One mole of monatomic ideal gas follows clockwise vertices (volume in 10⁻³ m³, pressure in kPa) (1, 100), (1, 300), (3, 300), (3, 100), (1, 100). (a) Find work on every leg and net work. (b) Find temperatures A, B and C using R = 8.31 J/(mol K). (c) Find ΔU and Q on A→B and B→C. (d) Verify net heat equals net work. (e) Explain orientation and heat rejection.

Question

Accepted Answer

Work 0, +600, 0, −200 J; net +400 J. Tₐ = 12.0 K, Tᵦ = 36.1 K, T𝒄 = 108 K. A→B: ΔU = Q = 300 J; B→C: ΔU = 900 J, Q = 1500 J. Constant-volume work is zero. B→C work = 300000(0.003 − 0.001) = +600 J; D→A work = 100000(0.001 − 0.003) = −200 J. Net is +400 J. Using pV = RT: Tₐ = 100000(0.001)/8.31 = 12.0 K; Tᵦ = 36.1 K; T𝒄 = 300000(0.003)/8.31 = 108 K. For monatomic gas ΔU = 3RΔT/2. A→B gives about 300 J and W = 0, so Q = 300 J. B→C gives ΔU about 900 J. With W = +600 J by gas, Q = ΔU + W = 1500 J. Cooling legs reject a total about 1400 J. Over a cycle ΔU = 0, so net Q = net W = +400 J. Clockwise orientation means positive enclosed work; heat rejection occurs on the falling-temperature C→D and D→A legs.

Ideal-gas cycle and first law

What a complete response should do

Rectangular gas cycle

Your self-assessment

Rectangular gas cycle — supplied values
Volume / 10⁻³ m³	Pressure / kPa
1	100
1	300
3	300
3	100
1	100