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

Concept: Work done by a gas during expansion at constant pressure is given by \[ W = -P\Delta V \] where \(P\) = external pressure, \(\Delta V = V_2 - V_1\). The negative sign indicates work done by the system on the surroundings.

Step 1: Calculate the change in volume. \[ V_1 = 0.2\,\text{dm}^3, \quad V_2 = 0.8\,\text{dm}^3 \] \[ \Delta V = 0.8 - 0.2 = 0.6\,\text{dm}^3 \] Convert to \(m^3\): \[ 1\,\text{dm}^3 = 10^{-3}\,\text{m}^3 \] \[ \Delta V = 0.6 \times 10^{-3}\,\text{m}^3 \]

Step 2: Convert pressure to SI units. \[ 1\,\text{bar} = 10^5\,\text{Pa} \] \[ P = 2 \times 10^5\,\text{Pa} \]

Step 3: Substitute into the formula. \[ W = -P\Delta V \] \[ W = -(2 \times 10^5)(0.6 \times 10^{-3}) \] \[ W = -120\,\text{J} \] Thus, the work done by the gas is \(-120\) J.