Question

Find the work done by a gas that expands from \(0.2\,\text{dm}^3\) to \(0.8\,\text{dm}^3\) against a constant pressure of \(2\) bar.

• A. \(-120\) J
• B. \(120\) J
• C. \(-60\) J
• D. \(60\) J

Hint:

For constant pressure processes, work done is simply \(W=-P\Delta V\). Always convert pressure and volume into SI units before calculating.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The task is to calculate the pressure-volume work done during an expansion process against a constant external pressure.
Step 2: Key Formula or Approach:
The formula for irreversible work is:
\[ W = -P_{ext} \Delta V \] Conversion factors: \( 1\ \text{bar} = 10^5\ \text{Pa} \) and \( 1\ \text{dm}^3 = 10^{-3}\ \text{m}^3 \).
Step 3: Detailed Explanation:
Change in volume \( \Delta V = V_2 - V_1 \):
\[ \Delta V = 0.8 - 0.2 = 0.6\ \text{dm}^3 = 0.6 \times 10^{-3}\ \text{m}^3 \] External pressure \( P = 2\ \text{bar} = 2 \times 10^5\ \text{Pa} \).
Calculate Work:
\[ W = -(2 \times 10^5\ \text{Pa}) \times (0.6 \times 10^{-3}\ \text{m}^3) \] \[ W = -1.2 \times 10^2\ \text{J} = -120\ \text{J} \] Step 4: Final Answer:
The work done by the gas is \(-120\) J.