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Question:

Helium gas goes through a cycle ABC (consisting of two isochoric and isobaric lines) as shown in figure. Efficiency of this cycle is nearly: (Assume the gas to be close to ideal gas)

9.1

15.4

10.5

12.5

Solution:

η=Work done by gas/Heat given to gas
Work done by gas = Area under curve = P₀V₀
Heat given by gas = Heat given by gas from A to B + Heat given by gas from B to C
Heat given by gas = nCᵥΔT + nCₚΔT
Cₚ and Cᵥ for monoatomic gas is 5R/2 and 3R/2 respectively.
Heat given by gas = (3/2)P₀V₀ + 5P₀V₀
η = P₀V₀/((3/2)P₀V₀ + 5P₀V₀) = 1/((3/2) + 5) = 1/(13/2) = 2/13 ≈ 0.154
Efficiency = 15.4%