A heat engine using 1 mole of a monatomic ideal gas uses the cycle:

  1. AB - isothermal
  2. BC - isobaric
  3. CA - isochoric

The volume at A is 50 liters. The volume at B is 100 liters. Step 1 is in equilibrium with a reservoir at 600 K. Step 2 is in contact with a reservoir at T꜀. Step 3 is in contact with the reservoir at 600 K. a) What is the temperature T꜀ at C? b) How much work is done by the engine in one cycle? c) What is the efficiency of the engine? (Be careful with the heat flows.) d) What is ΔS of the universe for one cycle of this engine? e) What is the efficiency of a Carnot engine operating between reservoirs at 600K and T꜀? Explain why your answer is either the same or different from your answer to part c. Question so kindly provided by Joe Kiskis

Part A Since step 2 is isochoric, for the beginning and the end of it, we know that: V₂⁄V₁ = T₂⁄T₁ Solve for T₂, and we get T꜀ = 300K

Part B To find the total work requires summing the work for the three seperate steps using:

  1. W₁ = Q = nRT ln(V₂⁄V₁) because it’s isothermal
  2. W₂ = PΔV because it’s isobaric
    • Note that this is going to be a negative number!
  3. W₃ = 0, because it’s isochoric

You end up with: Wₜₒₜₐₗ = 963 J

Part C Now that we know the total work produced by the engine, we need to find a Q to compare it to to find the efficiency of the engine.

  • Qₕₜₒₜₐₗ = Qₕ₁ + Qₕ₃
    • We calculated Q₁ in part B, because for an isothermal W=Q
    • Q₂ is in contact with T꜀, so isn’t of use here
  • Qₕₜₒₜₐₗ = 3456J + nCᵥΔT
  • Qₕₜₒₜₐₗ = 3456J + 1mole · 3⁄2 · 8.31J⁄molK · 300K
  • Qₕₜₒₜₐₗ = 7196J
    • Now we just plug that into the definition of e
  • e = W⁄Qₕ
  • e = 0.134

Part D The important thing to remember here is that the entropy of a cyclic process like an engine doesn’t change. But THAT IS NOT WHAT THIS IS ASKING! It wants to know ΔS for the universe. This is rather easy, because the two temperatures of the universe don’t change, so we cant just use ∑Q⁄T.

  • ΔSₜₒₜₐₗ = ΔS꜀ + ΔSₕ
  • ΔSₜₒₜₐₗ = [Qₕ−W]⁄T꜀ − 7196J⁄600K
    • Note the minus sign, because Q into the engine is Q out of the universe.
    • To be honest, I just knew that the two values had opposite signs, and just picked the one that gave me a positive answer. Understand the concept over memorizing the formulas.
  • ΔSₜₒₜₐₗ = 6233J⁄300K − 7196J⁄600K
  • ΔSₜₒₜₐₗ = 8.78 J⁄K

Part E

  • e꜀ₐᵣₙₒₜ = 1 − T꜀⁄Tₕ
  • e꜀ₐᵣₙₒₜ = 1 − 300K−600K
  • e꜀ₐᵣₙₒₜ = 0.50

As for why these two numbers are different, remember that the Carnot cycle is two pairs of isotherms and adiabats. This cycle only uses one isotherm, and an isobaric and isochoric step, so it is going to behave less than the theoretical maximum efficency possible.