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I'd like to understand whether a work interaction occurs during the thermodynamic process described below, and why or why not. My analysis seems to suggest that some sort of work should be occurring, but my intuition says there's no work done. So specifically I want to know where my approach fails, or why my intuition is off base.

Consider two equal volumes of an ideal gas, $A$ and $B$, separated by a fixed adiabatic wall. Both halves contain the same number of molecules: The only difference between the two halves is that they're at different temperatures, say $T_A$ and $T_B$. This is the initial equilibrium state of the system.

Now, suppose the adiabatic wall is replaced by a thermally conductive wall, also rigid. This wall keeps the volume of each half the same, but allows energy transfer between the two halves. After some time, the two halves reach a common temperature $(T_A+T_B)/2$.

I can say a few things about this process. Let's just focus on one half of the system, say $A$. First, assuming that the movement towards equilibrium after initial thermal contact is sufficiently slow, the subsystem $A$ passes through a succession of states with roughly well-defined thermodynamic variables. So, at every infinitesimal step of the process, we have:

$$dU=TdS-PdV,$$

where $U$, $T$, $S$, $P$, and $V$ are thermodynamic properties of $A$ alone. Second, by the first law of thermodynamics:

$$dU=\delta Q + \delta W.$$

Here, I've used the convention that $\delta Q$ is the heat flow into $A$ during the infinitesimal step, and $\delta W$ is the work done on A. Taking the difference between these two equations yields:

$$TdS - \delta Q = PdV + \delta W.$$

Now, it seems to me that the process is irreversible: Returning the total system $A$ and $B$ to its original state would require some sort of permanent alteration of its environment. We can't just make the temperatures of the two halves unequal again without any external effects (right? This is one part where I'm a little confused, so I'm being somewhat imprecise). So by the second law of thermodynamics, we know that $dS > \delta Q/T$, and so:

$$TdS - \delta Q = PdV + \delta W > 0.$$

But the volume of $A$ remains fixed during the process, so $dV=0$, and therefore we have $\delta W > 0$. So it seems like at each infinitesimal step in the process, some work is done on $A$. But I can't see a mechanism by which this work is being performed. So am I wrong in my conclusion that a work interaction is occurring, or I am not thinking about something else in the right way?

Thanks.

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The issue is one of terminology, and this particular issue is an extremely common one. (In the context of classical thermodynamics, almost all of the issues that students run into are ones of terminology.)

In this case, you are confusing the notions of reversible/irreversible and quasi-static/non-quasi-static. Quasi-static is a term that applies to an individual system, and reversible is a term that applies to a collection of systems. A system undergoes a quasi-static (or quasi-equilibrium) process when it moves through a sequence of equilibrium states. A quasi-static process can be irreversible, if the irreversibility occurs in some sense outside the system; in the context of elementary thermodynamics, irreversibility is usually a consequence of heat flow across a finite temperature difference between a system and a reservoir (or, as in the OP's case, between one system and another).

Therefore, the short answer is that the OP went wrong when saying that $dS \neq \delta Q/T$ for the process. By assumption, the individual processes undergone by systems A and B are both quasi-static, and so $$dS_A = \frac{\delta Q_A}{T_A},$$ and $$dS_B = \frac{\delta Q_B}{T_B}.$$


So here is how I would analyze the problem, based on what the OP has done. Suppose we have the same setup, in which both systems A and B are at finitely different temperatures $T_A$ and $T_B$ to begin with, and they are then brought into contact via a rigid, impermeable, immovable, conducting boundary. In this case, the work done on either gas must be zero, as the OP notes, due to the membrane being rigid and immovable. In addition, since the processes undergone are quasi-static, we can write that the change in entropy of each subsystem during a small chunk of the process is $$dS = \frac{\delta Q}{T},$$ and that the change in internal energy during that small chunk of the process is $$dU = \delta Q + \delta W = TdS.$$ By conservation of energy, the total change in energy of the combined system is given by $$dU = dU_A +dU_B = \delta Q_A + \delta Q_B=0,$$ since the systems are in thermal contact but are otherwise isolated from the rest of the universe.

Now, we do know that this process must be irreversible, since there is heat flow across a finite temperature difference, so let's compute the total change in entropy of the combined system and see if it's strictly positive. Without loss of generality, let's take $T_A > T_B$, in which case $$\delta Q_B = -\delta Q_A = |\delta Q_A|.$$ (That is, $\delta Q_B>0$ since heat must flow into system B, and $\delta Q_A<0$ since heat must flow out of system A.) Then, the combined change in entropy during one tiny chunk of the process (so that the temperatures can be treated as constant) is \begin{align*} dS &= dS_A + dS_B = \frac{\delta Q_A}{T_A} + \frac{\delta Q_B}{T_B}\\ & = \frac{-|\delta Q_A|}{T_A} + \frac{|\delta Q_A|}{T_B} = |\delta Q_A|\frac{T_A-T_B}{T_AT_B} >0, \end{align*} where the last inequality arises as a consequence of $T_A > T_B$. This shows that at every step along the way, the entropy increases until the systems have equilibrated.

