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ggrree
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Thermodynamics question:
http://gyazo.com/ae831c4c7ae3325cfce6f280cfa37b88
show that this is true.
 one year ago
 one year ago
ggrree Group Title
Thermodynamics question: http://gyazo.com/ae831c4c7ae3325cfce6f280cfa37b88 show that this is true.
 one year ago
 one year ago

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soty2013 Group TitleBest ResponseYou've already chosen the best response.1
easy.. use partial diffrentiation method...
 one year ago

Carl_Pham Group TitleBest ResponseYou've already chosen the best response.2
You need the "Antirecipe" for working with second derivatives of the fundamental relation, which you'll find in H. B. Callen "Thermodynamics", somewhere around Chapter 2 or 3. My copy of Callen is at work, so if you haven't figured it out by then, I'll see if I can dig up what you need when I get there.
 one year ago

Carl_Pham Group TitleBest ResponseYou've already chosen the best response.2
OK, it doesn't look like Callen's recipe is useful. But we can do it another way. First, A is a natural function of T,V and N. We'll ignore N here, because it's just held constant. So the total differential of A is, by definition:\[dA = \left(\frac{\partial A}{\partial T}\right)_V dT + \left(\frac{\partial A}{\partial V}\right)_T dV\]Divide through by dT and hold p constant:\[\left(\frac{\partial A}{\partial T}\right)_p = \left(\frac{\partial A}{\partial T}\right)_V + \left(\frac{\partial A}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_p\]Now there is a general theorem for partial variations of a function X(Y,Z) of several variables, which produces this general result:\[\left(\frac{\partial X}{\partial Y}\right)_Z =  \left(\frac{\partial X}{\partial Z}\right)_Y \left(\frac{\partial Z}{\partial Y}\right)_X\]It's derived in Callen Appendix A, and probably in some diff eq or calculus books. Using this on the second term above gives you your first equation. Maybe there's a way to get there in one step, but I'm not seeing it.
 one year ago

Carl_Pham Group TitleBest ResponseYou've already chosen the best response.2
Darn, I skipped a step. You have to use the chain rule first on the second term of the second equation:\[\left(\frac{\partial A}{\partial T}\right)_p = \left(\frac{\partial A}{\partial T}\right)_V + \left(\frac{\partial A}{\partial p}\right)_T \left(\frac{\partial p}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_p\]Now you can use that XYZ rule on the last term to get what you want.
 one year ago

Carl_Pham Group TitleBest ResponseYou've already chosen the best response.2
Now to get to the second equation, remember that the differential of A turns out to be:\[dA = S dT  p dV\]You get this by taking the Legendre transform of the canonical differential of internal energy U (dU = TdS  pdV) which you remember from intro thermo, since A is the Legendre transform of U. You can immediately identify:\[\left(\frac{\partial A}{\partial T}\right)_V =  S\] and\[\left(\frac{\partial A}{\partial V}\right)_T = p\]Using these in the second equation in the first post above:\[\left(\frac{\partial A}{\partial T}\right)_p = S  p \left(\frac{\partial V}{\partial T}\right)_p\]Now we use that XYZ theorem the other way, to bring the p inside the derivatives:\[\left(\frac{\partial A}{\partial T}\right)_p = S + p \left(\frac{\partial V}{\partial p}\right)_V \left(\frac{\partial p}{\partial T}\right)_V \]Rewrite the product as a fraction, inverting the derivative on the bottom, and you've got your second equation. This is obviously not the way the author did it, since he meant the second equation to follow from the first, but I don't have a quick isight into what he did.
 one year ago
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