# 5. Thermodynamic Processes

Although thermodynamics strictly speaking refers only to equilibria, by introducing the concept of work flow and heat flow, as discussed in chapter 1, we can discuss processes by which a system is moved from one state to another.

The concepts of heat and work are only meaningful because certain highly averaged variables are stable as a function of time. Energy changes related to such variables, like volume, are considered work. Microscopic variables, like the position of a single particle, are unstable and unpredictable as a function of time. If we treat the system classically, we would say such a variable has a Lyapunov coefficient $$L > 0$$. That is, its orbit diverges from prediction as $$\Delta X = \Delta X_0 \exp[-Lt]$$, where $$\Delta X_0$$ is the initial measurement error in the variable, and $$\Delta X$$ is the error at time $$t$$.

Here’s an example from meteorology: the Lyapunov coefficient for seasonal temperatures is 0; they have predictable averages (colder in the January, warmer in July in the northern hemisphere). Small weather patterns (e.g. motion of a cloud front) have Lyapunov coefficient of about (2 days)-1. This is not a question of better measurements, but a fundamental limitation: the error grows exponentially, so a multiplicative improvement in measurement buys you only a linear improvement in time. The situation is no better in quantum mechanics, where a particle initially started out in a position eigenstate $$\delta(x-x_0)$$ evolves to ever greater position uncertainty as a matter of Heisenberg’s principle. Thus both classical and quantum motions are inherently unpredictable, for different reasons; the corresponding energy flow is heat flow. But when one averages over enough degrees of freedom, the averaged variables may be well behaved; that energy flow is work flow.

Types of processes:

Definition: Quasistatic Processes

A quasistatic process lies on the surface $$S(U,x_i)$$

Note: this cannot be achieved in reality, but approximated by taking small steps whose endpoints lie on the fundamental surface.

Definition: Reversible Processes
A reversible process is a quasistatic process with $$S$$ = constant.

Note: according to postulate 2, upon change of constraints, any process must satisfy $$S_{final}>S_{initial}$$. The reverse of such a process would violate postulate 2 ($$S_{final}<S_{initial}$$), and real processes are therefore irreversible. A reversible process is the idealized quasistatic limit where $$S_{final} = Si_{nitial}$$. Fig. 5.1: Irreversible, quasistatic and reversible processes

Thermodynamics only makes statements about equilibrium states, when the fundamental equation is satisfied. However, by using quasistatic and reversible processes as idealized limits, we can derive inequalities satisfied by real processes.

As seen earlier $$dU = đQ + TdS$$ for small (quasistatic) heat transfers in absence of work. The best we can do in the presence of work is therefore that all of

$dU = Tds - pdV + \sum_i \mu_idn_i + \Gamma \,dA + H\,dM + ...$

goes into work except for the first term, which corresponds to the infinitesimal heat transfer. $$–PdV$$ would be simple bulk volume work (e.g. expansion of a gas), $$H dM$$ would be chemical work (e.g. electrochemical if $$n$$ refers to the mole number of an ion), $$\Gamma dA$$ would be surface tension work (e.g. blowing a soap bubble), $$H dM$$ would be magnetic work, etc.

The heat transfer cannot be reduced below $$TdS$$. However, part or all of the energy flow $$đW$$ can of course be converted to heat $$đQ_W$$. The entropy will rise further by $$T\,dS_W=đQ_W$$, and correspondingly less work is done:

$dU=TdS + đQ_W + đW_{\text{left over}}$

If all possible work is converted to heat, $$đW_{\text{left over}}=0$$ and heat flow is maximized. The following theorem tells us how much work at most we can extract from a system.

## Extracting Work from a Temperature Differential

Theorem: Reversible Work Theorem

The work that can be extracted form a system going from $$U_A(S_A) \rightarrow U_B(S_B) < U_A$$ is maximized by a reversible process, and is the same for all reversible processes $$U_A \rightarrow U_B$$.