To complete the calculation, we merely integrate each expression separately. Assuming that each system has the same mass and that the specific heats are temperature-independent, we have \begin{align*} \Delta S_A &= \int \frac{\delta Q_A}{T_A}\\ &=\int_{T_A}^{(T_A+T_B)/2} \frac{n C dT}{T_A}= nC\ln\left(\frac{(T_A+T_B)/2}{T_A}\right), \end{align*} where $n$ is the number of moles and $C$ is the molar specific heat. The same expression holds for B (with B switched with A), and when we add the two expressions, it simplifies to $$\Delta S = \Delta S_A + \Delta S_B = 2 n C\ln\left(\frac{(T_A+T_B)/2}{\sqrt{T_AT_B}}\right).$$

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  • $\begingroup$ This change in entropy is not calculated correctly. The correct result for the change in entropy is $$\Delta S=2MC_v\ln{\left[\frac{(T_A+T_B)/2}{\sqrt{T_AT_B}}\right]}$$where M is the number of moles of each gas and $C_v$ is the molar heat capacity. For the correct procedure for determining the change in entropy of a closed system experiencing an irreverisble process, see the following link: physicsforums.com/insights/grandpa-chets-entropy-recipe $\endgroup$ – Chet Miller Aug 26 '16 at 3:39
  • $\begingroup$ @ChesterMiller. My wording isn't clear, but I actually wasn't trying to calculate the total change in entropy for the entire process (which perhaps I should have) but rather calculating the change in entropy during a small chunk of the process, and left the rest as an exercise (since it wasn't really a necessary part of the answer anyway). My expression can be integrated to yield yours. $\endgroup$ – march Aug 26 '16 at 5:12
  • $\begingroup$ You cannot get the change in entropy correct for any part of this process by allowing the masses to be in contact with one another. Each and every part of the irreversible change involves finite temperature gradients, and, moreover, during the irreversible change, the temperatures are not even uniform or constant. To calculate the entropy change correctly, you need to totally disregard the actual irreversible process and focus only on the initial and final states. Then you need to devise a reversible path between these end states and calculate the integral of dq/T for the reversible path. $\endgroup$ – Chet Miller Aug 26 '16 at 20:18
  • $\begingroup$ In addition to all that, when applying the Clausius inequality, you are supposed to use, in the denominator of dq/T, the temperature at the interface (boundary) where the heat transfer is occurring, not the average temperature of the body. The temperature at the interface is neither TA nor TB, but a value somewhere in between these values. For more on this specific issue, see Moran et al, Fundamentals of Engineering Thermodynamics. $\endgroup$ – Chet Miller Aug 26 '16 at 20:46
  • $\begingroup$ @ChesterMiller. I am not using the Clausius inequality but rather the definition of the change in entropy of a system (A or B) during a very tiny part of a quasi-static process (so that the temperature of the system can be treated as constant). Your point about the systems not being allowed to be in contact is well-taken, however. The point, I think, is that when two objects at different temps are brought into contact, this necessarily sets up temperature gradients within the materials, so the system processes can no longer be even quasi-static, much less the whole process reversible... $\endgroup$ – march Aug 27 '16 at 3:34
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The problem starts from the first equation $dU=TdS-PdV$. When you assume this equation, $dS > \delta Q/T$ isn't valid, i.e. at one point, you assume it is irreversible and at another point you treated it as reversible.

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    $\begingroup$ thanks for the response. this is contrary to what I've read in multiple other sources. I've read that dU=TdS-PdV is valid for both reversible and irreversible processes. I've often seen it derived for reversible processes, but then it is noted that it also holds for irreversible processes, since it is expressed only in terms of path-independent quantities, state functions. for examples of sources, see: theory.physics.manchester.ac.uk/~judith/stat_therm/node38.html , en.m.wikipedia.org/wiki/Fundamental_thermodynamic_relation , and Principles of General Thermodynamics by Keenan $\endgroup$ – Wade Hodson Aug 23 '16 at 17:47
  • $\begingroup$ The second law is $dS \ge \delta Q/T$ where the equality is valid for reversible process. When it equals, the formula can be rewritten as $\delta Q=TdS $. The first law is $dU=\delta Q - PdV$ if other works can be ignored. Replacing $\delta Q$, you get the first equation $dU=TdS-PdV$. $\endgroup$ – user115350 Aug 23 '16 at 17:59

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