8: Thermodynamics

Mathematical introduction

If there exists a relation $f(x,y,z)=0$ between 3 variables, one can write: $x=x(y,z)$, $y=y(x,z)$ and $z=z(x,y)$. The total differential $dz$ of $z$ is than given by:

$$dz=\left(\frac{\partial z}{\partial x}\right)_{y}dx+\left(\frac{\partial z}{\partial y}\right)_{x}dy$$

By writing this also for $dx$ and $dy$ it can be shown that

$$\left(\frac{\partial x}{\partial y}\right)_{z}\cdot\left(\frac{\partial y}{\partial z}\right)_{x}\cdot\left(\frac{\partial z}{\partial x}\right)_{y}=-1$$

Because $dz$ is a total differential $\oint dz=0$.

A homogeneous function of degree $m$ obeys: $\varepsilon^m F(x,y,z)=F(\varepsilon x,\varepsilon y,\varepsilon z)$. For such a function Euler’s theorem applies:

$$mF(x,y,z)=x\frac{\partial F}{\partial x}+y\frac{\partial F}{\partial y}+z\frac{\partial F}{\partial z}$$

Definitions

• The isochoric pressure coefficient: $\displaystyle \beta_V=\frac{1}{p}\left(\frac{\partial p}{\partial T}\right)_{V}$
• The isothermal compressibility: $\displaystyle \kappa_T=-\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_{T}$
• The isobaric volume coefficient: $\displaystyle \gamma_p=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{p}$
• The adiabatic compressibility: $\displaystyle \kappa_S=-\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_{S}$

For an ideal gas it follows that: $\gamma_p=1/T$, $\kappa_T=1/p$ and $\beta_V=-1/V$.

Thermal heat capacity

• The specific heat at constant $X$ is: $\displaystyle C_X=T\left(\frac{\partial S}{\partial T}\right)_{X}$
• The specific heat at constant pressure: $\displaystyle C_p=\left(\frac{\partial H}{\partial T}\right)_{p}$
• The specific heat at constant volume: $\displaystyle C_V=\left(\frac{\partial U}{\partial T}\right)_{V}$

For an ideal gas : $C_{mp}-C_{mV}=R$. Further, if the temperature is high enough to thermalize all internal rotational and vibrational degrees of freedom: $C_V= \frac{1}{2} sR$. Hence $C_p= \frac{1}{2} (s+2)R$. From their ratio it now follows that $\gamma=(2+s)/s$. For a lower $T$ one needs only to consider the thermalized degrees of freedom. For a Van der Waals gas: $C_{mV}= \frac{1}{2} sR+ap/RT^2$.

In general:

$$C_p-C_V=T\left(\frac{\partial p}{\partial T}\right)_{V}\cdot\left(\frac{\partial V}{\partial T}\right)_{p}=-T\left(\frac{\partial V}{\partial T}\right)_{p}^2\left(\frac{\partial p}{\partial V}\right)_{T}\geq0$$

Because $(\partial p/\partial V)_T$ is always $<0$, the following is always valid: $C_p\geq C_V$. If the coefficient of expansion is 0, $C_p=C_V$, and this is true also at $T=0$K.

The laws of thermodynamics

The zeroth law states that heat flows from higher to lower temperatures. The first law is the conservation of energy. For a closed system: $Q=\Delta U+W$, where $Q$ is the total added heat, $W$ the work done and $\Delta U$ the difference in the internal energy. Notice that the work is taken as the work done by the system on the surroundings. In differential form this becomes: $d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt} Q=dU+d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}W$, where $d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}$ means that the it is not a differential of a state function. For a quasi-static process: $d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}W=pdV$. So for a reversible process: $d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}Q=dU+pdV$.

For an open (flowing) system the first law is: $Q=\Delta H+W_{\rm i}+\Delta E_{\rm kin}+\Delta E_{\rm pot}$. One can extract an amount of work $W_{\rm t}$ from the system or add $W_{\rm t}=-W_{\rm i}$ to the system. In chemistry, one uses the opposite convention, that positive work is work done on the system.

The second law states: for a closed system there exists an additive quantity $S$, called the entropy, the differential of which has the following property:

$$dS\geq\frac{d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}Q}{T}$$

If the only processes occurring are reversible: $dS=d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}Q_{\rm rev}/T$. So, the entropy difference after a reversible process is:

$$S_2-S_1=\int\limits_1^2 \frac{d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}Q_{\rm rev}}{T}$$

For a reversible cycle: $\displaystyle\oint\frac{d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}Q_{\rm rev}}{T}=0$.

