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For a general thermodynamic system, it is always possible to directly control its extensive variables discounting the entropy. For fluids, this is usually just the volume of the container. Say we have two finite containers in thermal contact and equilibrium. If I quasistatically expand one, the second will slowly push temperature into the first reaching some lower temperature for both. How do I know that when I push back to the original volume and the containers are in equilibrium again, they will be at the same temperature? All we know is that total energy is conserved and that the temperatures are the same but there is no reason for this to have to mean they are at the same temperature after. I know this may seem intuitive but I want to know which exact statement I am missing to make this conclusion.

Clarification: Although I mention containers here to help intuition, my question is for a general thermodynamic system so I am ignoring the special case of ideal gases where temperature and energy are very simply related.

Edit: A possible route I was thinking was that assume two containers in thermal contact have the exact same extensive variables discounting entropy. Then their final state at equilibrium is uniquely determined by the total energy between the two systems but I don't know how to show that either.

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3 Answers 3

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Energy, entropy and other quantities are state functions. That means that they are functions that depend solely on the state of the system. For example, energy (for an ideal gas) depends only on the temperature of the system. Work for example varies depending on the process that the system goes through and so work is not a state function. By definition a cyclic process is one in which the system is at the same state at the beginning and at the end, so if you know that the state functions didn't change then the process is cyclic.

In your example, if the pressure and the volume are the same as before then the temperature must be the same too (given that the same amount of the same gas is still present). Also, if energy was conserved then you also know that the temperature has to be the same too.

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  • $\begingroup$ So we already know temperature is the same to begin with because they are in thermal equilibrium. However, how do we know the pressures are the same if we are talking generally and not just for these gases?Pressure in the general case would just be the conjugate internal variable for the extensive variable being discussed $\endgroup$ Dec 19, 2018 at 22:27
  • $\begingroup$ The only thing we know that is the same as before is the volume of each. For energy, we only know the sum of the two is the same $\endgroup$ Dec 19, 2018 at 22:29
  • $\begingroup$ The way you phrased the example made me think of an adiabatic process. If you do it quasistatically then pressure will be the same on both containers. $\endgroup$
    – Kirtpole
    Dec 19, 2018 at 23:44
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I think I got the answer to my own question but correct me if I'm wrong. For a general thermodynamic system, I think it is true that an increase in temperature always result in an increase of energy. This means if the containers have the same extensive variables as before, the previous assignments of temperature is a possible end point. Any other temperatures would result in strictly more or less energy than the initial but we know energy is conserved in the system so this is not true. Hence, the same temperatures and, as a result, same state as before because we have sufficient number of equal state functions to say the whole state is equal.

Let me know if there is any fault in this.

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  • $\begingroup$ To return to the same state all thermodynamic variables have to the same, not just the temperature. All state functions will be the same as when they started: energy, entropy, enthalpy, etc. $\endgroup$
    – Kirtpole
    Dec 19, 2018 at 23:47
  • $\begingroup$ Are you assuming that the expansion and subsequent compression are adiabatic with respect to the overall combined system? $\endgroup$ Dec 19, 2018 at 23:54
  • $\begingroup$ @ChesterMiller Yes, in terms of the combined system, it is adiabatic. $\endgroup$ Dec 20, 2018 at 2:32
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OK. Let's see how it plays out in detail for an ideal gas undergoing adiabatic reversible expansion followed by adiabatic reversible compression.

Let:

$n_R$ = number of moles of gas in the rigid container

$n_V$ = number of moles of gas in variable volume container

$T_{init} =$ initial temperature in both containers

$V_{init}=$ initial volume of variable volume container

So, in the reversible expansion phase, we have:

$$dU=-PdV$$or $$(n_R+n_V)C_vdT=-\frac{n_VRT}{V}dV$$Integrating this equation between the initial and expanded states yields: $$\frac{T_{final}}{T_{init}}=\left(\frac{V_{init}}{V_{final}}\right)^{\frac{n_V}{(n_R+n_V)}\frac{R}{C_v}}\tag{1}$$And, the internal energy change between the initial and expanded states will be:$$\Delta U_{exp}=(n_R+n_V)C_vT_{init}\left[1-\left(\frac{V_{init}}{V_{final}}\right)^{\frac{n_V}{(n_R+n_V)}\frac{R}{C_v}}\right]$$ Now, if we apply the same methodology to the compression, we obtain: $$\Delta U_{compress}=(n_R+n_V)C_vT_{final}\left[1-\left(\frac{V_{final}}{V_{init}}\right)^{\frac{n_V}{(n_R+n_V)}\frac{R}{C_v}}\right]$$But, combining Eqn. 1 with this relationship yields: $$\Delta U_{compress}=(n_R+n_V)C_vT_{init}\left[\left(\frac{V_{init}}{V_{final}}\right)^{\frac{n_V}{(n_R+n_V)}\frac{R}{C_v}}-1\right]$$ So, $$\Delta U_{exp}+\Delta U_{compress}=0$$

The change in state of the system over the entire cycle is zero.

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  • $\begingroup$ Yes, I understand this. My question however was for a general thermodynamic system $\endgroup$ Dec 20, 2018 at 5:45
  • $\begingroup$ OK. I was wondering, then, the nature of the analysis it would take to convince you: (a) a similar analysis for a more general material like, say, a van der Waals gas, or (b) an entirely different kind of analysis? $\endgroup$ Dec 20, 2018 at 12:59

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