14 – Physics Help Please! Carnot Cycle Part 1?

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Refer to this image: http://session.masteringphysics.com/problemAsset/1

  1. Find the total work “W” done by the gas after it completes a single Carnot cycle.Express the work in terms of any or all of the quantities Qh, Th, Qc, and Tc. W=???

  2. Suppose there are “n” moles of the ideal gas, and the volumes of the gas in states A and B are, respectively, Va and Vb. Find , the heat absorbed by the gas as it expands from state A to state B.

Express the heat absorbed by the gas in terms of n, Va, Vb, the temperature of the hot reservoir, Th, and the gas constant R. Qh= ????

  1. The volume of the gas in state C is Vc, and its volume in state D is Vd. Find Qc, the magnitude of the heat that flows out of the gas as it proceeds from state C to state D.
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Express your answer in terms of n, Vc, Vd, TC (the temperature of the cold reservoir), and R. Qh= ???

There is a part 2 to this which I will label:

14 – Physics Help Please! Carnot Cycle Part 2

1 Answer

  • Dear Physics*,

    Let’s do the full cycle together, so you “see” how it works…

    First, I will assume, since you did not tell us, that the gas is an ideal monotomic gas.

    Th^(Cv/R) Vb = Tc^(Cv/R) Vc and

    Th^(Cv/R) Va = Tc^(Cv/R) Vd

    and Cv = (3/2) R for an ideal monotomic gas

    In the first step: Isothermal expansion…(From Va to Vb at Th)

    DeltaU = 0; -qa=wa=-nRTh * ln(Vb/Va)

    Step 2: (Adiabatic expansion) from Vb to Vc with a temperature change from Th to Tc:

    q=0; deltaU = w =

    n* Integral (from Tc on top and Th on bottom) Cv dT =

    n * int Tc->Th (3/2)R dT

    Third step: Isothermal compression from Vc to Vd at Tc

    delta U = 0 and

    -q=w=-nRTc * ln (Vd/Vc) and

    Step 4: Adiabatic compression (all of these processes are reversible…OK?) from Vd back to Va with a T change from Th to Tc


    DElta U = w= n * Int (Th on top/Tc on bottom) of Cv dT

    For the overall cycle, delta U = 0 and..

    -q total = w total (=area under enclosed PV curve) =

    -nR (Th-Tc) ln(Vb/Va)

    OK, that is the theory, now let’s apply it to your problem…

    1) ln(Vb/Va) = (for a Carnot engine) Qh/(nRTh)

    and solving for Vb/Va by exponentiating and substituting into…

    -nR (Th-Tc) ln(Vb/Va) gives W (cycle) in terms of the temps and heats (and n and R, of course).

    2) See step 1 above…(That’s why it’s great to do your analysis BRFORE you answer the questions…because you probably answered all the questions).

    -qa=wa=-nRTh * ln(Vb/Va)

    3) See Step 3 above.

    If you “approve” of my work, I will answer the second question on this subject (and all your other questions, too). I am so shocked that the most prolific contributors in physics don’t rush to answer thermo questions…I wonder why not? As the old bumpersticker says…”Honk if you PASSED Thermo”


    Source(s): *I’ll call you “SuperStar” when you can do thermo problems on your own 😉

    30 yr. prof. of physics

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