Isotherms and adiabats, part III

November 11, 2010 at 9:39 pm Leave a comment

Because parts I and II were just so wildly popular, here comes the third installment of Isotherms and adiabats.

Geo, may I offer a little piece of advice? First of all, you might try not posting about obscure topics from an obscure corner of the internet. Second, if you insist on doing this thing, you should use titles that draw people in, such as:


No mercy? What the hell does that even mean? Where the hell was I? Oh, yeah, I was busy showing that isotherms and adiabats must have a certain structure, independent of the substance under consideration.

Claim ~A. Adiabats can never cross each other (even on a bad day).

Proof. Assume the opposite of ~A, that is A, and pick an isotherm that crosses both of the crossing adiabats, as shown in the figure below.

Consider a process that follows the arrows in the diagram. The total work done by the system is the area between the curves. What is the total heat absorbed after a full cycle, starting, say, at the upper left corner and ending there? The only heat involved must be along the top curve, because the other two are adiabats. At the end of the cycle, the change in the internal energy is zero: U(end)-U(start) = 0. But by the first law, U(end)-U(start) = w + q, so -w = q is the work done by the system in one cycle. It is equal to the heat added along the isotherm! But there is no lower isotherm along which part of q can be transferred someplace else; hence, heat has been taken from a constant temperature reservoir and converted entirely into work, without affecting the thermodynamic state of any third body (the first two bodies being the reservoir and the working substance). This state of affairs is contrary to statement T (from Isotherms and adiabats, part I). Because statement T is true, statement A must be false: ~A. QED.


Entry filed under: Thermodynamics.

Isotherms and adiabats, part II Sanity control

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