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So I've been doing a course in Thermodynamics for a while and one of the questions that occasionally struck me was "How do I practically calculate the difference of internal energy between two arbitrary states, A and B?"

Thermodynamics by H.B. Callen provides a really good insight into this and according to him, Joule, after his experiments concluded that- "In an adiabatic system, any two equilibrium states can definitely be connected by some mechanical process and the work done in that mechanical process is the internal energy difference $\Delta U$ between them".

I'll start with a simple example of a "container with a piston" system with an "ideal gas" inside.

Let's start with some equilibrium state $(P_0,V_0)$ with adiabatic walls, so no exchange of heat is possible. So, I can expand or contract the system to take it to some state $(P,V)$ which satisfies $PV^{\gamma}=P_0 {V_0}^{\gamma}=\kappa$ (constant).

Now, if I talk about the equilibrium state $(P_0,2V_0)$, this state too is an equilibrium state but it lies on some different adiabat along which states $(P,V)$ satisfy $PV^{\gamma}=P_0 {(2V_0)}^{\gamma}=\kappa'$ (a different constant).

Suppose we have to find the $\Delta U$ between$(P_0,V_0)$ and $(P_0,2V_0)$. But, if we have an adiabatic system, we can't jump between different adiabats i.e. I can't find a mechanical process which takes $(P_0,V_0)$ to $(P_0,2V_0)$. I have explained this problem through a diagram too. enter image description here

If I can't find a mechanical process connecting these states, then I definitely can't find the internal energy difference between them which seems odd because we do that everytime in various thermodynamics problems. So, if I were to be an experimentalist, how would I go about finding the difference in internal energy between these two states?

Edit: I changed the example which I gave and added an image to further clarify the problem.

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  • $\begingroup$ This seems odd to me. For example, we know that the internal energy of an ideal gas does not change during an isothermal process. Why can't heat be added/taken to/from the system to determine internal energy differences? Does your text say more on this point later? $\endgroup$ – BioPhysicist Jul 11 at 11:59
  • $\begingroup$ You need to know the heat capacity of the gas from a separate experiment. $\endgroup$ – Chet Miller Jul 11 at 13:20
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    $\begingroup$ How would you define a "mechanical process"? Are you excluding the possibility of heat transfer during the process? $\endgroup$ – Bob D Jul 11 at 14:41
  • $\begingroup$ Definitely. Adiabatic walls so no heat transfer. $\endgroup$ – Tachyon209 Jul 11 at 18:15
  • $\begingroup$ @BioPhysicist I don't understand how heat added/taken from the system can help to determine internal energy differences. Do you have a way? $\endgroup$ – Tachyon209 Jul 11 at 20:35
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If you only can expand or contract the piston, indeed you cannot connect the two states with a mechanical process. But you can devise different mechanical ways to act on the system.

I borrow one from Callen himself, Thermodynamics and an Introduction to Thermostatistics, paragraph 1-7: through a very small hole in the adiabatic wall, you can pass a thin shaft carrying a propellor blade at the inner end and a crank handle at the outer end. Doing so without letting the gas escape and without perturbing the system would be very difficult, but not impossible in principle (anyway in practice it would be easier, and equivalent, to prepare the system with blade and handle already present from the beginning).

By turning the handle, you can now do a measurable mechanical work on the system, that increases its internal energy. If you keep the external pressure on the piston constant end equal to $P_0$, this energy increase will result in a gradual expansion. In this way you can move from $(P_0,V_0)$ to $(P_0,2V_0)$.


EDIT: The point is: we only need a process that connects the two states, during which all the energy transfers are directly under control and measurable. That's the whole point of having adiabatic walls (heat flow is not directly controllable or measurable) and of using only mechanical processes (measuring mechanical work is often as easy as measuring a displacement). When we know all the energy transfers involved in the process, then we know also the energy difference between initial state and final state.

As noted in a comment below, the process could be one-way (as it is in our example) but this makes no difference as long as all energy transfers are under control.

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  • $\begingroup$ So, you mean that as long as we have a process in which we can quantitatively determine how much energy we've introduced into the system, the difference in internal energies can be calculated? And does the introduction of a rotor introduce parameters in addition to P and V in the system? $\endgroup$ – Tachyon209 Jul 14 at 20:29
  • $\begingroup$ Also, does this mean that setting up an adiabatic wall and doing expansion or contraction is just another way of adding energy to the system in a certain manner? $\endgroup$ – Tachyon209 Jul 14 at 20:30
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    $\begingroup$ If this is the right interpretation, such a "link" would only be one-way: no matter how you twist your blade & handle, you can't go from $(P_0, 2V_0)$ back to $(P_0, V_0)$. Maybe this is what's intended, though. $\endgroup$ – Michael Seifert Jul 14 at 20:38
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    $\begingroup$ @Tachyon209 I am not sure I 100% understand your question. We use only $P$ and $V$ because they completely determine the equilibrium state of this simple system. Actually, we would also need the number of particles $N$, but most of the time we're assuming that it stays constant and we ignore it. Adding the propellor blade doesn't change this. In fact, it simply allows to explore the full $(P, V)$ space: as long as you are confined to an adiabat, only one parameter, e.g. $P$, would be enough to determine the state, with $V$ determined by $P$. Hope I've addressed your doubts. $\endgroup$ – HicHaecHoc Jul 15 at 7:06
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    $\begingroup$ @Tachyon209 I think atleast one of them should be the right interpretation of Joule's statement, since in adiabatic reversible processes, you can go both ways "mechanically" $\endgroup$ – netflix_and_physics Jul 15 at 12:22

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