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All the basic texts say $dQ = T dS$. Some say it is possible to define absolute entropy. That source gives an equation integrating C dT which should reduce to $Q=ST$, if $S(0) = 0$ and $Q$ is defined as all the heat that can possibly be extracted from the object. But it seems like sources overwhelmingly avoid writing $Q=ST$. If I could use that equation it should be simple to visualize the nature of entropy... (see below)

This would also seem to make entropy the number of different variables that have been able to take up thermal energy. But when taking actual entropy values, for example, 131 J/mol K for H2, 192 J/(mol K) for N2, and 193 J/(mol K) for NH3 at STP - the values are many times larger than the gas constant. Is there a way to conceptualize entropy to break down these numbers into 46 different degrees of freedom, or is this comparison intrinsically flawed? Why?

EDIT: in light on ongoing discussion, I'd like to quote a very basic exercise from LibreTexts that made me start thinking about this issue:

Calculate entropy change when 36.0 g of ice melts at 273 K and 1 atm. [enthalpy of fusion 6.01 kJ/mol].

Their solution is $\Delta S$ = (6.01 kJ mol-1)/272 K * (36 g)/(18 g mol-1) = 1.22 kJ / K

This illustrates we can have more than one entropy value for the same temperature, and also something of a surprising result, namely that the higher the melting point of the substance, the less melting it affects the entropy. Of course, this is countered by likely requiring more energy to do the actual melting as reflected in the enthalpy value, but still, it made me wonder what this denominator means. The interpretation I want to give is that there is a pool of heat energy which is larger at higher temperature, and the total entropy has been increased in proportion to the expanded capacity to store heat at the same temperature. But this would imply thinking of a relationship between these total quantities.

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    $\begingroup$ That's not really how integration works... $\int T \, dS$ is not the same thing as $T S$. $\endgroup$ – knzhou Nov 8 '20 at 20:10
  • $\begingroup$ What do you mean by "many times larger than the gas constant?" They have different dimensions. $\endgroup$ – Sandejo Nov 8 '20 at 20:12
  • $\begingroup$ @knzhou The integral from the source was S(0) + ∫CdT, not ∫TdS. Because C = dQ/dT and you're going from 0 to T, and assuming S(0) = 0, that should get you ST. $\endgroup$ – Mike Serfas Nov 9 '20 at 13:00
  • $\begingroup$ @Sandejo: Entropy can be expressed in J/K; when multiplied by temperature it gives energy - these are the units of the Boltzmann constant. But in chemistry problems we'll often see both entropy and the Boltzmann's constant multiplied by 6.02E23/mol and given as J/(mol K) $\endgroup$ – Mike Serfas Nov 9 '20 at 13:03
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Heat is not a thermodynamic state variable, while entropy is.

In other words, heat is a word that describes the spontaneous transfer of energy between a system and its environment (which we denote $dQ$), which depends on the path, or the way that you apply changes to the system's macroscopic variables. It is meaningless to talk about heat as a property of the system itself (there is no heat function $Q$ that is a function of the system's state).

Entropy, on the other hand, is a state variable, and a system in equilibrium does have a well defined value $S$.

Therefore, $Q=TS$ is meaningless, as $Q$ is meaningless.

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  • $\begingroup$ Obtaining $Q$ through integration is fine if a suitable path is specified. The problem is that the asker is not integrating correctly, as @knzhou notes. $\endgroup$ – Chemomechanics Nov 8 '20 at 21:39
  • $\begingroup$ The OP defined $Q$ as "all the heat that can possibly be extracted from the system," that definition is what I was using in my answer. I agree you can define the heat that is exchanged along a given path. $\endgroup$ – Andrew Nov 9 '20 at 6:19
  • $\begingroup$ I probably misused the term "system" there, so I changed it to "object". All I mean is that if you cool something down to absolute zero it will emit a specific amount of heat energy (if we neglect frozen entropy, mostly to avoid semantics) $\endgroup$ – Mike Serfas Nov 9 '20 at 13:09
  • $\begingroup$ @MikeSerfas The amount of heat lost will on the way in which you do the cooling. There is a tradeoff between work you do on the system and heat that is spontaneously transferred to the environment. The amount of work depends the process you use to go from state A to state B, and same for the heat exchange. However, the total amount of energy in state A and state B does not depend on the path you take. This is why "heat" is not a state variable, and you can't associate a given amount of "total heat" with an object or system in a given state. However, "total energy" is a state variable. $\endgroup$ – Andrew Nov 9 '20 at 13:22
  • $\begingroup$ We may be getting to the crux of the issue here, but I'm still not following. I'm thinking that if you're about to cool a stationary object to absolute zero, the total energy of that object (excluding chemical bonds and so on) will be heat energy. Each particle is moving or vibrating with a certain average kinetic energy and all of that energy is going to be extracted as heat energy to give a single value for the amount of heat emitted, regardless of the path taken. What am I missing here? $\endgroup$ – Mike Serfas Nov 9 '20 at 13:33
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The absolute entropy of a system in a given equilibrium state at temperature $T$ and other parameters specified is given by $$ S = S(0) + \int_0^T \frac{1}{T} dQ_{\rm rev} $$ where it is understood that the other parameters are being held constant and $dQ_{\rm rev}$ refers to heat entering the system by a reversible process, and the limits on the integral are temperatures. Taking $S(0) = 0$, a useful way to write the result is $$ S(V,T_f) = \int_0^{T_f} \frac{C_V(V,T)}{T} dT + \sum_i \frac{L_i}{T_i} $$ where $C_V$ is the heat capacity at constant volume (in the case of a simple mechanical system) and the sum gives the contribution from first order phase transitions. For the $i$th transition the heat entering is the latent heat $L_i$ and during the transition the temperature ($T_i$) does not change.

