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[Phys-l] PV question

If I have a PV diagram where the process travels diagonally along a straight
line from lower right to upper left (B to A), can a generic statement be
made where this shows NEGATIVE work BY the gas because the volume decreases.
...regardless of the change in pressure?

ASCII art below

| \
| \
| \
Po+ \B
Vo 2Vo

I cannot follow this "thread". The metaphor is inappropriate for such a tangled mess. I want to make some disjoint comments that may or may not answer some of the questions that have arisen during this discussion. I hope that thinking about these will help dispel misconceptions and misuse of conventional terminology.

1. The work done by the gas is the integral of PdV over the process (B to A, or 2Vo to Vo). In this hypothetical process that would be carried out over that portion of the curve

[ V ]
P = Po [ 3 - - ]
[ Vo ]

The answer is That the work done by the gas is - 3/2 PoVo.

I am sure everyone will agree on that.

I want to point out that the ASCII art above is drawn misleadingly from a pedagogical point of view. It would not have been much more difficult to draw it without a suppressed zero on the pressure axis. Then the answer can be read off the graph as an area:

2Po+ |\A
| | \
| | \
| | \
Po+ | \B
| | |
| | |
| | |
Vo 2Vo

2. If one insists that gas pressure must always be positive, then the work done by a gas being compressed will always be negative. So far as I know, only liquids and solids can exhibit negative pressures. Of course we are talking about absolute pressures! Confusion could certainly arise if gauge pressures are substituted here (a possibility in an engineering environment). In the engineering case one gets work out of a system by compressing and condensing a gas, steam. Hence the system does positive work while compressing. (Google Newcomen engine.)

3.The convention that work done by a gas is defined as positive comes from engineering, where the desired work is that done by an engine. I am old enough to have learned that convention first. The convention that is now in popular use among physicists is the opposite; we define the work done on a system as positive, and the increment of mechanical work is - PdV.

4. Somehow the term "reversible" showed up in the discussion. It does not appear in the question, so I couldn't figure out what relevance it had. It is implicit in the question that in this process the state of the system is always sufficiently near *mechanical* equilibrium that its pressure is uniform throughout. Note that this does not imply the stronger constraint that it be near *thermodynamic* throughout the process, though that is the usual case. If the process is reversible, the system must be near thermodynamic equilibrium at all points in the process, but that is not a necessary condition in this case.

5. Somehow the term "adiabatic" also showed up. That was neither specified in the question nor is it necessary to assume. If the gas is ideal then it is most unlikely that the process is adiabatic, but there is no reason to expect that the gas will be ideal; few gases are. Similarly the idea that this process must "cross isotherms" is irrelevant.

OK, throw rocks at me if you will. I just did that to assure myself that I still knew my classical thermodynamics. If I've said anything that is incorrect I would appreciate correction.