Proof

The system can moves from state $$A$$ to state $$B$$ by two paths, one reversible and one irreversible. Energy and entropy are state functions, so $$U_A$$, $$U_B­$$, $$S_A$$ and $$S_B$$ are the same for both paths, since the two paths connect the same initial and final states. The change in energy is $$U_B – U_A = Q + W$$. Then

1. Reversible case:

$\Delta S=0 \Rightarrow \Delta S_{sys} =S_B-S_A <0$

$\Delta S_{hs} = -\Delta S_{sys}$

by definition of the reversible heat source/sink

$\Rightarrow W_r = U_B-U_A -Q = U_B -U_A -\Delta S_{hs}T_{hs} = U_B-U_A + \Delta S_{sys}T_{hs}$

1. Irreversible:

$\Delta S>0 \Rightarrow \Delta S_{sys} =S_B-S_A <0$

Since $$U$$ and $$S$$ are state functions, so $$S_A$$ and $$S_B$$ are the same for the system as before

$\Delta S_{hs} = \Delta S -\Delta S_{sys}$

$\Rightarrow W_i = U_B-U_A - (\Delta S_{sys}- \Delta S)T_{hs}$

$$\Rightarrow W_i < W_r$$ by an amount $$\Delta ST_{hs}$$.

Basically, the theorem says you have a choice: you can either get the amount of work in case (a), or you can choose to turn some of it into more heat in case (b); but you cannot do the reverse. Fig. 5.2: For convenience, we split the reservoir=bath into a heat source/sink “hs” and a work source/sink “ws”. Or imagine your open system coupled to two reservoirs: on one side is a heat-insulated piston, so only work can be transferred; on the other side is a rigid diathermal wall, so only heat can be transferred. Of course both of these baths can be one and the same, but it is sometimes useful for bookkeeping to split the bath into a heat and a work source.

Example 1: Heat Engine and Refrigerator Efficiencies

Engine efficiency

Engines run in cycles to take the heat difference between a system and its environment, and use some of the heat flow from hot system to environment to do work. As you can imagine, the second law sets a limit on that: the smaller the temperature difference between system and environment (the reversible heat source in figure 5.2), the less work can be transferred into the reversible work source (which acts as a work sink in that case). So you need a hot system (e.g. car engine) and a cold bath (e.g. ambient air) if you want to do work with an engine. Of course, if you do it irreversibly, the Reversible Work Theorem says you will get even less work.

$dS = 0 = dS_{hs} + dS_{sys} = \dfrac{đQ_{hs}}{T_{hs}} - \dfrac{đQ_{hs}}{T_{sys}}$

$đW = đQ_{sys} - đQ_{hs} = đQ_{sys} - đQ_{sys} \dfrac{T_{hs}}{T_{sys}} = đQ \dfrac{T_{sys}-T_{hs}}{T_{sys}} \le đQ_{sys}$

The engine efficiency goes to zero as $$T_{hs} \rightarrow T_{sys}$$

Refrigerator efficiency

Refrigerators also run in cycles, but here the goal is to take a work source, and while dumping it into a warm heat source (which acts as a sink for that work converted into heat), use some of that work to draw heat from a cold system into your warm heat source (acting as a sink for heat despite it being warmer than the cold system). Again, the second law puts some limits: the cooler (smaller) the temperature of the system you try to get to, the faster the amount of work grows. Reaching $$T=0$$ in your system would require an infinite amount of work to be turned into heat, so that’s not going to happen.

$dS = 0 = dS_{hs} + dS_{sys} = \dfrac{đQ_{hs}}{T_{hs}} - \dfrac{đQ_{hs}}{T_{sys}}$

$đQ_{rhs} = đW + đQ_{sys} \Rightarrow đQ_{sys} \dfrac{T_{rhs}}{T_{sys}} = đW + đQ_{sys}$

$\Rightarrow đQ_{rhs} = đW \left( \dfrac{T_{rhs}}{T_{sys}} \right)^{-1} = đW \dfrac{T_{rhs}}{T_{rhs} - T_{sys}}$

Refrigerator efficiency approaches infinity as $$T_{rhs} \rightarrow T_{sys}$$ because no work is required to transfer an infinitesimal amount of heat between two systems at the same temperature

You can read about this in detail in advanced classical dynamics texts such as the ones by Lichtenberg and Lieberman, or Gutzwiller, or Arnol’d, or Ozorio de Almeida.