Monday, December 8, 2008

Physics 9B Practice Final Problem #4

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 TC . Step 3 is in contact with the reservoir at 600 K.
a) What is the temperature TC 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 TC? 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
TC = 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.
  • QH total = QH1 + QH3
    • We calculated Q1 in part B, because for an isothermal W=Q
    • Q2 is in contact with TC, so isn't of use here
  •  QH total = 3456J + nCVΔT
  • QH total = 3456J + 1mole ⋅ 3⁄2 ⋅ 8.31J⁄molK ⋅ 300K
  • QH total = 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 = ΔSC + ΔSH
  • ΔStotal = [QH−W]⁄TC − 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
  • eCarnot = 1 − TC⁄TH 
  • eCarnot = 1 − 300K−600K
  • eCarnot = 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.

2 comments:

  1. kenneth, when you get the chance, can you explain how you got parts c,d,and e. at the moment i'm stuck on c. i'm getting a value close to yours. i'm getting .139 for e.

    ReplyDelete