4
$\begingroup$

In Daniel Schroeder's book Thermal Physics, he discusses the amount of energy needed to create an object, or its internal energy. He makes the point that you don't need to supply the entire energy U, since the environment, assumed to be a large thermal reservoir at temperature T, can provide some of this energy:

"Often, however, we're not interested in the total energy needed or the total energy that can be recovered. If the environment is one of constant temperature, the system can extract heat from this environment for free so all we need to provide, to create the system from nothing, is any additional work needed."

Schroeder goes on to say that the amount able to be provided by the environment is given by $ T \Delta S = TS $, where S is the system's final entropy.

Although he is not explicit about this, I understand all this to mean that all we need to do to "create" an object is to supply enough energy so that it "exists" at absolute zero, and that the reservoir will spontaneously supply enough energy to bring it up to temperature T. I am not 100% sure that I've understood him correctly so perhaps this is all or one source of my misunderstanding.

In general, the formula for the entropy (assuming only heat is allowed to transfer back and forth) is

$$ dS = \frac{dQ}{T}$$

If we are trying to figure out how much heat it takes to bring a system at absolute zero, up to temperature $T$, then I would think that formula must be:

$$ Q = \int T dS $$

Since the object is being brought up from $0K$ up to temperature T, the temperature is not constant, so we are unable to remove it from the integral. However, as we said above, Schroeder gives this amount as $ TS $, which is exactly what I would expect to see if the temperature were constant, which its not.

Now it is true that the reservoir is at a constant temperature T, so therefore the heat that leaves the reservoir (the same heat that will enter into the system) is equal to $Q = T \Delta S $, where $\Delta S$ is the entropy lost by the reservoir. However, the entropy lost by the reservoir is in general not equal to the entropy gained by the system, since the system and the reservoir are at different temperatures (at least initially). So $\Delta S$, the entropy lost by the environment, is not equal to $S$, the final entropy of our system, and Schroeder specifically notes that S in the formula above refers to the (final) entropy of the system.

So what am I misunderstanding here?

$\endgroup$
1
$\begingroup$

Heat and work are path dependent. Internal energy change is not.

Propose to make a system from nothing, meaning $T_o = 0$ and $S_o = 0$. Heat to $T_f$. Ignore the heat required as the "infinite supply". Also, do this step in an isentropic manner so that system entropy remains at zero. On a diagram of $S-T$, this is a horizontal line. This step can be irreversible. We can take this step in a hypothetical manner in the same way that we also pretend the surroundings is an infinite supply of heat (it is not). One way to do this is to expand irreversibly against vacuum. Since $T$ is constant, internal energy does not change, and this free expansion gives $T\Delta S = p \Delta V$.

Next, change the entropy only but make no change in internal energy. In the case that $\Delta U = 0$, the mechanical work needed for this step is $w = q = T_f \Delta S$. Since the initial entropy was zero to start this step, the net result is $T_f\Delta S = w_{extra} =$ extra mechanical work needed = $T_f S_f$.

$\endgroup$
  • $\begingroup$ Thank you for your response...there are a few things I don't understand about your answer: Why are you assuming the change in internal energy is 0? I didn't see the author make this assumption. Second, is my understanding then wrong regarding what it means to "create" a system? I thought it meant creating it at 0K and then letting the environment warm it up. You seem to be saying that the system is doing mechanical work. Third, if the change in internal energy is zero, then it is isothermal, but then wouldn't that mean the temperature is constant, as opposed to the example I gave above? $\endgroup$ – Israel Apr 16 at 22:49
  • $\begingroup$ I'm afraid I'm still not following you...your first step in the "creation" process is to heat they system up to temperature T, then second, "expand" it out. But if you add heat to the system aren't you raising its entropy? And if so, once you carry out the second step, the work which you say is $T \Delta S $, is not starting from zero entropy, so we can't say that $T \Delta S = TS$. Obviously, what I'm saying is wrong -- I just don't see how. $\endgroup$ – Israel Apr 17 at 5:43
  • $\begingroup$ I'm still not following you...according to my understanding, the "creation" process has 2 steps: start with the raw ingredients at zero volume and zero energy, then 1) warm the system up, then 2) expand it out. Warming a system up brings in entropy, but you are saying that that the entropy does not change. In order to cancel out this entropy, it must be simultaneously removed. But how can you do this? I could always compress or expand the gas, but this wouldn't change the entropy of the system, right? So then how do you remove entropy from a system that's heating up? $\endgroup$ – Israel Apr 22 at 18:51
  • $\begingroup$ One approach is to balance $\int TdS$ with $\int pdV$. HOW this is done is eventually irrelevant. Reversible processes never exist in reality, yet we use them all the time conceptually. The same is true here. Draw a $S-T$ diagram. Move conceptually at zero $T$ horizontally to increase $S$. Then move *conceptually$ at that $S$ vertically to increase $T$. The first step may not have a practical way to happen, but the combination of the two steps provides a stronger conceptual understanding of the question. $\endgroup$ – Jeffrey J Weimer Apr 22 at 21:16
  • $\begingroup$ Yeah, but in the case of a reversible process (like heat flow), it is the limit of a real process (e.g. as the 2 temps get closer together, this gets better). Here, I don't see how you can do this even with the approach you gave above: $TdS = dU + PdV = dQ + dW + PdV = dQ - PdV + PdV = dQ.$ So we see here that work done by a gas is isentropic (right?), as contracting a gas adds energy and thus entropy, but also decreases the volume and thus entropy by the same amount. So if we want to keep the left side 0, we must have Q be 0, so what other way are we going to heat up the gas? $\endgroup$ – Israel Apr 29 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.