The integrals above are not the only way to obtain the absolute entropy, but other ways must be consistent with these. Sometimes it can be obtained by calculation from first principles, and sometimes by a judiciously chosen measurement combined with some thermodynamic reasoning. Here is a paper by myself on that:

https://iopscience.iop.org/article/10.1088/1367-2630/18/4/043022

The total heat that enters the system during such a warming is $Q = \int dQ_{\rm rev}$ and clearly one will not get $Q = ST$. You could if you like define a quantity $ST$ and denote it by the letter $Q$, but this $Q$ would not be the total heat that can be extracted from the system under some given constraint or path, unless by chance it might be for some carefully chosen path.

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  • $\begingroup$ This (and the paper) look like they will be very informative. I realize that in the free text I cited above, the T denominator had been omitted, and in my mind I added it back outside the integral. Still... is there a way to relate ST to some measure of the total thermal energy of an object? $\endgroup$ – Mike Serfas Nov 9 '20 at 16:32
  • $\begingroup$ @MikeSerfas In many simple cases, $ST$ is within an order of magnitude of the total internal energy, but there's no guarantee that will be the case in general. $\endgroup$ – knzhou Nov 9 '20 at 16:59
  • $\begingroup$ Thank you for an interesting paper, but I think the argument based on extensivity of $S(U,V,N)$ is not general - it seems limited to simple models of matter compatible with that idea of extensivity. Real bodies are complicated, they can have surface effects, zero-point disorder, etc which break that idea. Also, on a more hypothetical level, if there are singularities in $C_V$ at points $T_i$, do you know why (in the spirit of the paper, based on thermodynamics or experiments) there isn't one at $T=0$? If there was such a singularity, we wouldn't be able to define $S$ by integrating from $T=0$. $\endgroup$ – Ján Lalinský Nov 9 '20 at 22:22
  • $\begingroup$ @JánLalinský Thanks for your remarks. Complications such as surface effects obviously make it hard to get $S$ in some cases but they become negligible compared to the bulk in the thermodynamic limit. There is a very rich variety of cases that can be handled without such complication. Regarding singularities, you are right that one can never be sure there is not some further transition at low $T$ but again for the simpler systems one can be sure enough in a variety of cases. $\endgroup$ – Andrew Steane Nov 10 '20 at 9:39
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It's fine to wish to integrate $T\,dS$ and to plan for a lower limit of $S=0$ under the Third Law. It's not fine to bring $T$ outside the integral to obtain $T\Delta S=TS$ as if $T$ is constant, because it's not (unless you're integrating from 0 to 0, in which case no net energy is transferred).

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  • $\begingroup$ T doesn't need to be a constant; it just needs to not be a function of S. Is T a function of S? $\endgroup$ – Mike Serfas Nov 9 '20 at 13:25
  • $\begingroup$ Indeed it is... $\endgroup$ – Chemomechanics Nov 9 '20 at 17:28
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In an isothermal process, in which T is a constant, you can integrate and obtain $Q=T\Delta S$, in which Q is the heat exchanged to change the entropy by $\Delta S$. For non-isothermal processes you will obtain a different relationship though.

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  • $\begingroup$ Suppose we could do that and go on to extract all the entropy ΔS out of an object under isothermal conditions, and with it all the heat energy Q that it can possibly release. Now we cool the object to absolute zero. If it truly had no entropy to begin with, then would there be any heat emitted as it cooled? $\endgroup$ – Mike Serfas Nov 9 '20 at 13:21
  • $\begingroup$ the process is isothermal, so you cannot lower the temperature to zero, so the entropy cannot be zero either $\endgroup$ – Wolphram jonny Nov 9 '20 at 16:00

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