Could someone please check my work, I'm skeptical about it, but would like thoughts:
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Interesting. I own a copy of the second edition, but I didn't use the book for a course. I'll have to look through it to see why he omits R (I suspect he does something like make a choice of units for which R = 1). But it's the 'R' in PV = nRT, so if you are giving values for W, Q, etc., in any set of standard units (SI, cgs metric, British Engineering), R needs to be there.It seems I am not familiar with this R constant. I am using Thermal Physics by Charles Kittel and they did a similar calculation without using the R term.
For the Carnot cycle on a T-S diagram, the processes are isothermal, thus vertical lines [constant T], or reversible adiabatic, where Q = 0 and so dS = 0 [thus constant S]. (Note: there are also irreversible adiabatic processes, in which dS is not zero.)Also, I more or less guessed in some sense for the PV and TS graphs. I'm not sure why certain parts of the graph is curved. Is it safe to assume that a straight line in the PV graph correlates to a curved line on the TS graph? So for example, the Carnot cycle, the TS graph is a box, and this, the PV graph is made up of curved lines.
So the "heat in" would would only be from A->B since Q=0 for B->C in the Carnot cycle.For the Carnot cycle on a T-S diagram, the processes are isothermal, thus vertical lines [constant T], or reversible adiabatic, where Q = 0 and so dS = 0 [thus constant S]. (Note: there are also irreversible adiabatic processes, in which dS is not zero.)
This is true for the isothermal processes where dU = 0 (I presume you're using the form of the First Law, dU = dQ - dW , W being the work done by the system).I guess i'm not sure why I assumed that the heat taken up and the work done are equal (hence using dQ=dW)
Yes, because the cycle is isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression. Heat would have to be introduced during the expansions (and removed during the compressions), but the reversible adiabatic expansion involves no heat input. So that's one less term in the denominator of the efficiency quotient, thus the efficiency is higher.So the "heat in" would would only be from A->B since Q=0 for B->C in the Carnot cycle.
Yes. The isovolumic process requires a temperature increase to raise the gas pressure, so we have a heating step that doesn't occur in the Carnot cycle.But for the question at hand,we have to consider A->B->C because S always increases from A->B->C so for total heat in, we would consider A->B->C instead of just A->B in the Carnot cycle. Is this the right thought process?