \(\Delta V\)
Rearranging the equation gives:

\[ 101.325 \, \text{J/L} \cdot \Delta V = 500 \, \text{J} - 300 \, \text{J} \]

\[ 101.325 \, \text{J/L} \cdot \Delta V = 200 \, \text{J} \]

### Step 6: Solve for \(\Delta V\)
Now, divide both sides by \(101.325 \, \text{J/L}\):

\[ \Delta V = \frac{200 \, \text{J}}{101.325 \, \text{J/L}} \approx 1.975 \, \text{L} \]

### Step 7: Round to appropriate significant figures
Rounding this to two significant figures gives:

\[ \Delta V \approx 2 \, \text{L} \]

### Final Answer
The change in volume is approximately \(2 \, \text{liters}\).

---