For an irreversible cycle: $\displaystyle\oint\frac{d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}Q_{\rm irr}}{T}<0$.

The third law of thermodynamics is (Nernst's law):

$$\lim_{T\rightarrow0}\left(\frac{\partial S}{\partial X}\right)_{T}=0$$

From this it can be concluded that the thermal heat capacity $\rightarrow0$ if $T\rightarrow0$, so absolute zero temperature cannot be reached by cooling through a finite number of steps.

State functions and Maxwell relations

The state functions and their differentials are:

Internal energy:	$U$	$dU=TdS-pdV$
Enthalpy:	$H=U+pV$	$dH=TdS+Vdp$
Free energy:	$F=U-TS$	$dF=-SdT-pdV$
Gibbs free energy:	$G=H-TS$	$dG=-SdT+Vdp$

From this one can derive Maxwell’s relations:

$$\left(\frac{\partial T}{\partial V}\right)_{S}=-\left(\frac{\partial p}{\partial S}\right)_{V}~,~~\left(\frac{\partial T}{\partial p}\right)_{S}=\left(\frac{\partial V}{\partial S}\right)_{p}~,~~ \left(\frac{\partial p}{\partial T}\right)_{V}=\left(\frac{\partial S}{\partial V}\right)_{T}~,~~\left(\frac{\partial V}{\partial T}\right)_{p}=-\left(\frac{\partial S}{\partial p}\right)_{T}$$

From the total differential and the definitions of $C_V$ and $C_p$ it can be derived that:

$$TdS=C_VdT+T\left(\frac{\partial p}{\partial T}\right)_{V}dV~~\mbox{and}~~TdS=C_pdT-T\left(\frac{\partial V}{\partial T}\right)_{p}dp$$

For an ideal gas:

$$S_m=C_V\ln\left(\frac{T}{T_0}\right)+R\ln\left(\frac{V}{V_0}\right)+S_{m0}~~\mbox{and}~~ S_m=C_p\ln\left(\frac{T}{T_0}\right)-R\ln\left(\frac{p}{p_0}\right)+S_{m0}'$$

Helmholtz’ equations are:

$$\left(\frac{\partial U}{\partial V}\right)_{T}=T\left(\frac{\partial p}{\partial T}\right)_{V}-p~~,~~\left(\frac{\partial H}{\partial p}\right)_{T}=V-T\left(\frac{\partial V}{\partial T}\right)_{p}$$

For a macroscopic surface: $d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}W_{\rm rev}=-\gamma dA$, with $\gamma$ the surface tension. From this follows: 

$$\gamma=\left(\frac{\partial U}{\partial A}\right)_{S}=\left(\frac{\partial F}{\partial A}\right)_{T}$$

Processes

The efficiency $\eta$ of a process is given by: $\displaystyle\eta=\frac{\mbox{Work done}}{\mbox{Heat added}}$

The cold factor $\xi$ of a cooling down process is given by: $\displaystyle\xi=\frac{\mbox{Cold delivered}}{\mbox{Work added}}$

Reversible adiabatic processes

For adiabatic processes: $W=U_1-U_2$. For reversible adiabatic processes using Poisson’s equation with $\gamma=C_p/C_V$ one gets that $pV^\gamma=$constant. Also: $TV^{\gamma-1}=$constant. Also $T^\gamma p^{1-\gamma}=$constant. Adiabats are steeper on a $p$-$V$ diagram than isotherms because $\gamma>1$.

Isobaric processes

Here: $H_2-H_1=\int_1^2 C_pdT$. For a reversible isobaric process: $H_2-H_1=Q_{\rm rev}$.

Throttle processes

This is also called the Joule-Kelvin effect and is the result of an adiabatic expansion of a gas through a porous material or a small opening. Here $H$ is a conserved quantity, and $dS>0$. In general this is accompanied with a change in temperature. The quantity which is important here is the throttle coefficient:

$$\alpha_H=\left(\frac{\partial T}{\partial p}\right)_{H}=\frac{1}{C_p}\left[T\left(\frac{\partial V}{\partial T}\right)_{p}-V\right]$$

The inversion temperature is the temperature where an adiabatically expanding gas maintains the same temperature. If $T>T_{\rm i}$ the gas heats up, if $T<T_{\rm i}$ the gas cools down. $T_{\rm i}=2T_{\rm B}$, with for $T_{\rm B}$: $[\partial(pV)/\partial p]_T=0$. The throttle process is, for example, used in refrigerators.

