An engine operates by taking a monoatomic ideal gas through the cycle shown in the figure. The percentage efficiency of the engine is close to __________.

Solution

Physics sir · 3 steps

Work is the rectangle's area; for heat in, add up only the two legs where the gas heats up.

1. Work from area. The cycle is a rectangle on the P-V diagram, so the net work is its area: W = (3P₀ - P₀)(2V₀ - V₀) = 2P₀V₀.
2. Find heat-in legs. Heat enters only where PV (and so temperature) rises: A to B and B to C. For an ideal gas nRΔ T = Δ(PV), and a monoatomic gas has C_V = 3/2R and C_P = 5/2R.

   Process	Type	Δ(PV)	Heat
   A to B	isochoric	3P₀V₀ - P₀V₀ = 2P₀V₀	3/2 × 2P₀V₀ = 3P₀V₀ in
   B to C	isobaric	6P₀V₀ - 3P₀V₀ = 3P₀V₀	5/2 × 3P₀V₀ = 7.5P₀V₀ in
   C to D	isochoric	2P₀V₀ - 6P₀V₀ = -4P₀V₀	6P₀V₀ out
   D to A	isobaric	P₀V₀ - 2P₀V₀ = -P₀V₀	2.5P₀V₀ out

   η = W/Qᵢₙ = 2P₀V₀/(3P₀V₀ + 7.5P₀V₀) = 2/10.5 ≈ 0.19

3. So. The efficiency is about 19.05%, which is close to 19. As a check, heat in minus heat out is 10.5P₀V₀ - 8.5P₀V₀ = 2P₀V₀, the same as the work.

The common slip: 27. Using C_V instead of C_P for the isobaric leg B to C makes Qᵢₙ = 7.5P₀V₀ and gives about 27, not 19.

The idea to keep. For any cycle, η = W/Qᵢₙ. Take W from the enclosed area. For Qᵢₙ, use C_V on constant-volume legs and C_P on constant-pressure legs, and count only the legs that absorb heat.

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