1.1 A quick resumé of thermodynamics

• Thermodynamic state | State variables
  We can describe the state of a thermodymic system uniquely by specifying a small number of macroscopic observables (state variables) $X_{i}$:
  amount of substance (in mol) $n$; pressure $P$; volume $V$; temperature $T$; internal energy $E$; etc.
  
  Equations of state of the form Form $f(X_{1},X_{2},..)=0$ serve to reduce the number of independent state variables:
  $PV=nRT$; $E=n C_{V} T$ (ideal gas); $(P+an^{2}/V^{2})(V-bn)=nRT$ (real gas; Van der Waals); etc.
• First Law
  JOULE's experiment (heat of friction: stirring water with a spindle driven by a weight) proved that heat is a form of energy. Mechanical energy $(W)$ and heat $(Q)$ may be converted both ways. Following general custom we assign a positive sign to mechanical work performed by the system, while heat is counted positive when it is fed into the system. The law of energy conservation then reads, taking into account the internal energy $E$,
  

  $$dE=dQ-dW$$ (1.1)

  
  
  Frequently the mechanical energy occurs in the form of ''volume work'', i.e. work performed against the ambient pressure when the system expands: $dW= P dV$ (piston lifting weight).
  It should be noted that $E$ is a state variable, while $Q$ and $W$ may depend not only on the state of the system but also on the way it has reached that state; we may stress this by using the notation $dE=\bar{d}Q-\bar{d}W$.^1.1
  
  Internal energy of an ideal gas: The expansion experiment (also by JOULE: expand air from one bottle into an additional, evacuated bottle) demonstrates that $E_{idG}=C_{V} T$, meaning that the internal energy of an ideal gas depends only on the temperature $T$. (Do not confuse this with JOULE and THOMPSON'S throttling experiment, which relates to real gases.)
  
  Since $\Delta W=0$ we have $\Delta E=\Delta Q$; but as $T_{2}=T_{1}$ (empirical fact) there follows $\Delta Q \propto \Delta T=0$. Thus $\Delta E=0$. However, since the volumes of final and initial state differ, the internal energy $E$ cannot depend on $V$. In other words, for an ideal gas the formula $E=E(V,T)$ reads simply $E=E(T)$.
  
  We know even more about $E(T)$. Feeding energy to the gas in an isochoric way, i.e. with $V=const$, is possible only by heating: $\Delta E=\Delta Q = C_{V} \Delta T$. Therefore we find, using the empirical fact that $C_{V}=const$, that $E(T)=C_{V} T$.
  
  
• Thermal interaction | Thermal equilibrium | Temperature
  Bringing two bodies in contact such that they may exchange energy, we will in general observe a flow of a certain amount of energy in one direction. As soon as this flow has ebbed off, we have reached thermal equilibrium - by definition.
  A measurable quantity to characterize this equilibrium state is the temperature: $\left[ T_{1}=T_{2} \right]_{eq}$
  
  
• Second Law
  Not all transformations of energy that are in accordance with the First Law are really possible. The Second Law tells us which transformations are excluded. There are several logically equivalent formulations:

  Impossible are those processes in which nothing happens but a complete transformation of $\Delta Q$ into $\Delta W$. [Kelvin]

  Example: Expansion of a gas against a weighted piston; here we have a complete change of $\Delta Q$ into $\int P dV$, but the end state is different from the initial one (larger $V$).

  Impossible are those processes in which nothing happens but a transport of heat $\Delta Q$ from a colder to a warmer reservoir. [Clausius]

  Example: Heat pump; here we do transport $\Delta Q$, but in order to do so we have to feed in mechanical energy.
  
  
• Reversible processes
  If a process is such that the inversion of the arrow of time $t \longrightarrow -t$ will again refer to a possible process, this process is called reversible.
  
  It is a consequence of the Second Law that there are processes which are not reversible.
  
