Processing math: 27%

8. Thermodynamics

8.1 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=(zx)ydx+(zy)xdy

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

(xy)z(yz)x(zx)y=1

Because dz is a total differential dz=0.

A homogeneous function of degree m obeys: εmF(x,y,z)=F(εx,εy,εz). For such a function Euler’s theorem applies:

mF(x,y,z)=xFx+yFy+zFz

8.2 Definitions

For an ideal gas it follows that: γp=1/T, κT=1/p and βV=1/V.

8.3 Thermal heat capacity

For an ideal gas holds: CmpCmV=R. Further, if the temperature is high enough to thermalize all internal rotational and vibrational degrees of freedom: CV=12sR. Hence Cp=12(s+2)R. From their ratio it now follows that γ=(2+s)/s. For a lower T one needs only to consider the thermalized degrees of freedom. For a Van der Waals gas: CmV=12sR+ap/RT2.

In general holds:

CpCV=T(pT)V(VT)p=T(VT)2p(pV)T0

Because (p/V)T is always <0, the following is always valid: CpCV. If the coefficient of expansion is 0, Cp=CV, and this is true also at T=0K.

8.4 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=ΔU+W, where Q is the total added heat, W the work done and ΔU the difference in the internal energy. In differential form this becomes: đQ=dU+đW, where đ means that the it is not a differential of a state function. For a quasi-static process: đW=pdV. So for a reversible process: đQ=dU+pdV.

For an open (flowing) system the first law is: Q=ΔH+Wi+ΔEkin+ΔEpot. One can extract an amount of work Wt from the system or add Wt=Wi to 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:

dSđQT

If the only processes occurring are reversible: dS=đQrev/T. So, the entropy difference after a reversible process is:

S2S1=21đQrevT

So, for a reversible cycle: đQrevT=0.

For an irreversible cycle: đQirrT<0.

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

lim

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.

8.5 State functions and Maxwell relations

The state functions and their differentials are:

Internal energy: U dU=TdS-pdV
Enthalpy: H=U+pVdH=TdS+Vdp
Free energy: F=U-TSdF=-SdT-pdV
Gibbs free energy: G=H-TSdG=-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}'

Helmholtzequations 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: đW_{\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}

8.6 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 Poisson’s equation holds. With \gamma=C_p/C_V one gets that pV^\gamma=constant. Also: TV^{\gamma-1}=constant and 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 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 keeps 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, applied in refrigerators.

The Carnot process

The system undergoes a reversible cycle with 2 isothemics and 2 adiabats:

  1. Isothermic expansion at T_1. The system absorbs a heat Q_1 from the reservoir.
  2. Adiabatic expansion with a temperature drop to T_2.
  3. Isothermic compression at T_2, removing Q_2 from the system.
  4. Adiabatic compression 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 machine 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
Stirling’s cycle consists of 2 isotherms and 2 isochorics. The efficiency in the ideal case is the same as for a Carnot cycle.

8.7 Maximal 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 minimal work needed to attain a certain state is: W_{\rm min}=-W_{\rm max}.

8.8 Phase transitions

Phase transitions are isothermic 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 transition heat of phase \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 exist also phase transitions with r_{\beta\alpha}=0. For those there will occur only be a discontinuity in the second derivatives of G_m. These second-order transitions appear at organization phenomena.

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

8.9 Thermodynamic potential

When the number of particles within a system changes this number becomes a third quantity of state. 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 Gibb's phase rule. .

8.10 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 ith component in liquid phase and y_i the fraction of the ith 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

8.11 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.

8.12 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 for a discrete system the Maxwell-Boltzmann distribution. 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)

8.13 Application to other systems

Thermodynamics can be applied to other systems than gases and liquids. To do this the term đW=pdV has to be replaced with the correct work term, like đW_{\rm rev}=-Fdl for the stretching of a wire, đW_{\rm rev}=-\gamma dA for the expansion of a soap bubble or đW_{\rm rev}=-BdM for a magnetic system.

A rotating, non-charged black hole has a temperature of 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.