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Is it possible to represent an adiabatic process for an ideal gas by a formula other than $PV^\gamma=Const$?:

Relevant Considerations:

We always need to connect a pair of arbitrary points/states $A:(P_1,V_1)$ and$B:(P_2,V_2)$ on the P-V indicator diagram to define the Internal Energy function(U)

If the two points do not lie on some $PV^\gamma=Const$, we may think of some abrupt process connecting the two states. Such a process cannot be represented by a continuous curve on the diagram since the intermediate states are non-equilibrium states. Pressure will not have a unique value wrt the system for such processes.It will be different from point to point.We have a pair of points A and B without any curve connecting them

Now we connect the points A and B by an arbitrary "continuous" path on our diagram(on paper) and select close discrete points on it. The consecutive "close" discrete points may be connected by abrupt adiabatic "processes" as described previously. Therefore in the pragmatic sense we can have an arbitrary quasi-static "adiabatic" path connecting the points A and B : The "close" points ie physical; states,will lie on a curve different from $PV^\gamma=Const$

[In the physical/practical implementation of the $PV^\gamma=Const$ curve we do consider/implement a chain of discrete equilibrium states]

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up vote 2 down vote accepted

Yes, this is possible, but the word "adiabatic" is not usually used to describe this. I will call a process that you describe a no-external-heat-flow process, to distinguish from an adiabatic, or constant entropy, process.

The only restriction is that at at any two successive stopping points, the entropy must go up. To realize this, first do a true adiabatic transformation to get you to the volume you want, then quickly jiggle the piston, making soundwaves that dissipate heat into the gas until you get it to the temperature you want. The combinations of adiabatic lines and heat-generating volume-preserving lines let you approximate any curve that links points such that successive points are always higher entropy.

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One doesn't have to "jiggle" the piston. All one needs to do is insert a stirrer into the system and turn it. No heat is transferred to the system so the process is adiabatic. The work done to move the stirrer appears as heat inside the system. Conceptually this can be done reversibly by moving the stirrer very very very (etc.) slowly. Of course "jiggling" the piston is really the same thing, but working out what happens is a bit more complex. – Paul J. Gans Sep 16 '12 at 0:51

Classical thermodynamics is only valid if there is a local equilibrium. If it's not the case, nothing much can be said. Transport coefficients are sometimes not even local. So yes, it is possible to have an adiabatic process (no heat transferred) without having PV^γ=Const. However you can not treat these problems by using standard thermodynamics. You need to go kinetics. The condition for the transport coefficients to be valid is that the mean free path must be shorter than the characteristic gradient lenght.

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