The Carnot Cycle

The system undergoes a reversible cycle consisting of two isotherms and two adiabats

1. Isothermal expansion at $T_{1}$. The system absorbs an amount of heat $Q_{1}$ from the reservoir.
2. Adiabatic expansion with a temperature drop to $T_{2}$
3. Isothermal compression at $T_{2}$ removing an amount of heat $Q_{2}$ from the system.
4. Adiabatic compression with the temperature increasing to $T_{1}$

The efficiency for a Carnot cycle is:

$$\eta=1-\frac{|Q_2|}{|Q_1|}=1-\frac{T_2}{T_1}:=\eta_{\rm C}$$

The Carnot efficiency $\eta_{\rm C}$ is the maximal efficiency at which a heat engine can operate. If the process is applied in reverse order and the system performs a work $-W$ the cold factor is given by:

$$\xi=\frac{|Q_2|}{W}=\frac{|Q_2|}{|Q_1|-|Q_2|}=\frac{T_2}{T_1-T_2}$$

The Stirling cycle

The Stirling cycle consists of 2 isotherms and 2 isochoric processes. The efficiency in the ideal case is the same as for a Carnot cycle.

Maximum work

Consider a system that changes from state 1 into state 2, with the temperature and pressure of the surroundings given by $T_0$ and $p_0$. The maximum work which can be obtained from this change is, when all processes are reversible:

1. Closed system: $W_{\rm max}=(U_1-U_2)-T_0(S_1-S_2)+p_0(V_1-V_2)$.
2. Open system: $W_{\rm max}=(H_1-H_2)-T_0(S_1-S_2)-\Delta E_{\rm kin}-\Delta E_{\rm pot}$.

The minimum work needed to attain a certain state is: $W_{\rm min}=-W_{\rm max}$.

Phase transitions

Phase transitions are isothermal and isobaric, so $dG=0$. When the phases are indicated by $\alpha$, $\beta$ and $\gamma$: $G_m^\alpha=G_m^\beta$ and

$$\Delta S_m=S_m^\alpha - S_m^\beta=\frac{r_{\beta\alpha}}{T_0}$$

where $r_{\beta\alpha}$ is the heat of transition from $\beta$ to phase $\alpha$ and $T_0$ is the transition temperature. The following holds: $r_{\beta\alpha}=r_{\alpha\beta}$ and $r_{\beta\alpha}=r_{\gamma\alpha}-r_{\gamma\beta}$. Further

$$S_m=\left(\frac{\partial G_m}{\partial T}\right)_{p}$$

so $G$ has a kink in the transition point and the derivative is discontinuous. In a two phase system Clapeyron’s equation is valid:

$$\frac{dp}{dT}=\frac{S_m^\alpha-S_m^\beta}{V_m^\alpha-V_m^\beta}= \frac{r_{\beta\alpha}}{(V_m^\alpha-V_m^\beta)T}$$

For an ideal gas one finds for the vapor line at some distance from the critical point:

$$p=p_0{\rm e}^{-r_{\beta\alpha/RT}}$$

There also exist also phase transitions with $r_{\beta\alpha}=0$. For those there will only be a discontinuity in the second derivatives of $G_m$. These second-order transitions appear as organization phenomena.

A phase-change of the 3rd order, with e.g. $[\partial^3 G_m/\partial T^3]_p$ non continuous arises e.g. when ferromagnetic iron changes to the paramagnetic state.

Thermodynamic potential

When the number of particles within a system changes this number becomes a third state function. Because addition of matter usually takes place at constant $p$ and $T$, $G$ is the relevant quantity. If a system has many components this becomes:

$$dG=-SdT+Vdp+\sum_i\mu_idn_i$$ where $\displaystyle\mu=\left(\frac{\partial G}{\partial n_i}\right)_{p,T,n_j}$

is called the thermodynamic potential. This is a partial quantity. For $V$:

$$V=\sum_{i=1}^c n_i\left(\frac{\partial V}{\partial n_i}\right)_{n_j,p,T}:=\sum_{i=1}^c n_i V_i$$

where $V_i$ is the partial volume of component $i$. The following holds:

$$\begin{aligned} V_m&=&\sum_i x_i V_i\\ 0&=&\sum_i x_i dV_i\end{aligned}$$

where $x_i=n_i/n$ is the molar fraction of component $i$. The molar volume of a mixture of two components can be a concave line in a $V$-$x_2$ diagram: the mixing leads to a contraction of the volume

The thermodynamic potentials are not independent in a multiple-phase system. It can be derived that $\sum\limits_i n_i d\mu_i=-SdT+Vdp$, this gives at constant $p$ and $T$: $\sum\limits_i x_i d\mu_i=0$ (Gibbs-Duhmen).

