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Step-by-Step Solution
Step 1: Identify the Given Data
• External pressure, $P_{ext} = 2.5 \text{ atm}$
• Initial volume, $V_i = 2.50 \text{ L}$
• Final volume, $V_f = 4.50 \text{ L}$
• The process takes place in a well-insulated container, so no heat is exchanged: $q = 0$.
Step 2: Write the Formula for Work Done
In thermodynamics, when a gas expands (or is compressed) against a constant external pressure, the work done ($w$) by the gas is given by:
$$
w = -P_{ext} \Delta V
$$
where $\Delta V = V_f - V_i$ is the change in volume. The negative sign indicates work done by the system on the surroundings.
Step 3: Calculate Work Done
First, compute the change in volume:
$$
\Delta V = V_f - V_i = 4.50 \text{ L} - 2.50 \text{ L} = 2.00 \text{ L}.
$$
Hence,
$$
w = -P_{ext} \Delta V = - (2.5 \text{ atm}) \times (2.00 \text{ L}).
$$
This equals:
$$
w = -5.0 \text{ L atm}.
$$
Next, convert L·atm to joules using the conversion factor $1 \text{ L atm} \approx 101.325 \text{ J}$:
$$
w = -5.0 \; (\text{L atm}) \times 101.325 \;\text{J (per L atm)}
= -506.625 \;\text{J}.
$$
For simplicity, we often round to a suitable number of significant figures, giving approximately $-505 \text{ J}$.
Step 4: Relate Work to Change in Internal Energy
Using the first law of thermodynamics,
$$
\Delta U = q + w.
$$
Since the container is well-insulated, $q = 0$. Therefore,
$$
\Delta U = 0 + w = w = -506.625 \text{ J} \approx -505 \text{ J}.
$$
Step 5: Conclusion
The change in internal energy, $\Delta U$, is about $-505 \text{ J}$. Hence, the correct answer is:
$-505 \text{ J}$.
