# Reversibility/quasistatic and the first law

1. Nov 5, 2014

### Avatrin

Hi
The thing I am struggling with is why for reversible or quasistatic (depending on what book you use):
Q = TdS and W = -PdV
Thus, dU = TdS - PdV

What about reversible/quasistatic processes makes -PdV the only work the system can do?
While I know $\frac{1}{T} = (\frac{dS}{dU})_{N,V}$, I dont understand why Q = TdS in quasistatic/reversible processes. I also understand that $dS = \frac{Q}{T}$ was the original definition of entropy, but how am I supposed to understand this in terms of quasistatic processes?

What makes quasistatic/reversible processes so fundamentally different than other types of processes? I understand that they are always in equilibrium, but I cannot connect the dots.

2. Nov 5, 2014

### Staff: Mentor

If a process is reversible, it means that no additional entropy, over $Q/T$, is produced. The general formula is
$$dS \ge \frac{Q}{T}$$
for which you get an equality only for a reversible process. This is what makes it special.

Another way to see it is that it is the extra entropy created by an irreversible process that makes it such that it is not reversible.

3. Nov 5, 2014

### Avatrin

Okay, but why? Also, what about W = -PdV? I get why PdV does work on the system, but why, in quasistatic processes, is it the only thing that does work?

4. Nov 5, 2014

### Andrew Mason

By definition dS = dQ/T where dQ is the reversible heat flow. So we can write dQrev = TdS. That just follows from the definition.

The heat flow in a reversible process is not necessarily the same as the heat flow in a quasi-static process. A reversible process is one that can be reversed by an infinitesimal change of conditions: ie. the system and surroundings are in equilibrium. A quasi-static process is one that proceeds infinitely slowly, but the system and surroundings need not be in a state of equilibrium.

Work is force x distance through which the force acts or pressure x change in volume through which the pressure acts. For a gas expanding against an external pressure, the work done on the surroundings is PextdV where Pext is the pressure in the surroundings not the gas pressure. If W = -PdV where W is the work done on the gas and P is the gas pressure, then the gas pressure and surrounding pressure must be the same. ie. the system and surroundings are in equilibrium (reversible).

AM

5. Nov 5, 2014

### Staff: Mentor

Let's see if I can paraphrase what you're asking. Are you asking "why is it that the statistical thermodynamic definition of entropy is quantitatively the same as the classical thermodynamic definition of entropy?"

Chet

6. Nov 6, 2014

### Avatrin

No, not really.. I am asking why Q = TdS and W = -PdV in the case of reversible processes. I think the main problem is that I have a problem understanding reversible and quasistatic processes. The fact that the two terms are sometimes used interchangeably only compounds the confusion. So, going to the library to read some other book on thermodynamics did not really help.

Also, most explanations, like the replies above, are not really explanations; They are assertions. They just tell me that something is true without explaining why. I have already read it is true.

So, in the case of Q and W, why do we have equalities in the case of reversible processes?

7. Nov 6, 2014

### Staff: Mentor

I think that the following might help.

FIRST LAW OF THERMODYNAMICS

Suppose that we have a closed system that at initial time ti is in an initial equilibrium state, with internal energy Ui, and at a later time tf, it is in a new equilibrium state with internal energy Uf. The transition from the initial equilibrium state to the final equilibrium state is brought about by imposing a time-dependent heat flow across the interface between the system and the surroundings, and a time-dependent rate of doing work at the interface between the system and the surroundings. Let $\dot{q}(t)$ represent the rate of heat addition across the interface between the system and the surroundings at time t, and let $\dot{w}(t)$ represent the rate at which the system does work on the surroundings at the interface at time t. According to the first law (basically conservation of energy),
$$\Delta U=U_f-U_i=\int_{t_i}^{t_f}{(\dot{q}(t)-\dot{w}(t))dt}=Q-W$$
where Q is the total amount of heat added and W is the total amount of work done by the system on the surroundings at the interface.

The time variation of $\dot{q}(t)$ and $\dot{w}(t)$ between the initial and final states uniquely characterizes the so-called process path. There are an infinite number of possible process paths that can take the system from the initial to the final equilibrium state. The only constraint is that Q-W must be the same for all of them.

If a process path is irreversible, then the temperature and pressure within the system are inhomogeneous (i.e., non-uniform, varying with spatial position), and one cannot define a unique pressure or temperature for the system (except at the initial and the final equilibrium state). However, the pressure and temperature at the interface can be measured and controlled using the surroundings to impose the temperature and pressure boundary conditions that we desire. Thus, TI(t) and PI(t) can be used to impose the process path that we desire, where the subscript I refers to the parameter values at the interface. Alternately, and even more fundamentally, we can directly control, by well established methods, the rate of heat flow and the rate of doing work at the interface $\dot{q}(t)$ and $\dot{w}(t)$).

Both for reversible and irreversible process paths, the rate at which the system does work on the surroundings is given by:
$$\dot{w}(t)=P_I(t)\dot{V}(t)$$
where $\dot{V}(t)$ is the rate of change of system volume at time t. However, if the process path is reversible, the pressure P within the system is uniform, and

$P_I(t)=P(t)$ (reversible process path)

Therefore, $\dot{w}(t)=P(t)\dot{V}(t)$ (reversible process path)

Another feature of reversible process paths is that they are carried out very slowly, so that $\dot{q}(t)$ and $\dot{w}(t)$ are both very close to zero over the entire process path. However, the amount of time between the initial equilibrium state and the final equilibrium state (tf-ti) becomes exceedingly large. In this way, Q-W remains constant and finite.

