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I am confusing myself with this topic and I would appreciate some insight into this. You see, one takes the area of a PV graph to obtain the work done on the system (for compression) or by the system (for expansion). However, I am learning that in the irreversible path, compression takes the area above the PV curve while expansion is below it. For representational purposes something like the graph on this paper:http://www.csrc.ac.cn/~suncp/papers/PR/PRE08.pdf

a1) Does this mean that the amount of work done in compression is always greater than expansion (because of the "extra" area? I don't find this intuitive because I am thinking that for compression, the external pressure (outside of the system) approaches that of the gas(boundary) plus the atmospheric pressure and, aided by the gravitational force, this reduces the amount of Pexternal necessary to apply unto the system to achieve compression.

a2) I realized that perhaps the reason for a1 may be due to the starting Pext which changes depending on compression (high P to low P, or left to right) or expansion (low P to high P, or right to left). But how does this reflect on the work (i.e. area below the graph?) Please help! I think I am going in circles! =S

On another note, one can determine what the maximum or minimum Pexternal and work can be used to achieve compression or expansion in the reversible process. Compression corresponds to minimum Pext and work that the surroundings has to do to compress the system, whereas expansion corresponds to maximum Pext and work that the system can perform on the surroundings to achieve the expansion.

b) This approach to infinitesimal changes sort of makes sense if I follow my reasoning on (a), but i am not sure if this is correct at all. In addition, why is w(irreversible)>w(reversible) in compression, and vice versa for expansion? Does it have to do with the "extra" or "less" area acquired in the irreversible process in comparison to the closer approximation ot the PV curve by the infinitesimal changes in the reversible process?

Finally (as in my last question for this thread), when calculating the reversible work for some expansion or compression one has to make the following integration:

dw=-∫PdV, where P is Pexternal but for an ideal gas P=nRT/V,

So, dw = -nRT∫ (1/V) dV => w= -nRTln(V2-V1)

c) When is it possible to substitute P1V1 for nRT in the above result, such that w= -P1V1ln(V2-V1)? I would simply retain the previous solution; and I assume that for an ideal gas, both solutions should give the same answer?

I hope that someone could take the time to read through my (rather long and pesky) entry and provide some helpful insight... it would certainly help me get rid of many doubts and correct some of my reasoning mistakes as well. Thanks! =D

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# Thermodynamics - expansion, compression, work

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