Does $Q=\int T dS$ for any quasistatic process?

Question

According to the first law of thermodynamics any thermodynamic process whatsoever, $$dU=\bar{d}Q-\bar{d}W_{by}$$ Moreover, is we also have $$dU=TdS-PdV$$ for any process of a system that does only P-V work provided the state functions $T,P,S$ are well defined. Now, according to several answers I've seen on this site, for any process( quasistatic or otherwise), we have $dW_{by}=P_{ext}dV$ and in the special case where the process is also quasistatic, $P_{ext}=P$ and thus $dW_{by}=PdV$. Does this, along with the first law and fundmental thermodynamic relation, imply that $$\bar{d}Q=TdS\implies Q=\int TdS$$ for any quasistatic process, reversible or otherwise? I can't find a fault in my reasoning, yet Clausius' theorem states that for an irreversible change, we have $dS>dQ/T$. Is this not a contradiction? I have a feeling that this has something to do with the fact that $T$ in the fundamental thermodynamic relation is the temperature of the system in one case and the reservoir in the other, but I am not sure about it at all. I would really appreciate if someone can help me sort this out.

Edit

I have tested the proposition for the irreversible, quasistatic isochoric heating of an ideal gas connected to a reservoir at constant temperature $T_r>T_0$. This gives $$\int_{S_0}^{S_f} TdS=C_v (T_r-T_0)=Q$$ using the expression $S=C_v\ln(T)+nR\ln(V)+C$ for an ideal gas.

The proposition appears to hold for this special case at least. I believe the expression is, in fact, true in general, because a quasistatic process can be irreversible only if there is heat transfer at a finite temperature gradient. In such a case, however, we may carry out the same change by transferring the same amount of heat at each step but reversibly (using reservoirs at slightly higher temperatures than the instantaneous temperature of the system). Then we have $$dQ=TdS$$ by Clausius' theorem, where $T$ is the common instantaneous temperature of the system and the reservoir. Clearly, however, $TdS$ here equals that in the irreversible process because each step is b/w the same two equilibrium states and at the same temperature. Does this reasoning not work?

Answer 1

I strongly recommend that you avoid using differentials in S for processes that are irreversible, since S is an equilibrium state function and S along an irreversible path is ambiguous to define.

For all process paths of a closed system, the Clausius inequality should read $$\Delta S\geq\int{\frac{dq_{rev}}{T_{Ext}}}$$where $dq_{rev}$ represents the heat added increments over a reversible path (between the two equilibrium end states) and $T_{Ext}$ represents the temperature at the interface boundary between the system and its surroundings.

The procedure for deterring the change in entropy for a closed system that experiences an irreversible process path should be as given in the following reference (which includes worked examples of how to apply the procedure: www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Answer 2

Does $Q=\int T dS$ for any quasistatic process?

No.

For example, in a quasistatic adiabatic compression of a gas involving mechanical friction, as between piston and cylinder surfaces, both $\Delta T\gt 0$ snd $\Delta S\gt 0$ due to entropy generation, yet $Q=0$.

It is not sufficient for a process to be quasistatic for it to be reversible. It must also not involve mechanical friction.

Even if I accept your counterexample, however, my question was more about the error in my reasoning that =∫

$Q$ (heat) in the first law equation refers to energy transfer crossing the boundary between the system and its surroundings due to a temperature difference.

Although a change in entropy of the system is defined for a reversible transfer of heat divided by the temperature at the boundary between the system and its surroundings, that doesn't mean a change in entropy requires heat. As my example shows, entropy can be generated within the system due to internal friction not involving heat crossing the boundary.

However, what mechanical friction, as between the piston and cylinder walls, does is raise the temperature of the gas near the sliding surfaces above the interior temperature of the gas. This in effect results in irreversible heat transfer due to temperature gradients internal to the system, generating entropy. This "heat" is not the same as the $Q$ in the first law equation, or in the definition of the change in entropy of the system.

Hope this helps.

Answer 3

IMHO this kind of introduction to thermodynamics is often quite unfortunate and confusing.

Let's quickly review it starting from the first principle and theorem of kinetic energy known from classical mechanics.

• First principle of thermodynamics is the total energy balance equation for closed systems, namely

  $$d E^{tot} = \delta L^{e} + \delta Q^e \ ,$$

  i.e. change in total energy of a closed system is due to work done on the system by external actions and heat from the environment to the system.

• Theorem of kinetic energy from classical mechanics tells us that

  $$d K = \delta L^{tot} = \delta L^e + \delta L^i \ ,$$

  i.e. change in kinetic (macroscopic) energy is due to total work (both internal and external) done on the system.

Following Gibbs, the first who provided a reasonable mathematical formalism to thermodynamics, internal energy of the system is a state variable of the system defined as the difference of total and kinetic energy, $E = E^{tot} - K$ and balance equation for internal energy reads

$$d E = \delta Q^e - \delta L^i \ .$$

Now, let's do the last step: internal work can be divided as the sum of reversible and irreversible contributions (e.g. work done by viscous stress in fluids) that is always non-positive,

$$\delta L^i = \delta L^{i,rev} - \delta^+ D \ ,$$

so that the balance equation of internal energy can be written as

$$d E = - \delta L^{i,rev} + \delta Q^e + \delta^+ D \ .$$

Now, $dE$ is reversible since $E$ is a state function; $\delta L^{i,rev}$ is reversible by definition. It follows that $\delta Q^{e} + \delta^+ D$ is reversible. This contribution is recognized to be $T \, dS$.

Doing so,

• the inequality

  $$T dS = \delta Q^e + \delta^+ D \ge \delta Q^e \ ,$$

  holds and Clausius statement of the second principle of thermodynamics immediately follows,

  $$dS \ge \frac{\delta Q^e}{T} \ .$$

• For a fluid system whose reversible contribution to internal work reads $\delta L^{i,rev} = P dV$, the balance of internal energy reads

  $$\begin{aligned} dE & = \delta Q^{e} - \delta L^{i} = \\ & = \delta Q^{e} + \delta^+ D - \delta L^{i,rev} = \\ & = T dS - P dV \end{aligned}$$

  If macroscopic kinetic energy is negligible w.r.t. other contributions, $d K \sim 0$, the following approximations hold $d E^{tot} \sim d E$, $\delta L^e \sim - \delta L^i$, so that it's possible to directly relate total energy and internal energy balance equations

Term $T dS$ contains also heat transfer between parts of the system, if it's not homogeneous.

Now, let's deal with the definition of quasi-static and reversible. In quasi-static processes it's possible to neglect macroscopic kinetic energy contributions. Reversible processes can be thought to be those processes where $\delta^+D = 0$, and no heat transfer between regions of a non-homogeneous system occurs (or the system is homogeneous).