SECOND LAW OF THERMODYNAMICS

In the previous section, we focused on the infinite number of process paths that are capable of taking a closed thermodynamic system from an initial equilibrium state to a final equilibrium state. Each of these process paths is uniquely determined by specifying the heat transfer rate $\dot{q}(t)$ and the rate of doing work $\dot{w}(t)$ as functions of time at the interface between the system and the surroundings. We noted that the cumulative amount of heat transfer and the cumulative amount of work done over an entire process path are given by the two integrals:
$$Q=\int_{t_i}^{t_f}{\dot{q}(t)dt}$$
$$W=\int_{t_i}^{t_f}{\dot{w}(t)dt}$$
In the present section, we will be introducing a third integral of this type (involving the heat transfer rate $\dot{q}(t)$) to provide a basis for establishing a precise mathematical statement of the Second Law of Thermodynamics.

The discovery of the Second Law came about in the 19th century, and involved contributions by many brilliant scientists. There have been many statements of the Second Law over the years, couched in complicated language and multi-word sentences, typically involving heat reservoirs, Carnot engines, and the like. These statements have been a source of unending confusion for students of thermodynamics for over a hundred years. What has been sorely needed is a precise mathematical definition of the Second Law that avoids all the complicated rhetoric. The sad part about all this is that such a precise definition has existed all along. The definition was formulated by Clausius back in the 1800's.

Clausius wondered what would happen if he evaluated the following integral over each of the possible process paths between the initial and final equilibrium states of a closed system:
$$I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}$$
where TI(t) is the temperature at the interface with the surroundings at time t. He carried out extensive calculations on many systems undergoing a variety of both reversible and irreversible paths and discovered something astonishing. He found that, for any closed system, the values calculated for the integral over all the possible reversible and irreversible paths (between the initial and final equilibrium states) was not arbitrary; instead, there was a unique upper bound (maximum) to the value of the integral. Clausius also found that this result was consistent with all the "word definitions" of the Second Law.

Clearly, if there was an upper bound for this integral, this upper bound had to depend only on the two equilibrium states, and not on the path between them. It must therefore be regarded as a point function of state. Clausius named this point function Entropy.

But how could the value of this point function be determined without evaluating the integral over every possible process path between the initial and final equilibrium states to find the maximum? Clausius made another discovery. He determined that, out of the infinite number of possible process paths, there existed a well-defined subset, each member of which gave the same maximum value for the integral. This subset consisted of what we call today the reversible process paths. So, to determine the change in entropy between two equilibrium states, one must first conceive of a reversible path between the states and then evaluate the integral. Any other process path will give a value for the integral lower than the entropy change.

So, mathematically, we can now state the Second Law as follows:

$$I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}\leq\Delta S=\int_{t_i}^{t_f} {\frac{\dot{q}_{rev}(t)}{T(t)}dt}$$
where $\dot{q}_{rev}(t)$ is the heat transfer rate for any of the reversible paths between the initial and final equilibrium states, and T(t) is the system temperature at time t (which, for a reversible path, is equal to the temperature at the interface with the surroundings). This constitutes a precise mathematical statement of the Second Law of Thermodynamics.

8. Nov 6, 2014

### Andy Resnick

This is not as simple a question as it may first appear, and I may be over-interpreting your actual question. Part of the difficulty is that the heat Q, unlike the work W, is not a state variable. One unfortunate 'side effect' of this is that introductory presentations tacitly change thermo*dynamics* into thermo*statics*, which concerns equilibrium systems and process and can be directly related to statistical (equilibrium) mechanics. Even going from thermostatics to thermokinetics is complicated (Onsager's relations), and AFAIK has not yet been completely developed in the context of statistical mechanics (fluctuation-dissipation theory). Going to full thermodynamic models can be done, but AFAIK has not yet been done in terms of statistical mechanics- dissipation remains stubbornly unaccounted for in a mechanical theory.

So... quasi-static (reversible) processes are used to recover equilibrium conditions, and there are three special types typically encountered: adiabatic, isentropic, and isothermal. Under those conditions (no dissipation), dQ can be treated as an exact differential (like dW) and the math is simpler.

Alternatively- you could simply be asking (for example) why there is no VdP term in the work- that's because the pressure is defined as P = P(V,T). The characterization of work in terms of volume expansion and contraction was originally done by Carnot.

9. Nov 6, 2014

### Staff: Mentor

The work W is not a state variable either, even if the process is reversible. The internal energy U is the only state variable involved. ΔU = Q - W, so if Q is not a state variable, neither can W be a state variable.

Chet

10. Nov 6, 2014

### Andy Resnick

quite right- I meant the work is a path-independent quantity. Call it a typo....

11. Nov 6, 2014

### Staff: Mentor

I respectfully disagree. The work between an initial thermodynamic equilibrium state and a final thermodynamic equilibrium state of a closed system is not path-independent. Of course, since ΔU is path independent, if Q is path dependent, so must W be path dependent. The key constraint is that the difference between Q and W must be path independent.

You can easily prove that the work W is path dependent by identifying two thermodynamic equilibrium states of a system, and showing that there is more than one path between the initial state to the final state in which W (and Q) differs for the two paths. Just define each path using a different sequence of reversible adiabatic and reversible isothermal segments.

Chet

12. Nov 7, 2014

### Andy Resnick

Bleah- this is what I get for responding while I'm watching a football game (Go Browns!!!!!)

"The law of conservation of energy dE = dQ + dW means that changes to a path-independent property of the state of the system (dE) can be calculated by in terms of path-dependent transfers of energy that are not themselves state functions."

13. Nov 7, 2014

### Staff: Mentor

Much better.

I watched the game also. It was great. Total domination.

Chet

14. Nov 8, 2014

### Andy Resnick

Getting back to the OP, this whole discussion has focused on homogenous systems. Heterogeneous systems can have 'non-PV' forms of work: deformation of an interface and wetting (interfacial energy), addition and subtraction of particles (chemical potnetial), etc.