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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=???
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= ????
- 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.
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
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”
-FredSource(s): *I’ll call you “SuperStar” when you can do thermo problems on your own 😉
30 yr. prof. of physics