A gas is allowed to expand in a well insulated container against a constant external pressure of `2.5 atm` from an initial volume of `2.50 L` to a final volume of `4.50 L`. The change in internal energy `Delta U` of the gas in joules will be:

Options

A. `-500 J`
B. `-505 J`
C. `+ 505 J`
D. `1136.25 J`

Explanation

To find the change in internal energy (ΔU) of the gas, we can follow these steps:

### Step 1: Calculate the change in volume (ΔV)
The change in volume (ΔV) can be calculated using the formula:
\[ \Delta V = V_f - V_i \]
Where:
- \(V_f\) = final volume = 4.50 L
- \(V_i\) = initial volume = 2.50 L

Calculating ΔV:
\[ \Delta V = 4.50 \, \text{L} - 2.50 \, \text{L} = 2.00 \, \text{L} \]

### Step 2: Calculate the work done (W)
The work done by the gas during expansion against a constant external pressure is given by the formula:
\[ W = -P_{\text{external}} \times \Delta V \]
Where:
- \(P_{\text{external}} = 2.5 \, \text{atm}\)

Substituting the values:
\[ W = -2.5 \, \text{atm} \times 2.00 \, \text{L} = -5.00 \, \text{atm} \cdot \text{L} \]

### Step 3: Convert work from atm·L to joules
To convert work from atm·L to joules, we use the conversion factor:
\[ 1 \, \text{atm} \cdot \text{L} = 101.3 \, \text{J} \]
Thus,
\[ W = -5.00 \, \text{atm} \cdot \text{L} \times 101.3 \, \text{J/atm·L} = -506.5 \, \text{J} \]

### Step 4: Apply the First Law of Thermodynamics
According to the First Law of Thermodynamics:
\[ \Delta U = Q + W \]
Where:
- \(Q\) = heat exchange (since the container is insulated, \(Q = 0\))

Substituting the values:
\[ \Delta U = 0 + (-506.5 \, \text{J}) = -506.5 \, \text{J} \]

### Final Answer
The change in internal energy (ΔU) of the gas is:
\[ \Delta U = -506.5 \, \text{J} \]