Each component has as many $\mu$’s as there are phases. The number of free parameters in a system with $c$ components and $p$ different phases is given by $f=c+2-p$ which is called the Gibbs phase rule. 

Ideal mixtures

For a mixture of $n$ components (the index $^0$ is the value for the pure component):

$$U_{\rm mixture}=\sum_i n_i U^0_i~~,~~H_{\rm mixture}=\sum_i n_i H^0_i~~,~~ S_{\rm mixture}=n\sum_i x_i S^0_i+\Delta S_{\rm mix}$$

where for ideal gases: $\Delta S_{\rm mix}=-nR\sum\limits_i x_i\ln(x_i)$.

For the thermodynamic potentials: $\mu_i=\mu_i^0+RT\ln(x_i)<\mu_i^0$. A mixture of two liquids is rarely ideal: this is usually only the case for chemically related components or isotopes. In spite of this Raoult’s law holds for the vapour pressure for many binary mixtures: $p_i=x_ip^0_i=y_ip$. Here $x_i$ is the fraction of the $i$th component in liquid phase and $y_i$ the fraction of the $i$th component in gas phase.

A solution for one component in a second gives rise to an increase in the boiling point $\Delta T_{\rm k}$ and a decrease of the freezing point $\Delta T_{\rm s}$. For $x_2\ll1$:

$$\Delta T_{\rm k}=\frac{RT_{\rm k}^2}{r_{\beta\alpha}}x_2~~,~~ \Delta T_{\rm s}=-\frac{RT_{\rm s}^2}{r_{\gamma\beta}}x_2$$

with $r_{\beta\alpha}$ the heat of evaporation and $r_{\gamma\beta}<0$ the melting heat. For the osmotic pressure $\Pi$ of a solution: $\Pi V_{m1}^0=x_2RT$.

These are called collegative properties

Conditions for equilibrium

When a system evolves towards equilibrium the only changes that are possible are those for which: $(dS)_{U,V}\geq0$ or $(dU)_{S,V}\leq0$ or $(dH)_{S,p}\leq0$ or $(dF)_{T,V}\leq0$ or $(dG)_{T,p}\leq0$. In equilibrium for each component: $\mu_i^\alpha=\mu_i^\beta=\mu_i^\gamma$.

Statistical basis for thermodynamics

The number of possibilities $P$ to distribute $N$ particles on $n$ possible energy levels, each with a $g$-fold degeneracy is called the thermodynamic probability and is given by:

$$P=N!\prod_i\frac{g_i^{n_i}}{n_i!}$$

The most probable distribution, that with the maximum value for $P$, is the equilibrium state. When Stirling’s equation, $\ln(n!)\approx n\ln(n)-n$ is used, one finds the Maxwell-Boltzmann distribution for a discrete system . The occupation numbers in equilibrium are then given by:

$$n_i=\frac{N}{Z}g_i\exp\left(-\frac{W_i}{kT}\right)$$

The state sum $Z$ is a normalization constant, given by: $Z=\sum\limits_ig_i\exp(-W_i/kT)$. For an ideal gas:

$$Z=\frac{V(2\pi mkT)^{3/2}}{h^3}$$

The entropy can then be defined as:  $S=k\;ln\left ( P \right )$. For a system in thermodynamic equilibrium this becomes:

$$S=\frac{U}{T}+kN\ln\left(\frac{Z}{N}\right)+kN\approx\frac{U}{T}+k\ln\left(\frac{Z^N}{N!}\right)$$

For an ideal gas, with $U=\frac{3}{2}kT$ then: $\displaystyle S=kN+kN\ln\left(\frac{V(2\pi mkT)^{3/2}}{Nh^3}\right)$

Application to other systems

Thermodynamics can be applied to other systems than gases and liquids. To do this the term $d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}W=pdV$ has to be replaced with the correct work term, for example $d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}W_{\rm rev}=-Fdl$ for the stretching of a wire, $d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}W_{\rm rev}=-\gamma dA$ for the expansion of a soap bubble or $d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}W_{\rm rev}=-BdM$ for a magnetic system.

A rotating, non-charged black hole has a temperature $T=\hbar c/8\pi km$. It has an entropy $S=Akc^3/4\hbar\kappa$ with $A$ the area of its event horizon. For a Schwarzschild black hole $A$ is given by $A=16\pi m^2$. Hawkings area theorem states that $dA/dt\geq0$.

Hence, the lifetime of a black hole $\sim m^3$.