  Example 1: Reversible expansion
  
  
  
  
  
  Let an ideal gas expand isothermally - i.e. with appropriate input of heat - against a (slowly diminishing!) weight. The internal energy of the gas remains constant, $(E_{idG}=n C_{V} T)$, and we have
  

  $$\Delta Q = \Delta W = \int P(V) dV = nRT \int \frac{1}{V} dV = nRT \ln \frac{V_{2}}{V_{1}}$$ (1.2)

  
  
  This process may be reversed: by slowly increasing the weight load on the piston and removing any excess heat to a thermostatted heat bath we may recompress the gas isothermally, to reach the initial state again.
  
  Example 2: Irreversible expansion
  Let an ideal gas expand into vacuum (JOULE's expansion experiment). Result: $T_{2}=T_{1}$, therefore no change of internal energy. In this case no energy has been exerted and stored, but the volume has increased.
  Changing the direction of time does not produce a possible process: the shrinking of a gas to a smaller volume without input of mechanical energy does not occur.
  
  
• Entropy
  A mathematically powerful formulation of the Second Law relies on a quantity which is defined as follows:
  
  Let $1 \longrightarrow 2$ be a reversible path from state $1$ to $2$. The difference of entropies is then given by
  

  $$\Delta S \equiv S(2)-S(1) = \int_{1}^{2}\frac{dQ}{T}$$ (1.3)

  
  
  
  
  $S$ has the following important properties:
  

  $\circ$

  $S$ is a state variable, i.e. in any particular physical state of a system the entropy $S$ has a well-defined value, independent of what happened to the system before. This value, however, is defined only up to an additive constant; the Third Law removes this ambiguity by stating that $S(T \rightarrow 0) =0$.

  $\circ$

  $\Delta S > \int_{1}^{2} dQ / T$ in all irreversible processes.
  

  $\circ$

  When a thermally isolated system is in equilibrium the entropy is at its maximum.
  

  $\circ$

  Thus in a thermally isolated system (the universe being a notable special case) we have $dQ=0$ and therefore $\Delta S \geq 0$, meaning that the entropy will never decrease spontaneously.
  

  $\circ$

  As a consequence of the First Law $dE=T dS -P dV$ we have
  

  $$\left. \frac{dE}{dS}\right\vert _{V}=T$$ (1.4)

  
  

  But this means that we can describe thermal equilibrium in yet another way:
  The flow of energy between two bodies in contact will come to an end when
  

  $$\left. \frac{dE}{dS}\right\vert _{1}=\left. \frac{dE}{dS}\right\vert _{2}$$ (1.5)

  
  

  $\circ$

  If two systems are in thermal equilibrium then $S_{tot}=S_{1}+S_{2}$; entropy is additive (extensive).

  
  
  
• Entropy applied:
  By scrutinizing the entropy balance in a thermodynamic process we may find out whether that process is, according to the Second Law, a possible one. In particular, we can easily decide whether the change of state is reversible or not.
  
  Example 1: Reversible expansion
  In the former experiment the entropy of the gas increases by the amount
  

  $$\Delta S_{gas} = \frac{\Delta Q}{T} = \frac{\Delta W}{T} = nR \ln \frac{V_{2}}{V_{1}}$$ (1.6)

  
  
  At the same time the heat reservoir delivers the energy $\Delta Q$ to the gas; this decreases the reservoir's entropy by
  

  $$\Delta S_{res} = - \frac{\Delta Q}{T} = - nR \ln \frac{V_{2}}{V_{1}}$$ (1.7)

  
  
  Thus the entropy of the combined system remains the same, and the reverse process is possible as well. $\Longrightarrow$reversible.
  
  Example 2: Irreversible expansion:
  Let the initial and final states of the gas be the same as above. Since $S$ is a state variable we have as before
  

  $$\Delta S_{gas} = nR \ln \frac{V_{2}}{V_{1}}$$ (1.8)

  
  
  This time, however, no energy is fed in, which makes $\Delta S_{res} = 0$ and thus $\Delta S_{tot} > 0$. The reverse process would therefore give rise to a decrease in entropy - which is forbidden: $\Longrightarrow$irreversible.
  
