An ideal gas has undergone through the cyclic process as shown in the figure. Work done by the gas in the entire cycle is ________ × 10⁻¹ J. (Take π=3.14 )

Solution

Physics sir · 4 steps

Work in a cycle is the area enclosed by the loop, so find this circle's area in real units.

1. Work equals area. For any cyclic process, the net work done by the gas equals the area enclosed by the loop. This holds on a V–P graph as well as on a P–V graph.
2. Read the semi-axes. Along P the loop runs from 300 to 500 kPa, so its half-width is a = 100 kPa = 10⁵ Pa. Along V it runs from 150 to 350 cm³, so its half-height is b = 100 cm³ = 10⁻⁴ m³.
3. Use ellipse area. The two half-lengths carry different units, so treat the loop as an ellipse with area π a b, not as π r².

   W = π a b = 3.14 × 10⁵ Pa × 10⁻⁴ m³ = 31.4 J

4. So. W = 31.4 J = 314 × 10⁻¹ J, so the answer is 314.

The common slip: 31.4. Stopping at the work in joules gives 31.4. The blank is in units of 10⁻¹ J, so it needs 314.

The idea to keep. Net work in a cycle is the area enclosed by the loop. A circle drawn on a P–V graph is really an ellipse of area π × Δ P_(half) × Δ V_(half), with both in SI units.

Answer from the NTA final answer key. Solution by PrepWiser.