Question

A gas expands by \(2500\text{ cm}^{3}\) against constant external pressure \(1.5\text{ atm}\) at constant temperature of \(300\text{ K}\). Calculate the work of expansion in kJ.

• A. \(-0.380\)
• B. \(-0.350\)
• C. \(-0.42\)
• D. \(-0.45\)

Hint:

Use work equals minus external pressure times change in volume, then convert litre atmosphere into joules.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Against a constant external pressure, \(w = -P_{ext}\Delta V\).

Step 2: Convert the Volume
\(\Delta V = 2500\text{ cm}^3 = 2.5\text{ L}\)

Step 3: Calculate
\(w = -1.5\text{ atm} \times 2.5\text{ L} = -3.75\text{ L atm}\)

Step 4: Convert Units
\(1\text{ L atm} = 101.3\text{ J}\), so \(w = -3.75 \times 101.3 \approx -380\text{ J} = -0.380\text{ kJ}\).

Final Answer:
The work of expansion is about \(-0.380\text{ kJ}\). \[ \boxed{\text{(A) }-0.380} \]