A system undergoes a process in which `DeltaU = + 300 J` while absorbing 400 J of heat energy and undergoing an expansion against 0.5 bar. What is the change in the volume (in L)?

Options

A. 4
B. 5
C. 2
D. 3

Explanation

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

\[ \Delta U = Q + W \]

Where:
- \(\Delta U\) is the change in internal energy,
- \(Q\) is the heat added to the system,
- \(W\) is the work done by the system.

### Step 1: Identify the given values
From the problem, we have:
- \(\Delta U = +300 \, \text{J}\)
- \(Q = +400 \, \text{J}\)
- \(P_{\text{external}} = 0.5 \, \text{bar}\)

### Step 2: Express work done in terms of volume change
The work done by the system during expansion against an external pressure is given by:

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

### Step 3: Substitute the expression for work into the first law of thermodynamics
Substituting \(W\) into the first law equation gives:

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

### Step 4: Rearrange the equation to solve for \(\Delta V\)
Rearranging the equation, we have:

\[ \Delta V = \frac{Q - \Delta U}{P_{\text{external}}} \]

### Step 5: Substitute the known values into the equation
Substituting the known values:

\[ \Delta V = \frac{400 \, \text{J} - 300 \, \text{J}}{0.5 \, \text{bar}} \]

### Step 6: Calculate the numerator
Calculating the numerator:

\[ 400 \, \text{J} - 300 \, \text{J} = 100 \, \text{J} \]

### Step 7: Convert pressure to appropriate units
We know that \(1 \, \text{L} \cdot \text{bar} = 100 \, \text{J}\). Therefore, we can convert \(0.5 \, \text{bar}\) to \(J/L\):

\[ 0.5 \, \text{bar} = 0.5 \, \text{L} \cdot \text{bar} = 50 \, \text{J/L} \]

### Step 8: Substitute the values into the equation
Now substituting back into the equation:

\[ \Delta V = \frac{100 \, \text{J}}{0.5 \, \text{bar}} = \frac{100 \, \text{J}}{50 \, \text{J/L}} = 2 \, \text{L} \]

### Final Answer
Thus, the change in volume \(\Delta V\) is:

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

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