  
• Macroscopic conditions and thermodynamic potentials
  The thermodynamic state - the ``macrostate'' - of a gas is uniquely determined by two independent state variables. Any of the two mechanical variables - $P$ or $V$ - may be combined with one of the two thermal variables - $T$ or $S$ - to describe the state. Of course, the natural choice will be that pair of variables that is controlled, or monitored, by the experimenter. For example, if a sample of gas is contained in a thermostatized ($T$) cylinder with a piston whose position ($V$) is directly observed, the description in terms of ($V,T$) is indicated. Then again, it may be the force on the piston, and not its position, which we measure or control; the natural mechanical variable is then the pressure $P$. And instead of holding the temperature constant we may put a thermal insulation around the system, thus conserving its thermal energy or, by $0=dQ=T dS$, its entropy. To describe this experimental situation we will of course choose the pair ($P,S$). Let us consider the possible choices:

  *

  $(S,V)$: The system's volume and heat content are controlled. This may be done by a thermally insulating the sample (while allowing for controlled addition of heat, $dQ=T dS$) and manipulating the position of the piston, i. e. the sample volume $V$. The internal energy is a (often unknown) function of these variables: $E=E(S,V)$.
  
  Now let the macrostate $(S,V)$ be changed reversibly, meaning that we introduce small changes $dS$, $dV$ of the state variables. According to the First Law, the change $dE$ of internal energy must then be the net sum of the mechanical and thermal contributions: $dE(S,V)= (\partial E/ \partial S)_{V} dS + ( \partial E/ \partial V)_{S} dV$. Comparing this to $dE=T dS -P dV$ we derive the thermodynamic (``Maxwell's'') relations $P=-(\partial E/ \partial V)_{S}$ and $T=(\partial E/ \partial S)_{V}$.
  
  Special case $dS=0$: For adiabatic expansion we see that the internal energy of the gas changes by $\Delta E = -\int P dV$ - just the amount of mechanical work done by the gas.
  
  The internal energy $E(S,V)$ is our first example of a Thermodynamic Potential. Its usefulness stems from the fact that the thermodynamic observables $P$ and $T$ may be represented as derivatives of $E(S,V)$. We will now consider three more of these potentials, each pertaining to another choice of state variables.
  
  

  *

  $(V,T)$: The system's volume and its temperature are controlled.
  The appropriate thermodynamic potential is then the Helmholtz Free Energy defined by $A(V,T) \equiv E-TS$. Using $dA = dE-T dS-S dT$ $=-P dV-S dT$ we may derive the relations $P=-(\partial A/ \partial V)_{T}$ and $S=-(\partial A / \partial T)_{V}$.
  
  Special case $dT=0$: Considering a process of isothermal expansion we find that now the free energy of the gas changes by $\Delta A = -\int P dV$ - again, the amount of mechanical work done by the gas.
  
  This is worth remembering: the mechanical work done by a substance (not just an ideal gas!) that is allowed to expand isothermally and reversibly from volume $V_{1}$ to $V_{2}$ is equal to the difference between the free energies of the final and initial states, respectively.
  
  

  *

  $(T,P)$: Heat bath and force-controlled piston.
  
  Gibbs Potential $G(P,T) \equiv A+PV$: $dG = dA+P dV+V dP$ $=-S dT + V dP$ $\Longrightarrow$ $S=-(\partial G/ \partial T)_{P}$ and $V=(\partial G/ \partial P)_{T}$.
  
  Special case $dT=0$: The basic experiment described earlier, isothermal expansion against a controlled force (= load on the piston) is best discussed in terms of $G$: the change $\Delta G$ equals simply $\int_{1}^{2}V(P;T=const) dP$, which may be either measured or calculated using an appropriate equation of state $V(P,T)$.
  
