MCQ Quiz - NN Education

💡 Explanation

From the diagram, the cycle goes B(3 cm³, 2 Pa) → A(3 cm³, 6 Pa) → C(6 cm³, 2 Pa) → B.

Work done in each step (work = area under P-V curve, positive when V increases):

B → A (constant volume, [latex]V=3\text{ cm}^3[/latex]): [latex]W_{BA} = 0[/latex] (no volume change)

A → C (diagonal line from (3, 6) to (6, 2)): area under this line (a trapezoid):
[latex]W_{AC} = \frac{1}{2}(6+2)(6-3) = 12\text{ Pa·cm}^3[/latex]

C → B (constant pressure, [latex]P=2\text{ Pa}[/latex], V decreasing from 6 to 3):
[latex]W_{CB} = P\Delta V = 2\times(3-6) = -6\text{ Pa·cm}^3[/latex]

Total work done by the gas over the cycle:
[latex]W_{net} = 0+12-6 = 6\text{ Pa·cm}^3[/latex]

Converting units: [latex]1\text{ Pa·cm}^3 = 1\text{ Pa}\times10^{-6}\text{ m}^3 = 10^{-6}\text{ J}[/latex]

[latex]W_{net} = 6\times10^{-6}\text{ J}[/latex]