  

  *

  $(P,S)$: As in the first case, the heat content is controlled; but now the force acting on the piston (and not the latter's position) is the controllable mechanical parameter.
  
  Define the Enthalpy $H\equiv E+PV$: $dH = dE + P dV+V dP$ $=T dS + V dP$ $\Longrightarrow$ $T=(\partial H/ \partial S)_{P}$ and $V=(\partial H/ \partial P)_{S}$.
  Special case $dS=0$: Again, the system may be allowed to expand, but without replacement of the exported mechanical energy by the import of heat (adiabatic expansion). The enthalpy difference is then just $\Delta H = \int_{1}^{2}V(P;S=const) dP$, which may be measured. (In rare cases, such as the ideal gas, there may be an equation of state for $V(P;S=const)$, which allows us to calculate the integral.)

  
  
  The multitude of thermodynamic potentials may seem bewildering; it should not. As mentioned before, they were actually introduced for convenience of description by the ``natural'' independent variables suggested by the experiment - or by the machine design: we should not forget that the early thermodynamicists were out to construct thermomechanical machines.
  
  There is a extremely useful and simple mnemonic scheme, due to Maxwell, that allows us to reproduce the basic thermodynamic relations without recourse to longish derivations: Maxwell's square. Here it is:
  
  

  Thermodynamics in a square: Maxwell's mnemonic device

• First of all, it tells us that $A=A(V,T)$, $E=E(S,V)$, etc.
• Second, the relations between the various thermodynamic potentials are barely hidden - discover the rules yourself: $A=E-TS$; $H=E+PV$; $G=A+PV$.
• Next, the eight particularly important Maxwell relations for the state variables $V,T,P,S$ may be extracted immediately:
  

  $$V = \left. \frac{\partial H}{\partial P} \right\vert _{S} = \left. \frac{\partial G}{\partial P} \right\vert _{T}$$ (1.9)

  $$T = \left. \frac{\partial H}{\partial S} \right\vert _{P} = \left. \frac{\partial E}{\partial S} \right\vert _{V}$$ (1.10)

  $$P = - \left. \frac{\partial A}{\partial V} \right\vert _{T} = - \left. \frac{\partial E}{\partial V} \right\vert _{S}$$ (1.11)

  $$S = - \left. \frac{\partial A}{\partial T} \right\vert _{V} = - \left. \frac{\partial G}{\partial T} \right\vert _{P}$$ (1.12)

  
  

• Using these relations we may easily write down the description of infinitesimal state changes in the various representations given above. To repeat,
  

  $$dE(V,S) = -P dV + T dS$$ (1.13)

  $$dA(V,T) = -P dV - S dT$$ (1.14)

  $$dG(P,T) = V dP - S dT$$ (1.15)

  $$dH(P,S) = V dP + T dS$$ (1.16)

  
  

Chemical Potential: So far we have assumed that the mass (or particle) content of the considered system was constant. If we allow the number $n$ of moles to vary, another important parameter enters the scene. Generalizing the internal energy change to allow for a substance loss or increase, we write

$$dE(S,V,n)=-PdV+TdS+\mu dn$$ (1.17)

In other words, the chemical potential $\mu$ quantifies the energy increase of a system upon addition of substance, keeping entropy and volume constant: $\mu = \partial E/\partial n\vert _{S,V}$. Since $S$ is itself dependent on $n$ another, equally valid relation is usually preferred:

$$\mu = \left. \frac{\partial G}{\partial n}\right\vert _{T,P}$$ (1.18)

The name given to this state variable stems from the fact that substances tend to undergo a chemical reaction whenever the sum of their individual chemical potentials exceeds that of the reaction product. However, the importance of $\mu$ is not restricted to chemistry. We will encounter it several times on our way through statistical/thermal physics.



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