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General Physics part 1 / PhysPraktikum / Woan G. The Cambridge Handbook of Physics Formulas (CUP, 2000)(ISBN 0521573491)(230s)_PRef_.pdf

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Chapter 5 Thermodynamics

5.1Introduction

The term thermodynamics is used here loosely and includes classical thermodynamics, statistical thermodynamics, thermal physics, and radiation processes. Notation in these subjects can be confusing and the conventions used here are those found in the majority of modern treatments. In particular:

• The internal energy of a system is deﬁned in terms of the heat supplied to the system plus

the work done on the system, that is, dU = d Q+ d W .

•The lowercase symbol p is used for pressure. Probability density functions are denoted by pr(x) and microstate probabilities by pi.

•With the exception of speciﬁc intensity, quantities are taken as speciﬁc if they refer to unit mass and are distinguished from the extensive equivalent by using lowercase. Hence speciﬁc

volume, v, equals V /m, where V is the volume of gas and m its mass. Also, the speciﬁc heat capacity of a gas at constant pressure is cp = Cp/m, where Cp is the heat capacity of mass m of gas. Molar values take a subscript “m” (e.g., Vm for molar volume) and remain in upper case.

• The component held constant during a partial di erentiation is shown after a vertical bar;

hence ∂V is the partial di erential of volume with respect to pressure, holding temperature constant∂p. T

The thermal properties of solids are dealt with more explicitly in the section on solid state physics (page 123). Note that in solid state literature speciﬁc heat capacity is often taken to mean heat capacity per unit volume.

106	Thermodynamics

5.2 Classical thermodynamics

Thermodynamic laws

Thermodynamic

lim(pV )

(5.1)

temperaturea

→

Kelvin

lim(pV )T

T /K = 273.16

p→0

(5.2)

temperature scale

lim(pV )tr

p→0

First lawb

(5.3)

dU = d

Q+ d

Qrev

Entropyc

dS =

≥ T

(5.4)

Tthermodynamic temperature

Vvolume of a ﬁxed mass of gas

pgas pressure

Kkelvin unit

tr	temperature of the triple point
	of water

dU change in internal energy

d W work done on system

d Q heat supplied to system

Sexperimental entropy

Ttemperature

rev reversible change

aAs determined with a gas thermometer. The idea of temperature is associated with the zeroth law of thermodynamics: If two systems are in thermal equilibrium with a third, they are also in thermal equilibrium with each other.

bThe d notation represents a di erential change in a quantity that is not a function of state of the system.

cAssociated with the second law of thermodynamics: No process is possible with the sole e ect of completely converting heat into work (Kelvin statement).

Thermodynamic worka

Hydrostatic				p		(hydrostatic) pressure
Hydrostatic	d W = −p dV		(5.5)	p		(hydrostatic) pressure
pressure	d W = −p dV		(5.5)	dV		volume change
					W work done on the system
				d	W work done on the system
Surface tension		W = γ dA	(5.6)
Surface tension	d	W = γ dA	(5.6)	γ		surface tension
				dA		change in area
				E		electric ﬁeld

Electric ﬁeld	d W = E · dp		(5.7)
Electric ﬁeld	d W = E · dp		(5.7)	dp		induced electric dipole moment
				B		magnetic ﬂux density
Magnetic ﬁeld	d W = B · dm		(5.8)	B		magnetic ﬂux density
Magnetic ﬁeld	d W = B · dm		(5.8)	dm		induced magnetic dipole moment
				∆φ		potential di erence
Electric current	d W = ∆φ dq		(5.9)	∆φ		potential di erence
Electric current	d W = ∆φ dq		(5.9)	dq		charge moved
				dq		charge moved

aThe sources of electric and magnetic ﬁelds are taken as being outside the thermodynamic system on which they are working.

5.2 Classical thermodynamics	107

Cycle e ciencies (thermodynamic)a

work extracted

Th − Tl

e ciency

Heat engine

η =

(5.10)

higher temperature

heat input

≤

lower temperature

Refrigerator

heat extracted

η =

work done

≤ Th

−

(5.11)

Heat pump

heat supplied

η =

work done

≤ Th

−

(5.12)

γ−1

compression ratio

work extracted

Otto cycle

γ ratio of heat capacities

η =

heat input

= 1 − V1

(5.13)

(assumed constant)

aThe equalities are for reversible cycles, such as Carnot cycles, operating between temperatures Th and Tl.

bIdealised reversible “petrol” (heat) engine.

Heat capacities

heat capacity, V constant

heat

Constant

d Q

∂U

∂S

(5.14)

volume

CV = dT V = ∂T V

= T ∂T V

volume

temperature

internal energy

p = T

entropy

Constant

∂H

∂S

heat capacity, p constant

(5.15)

pressure

Cp = dT

p = ∂T

∂T

p p

pressure

enthalpy

Cp − CV =

∂U

+ p

∂V

(5.16)

Di erence in

∂V

∂T

βp

isobaric expansivity

heat capacities

V T βp

(5.17)

κT

isothermal compressibility

κT

Ratio of heat

γ =

κT

(5.18)

ratio of heat capacities

capacities

κS

adiabatic compressibility

Thermodynamic coe cients

Isobaric

∂V

βp

isobaric expansivity

(5.19)

expansivitya

βp = V ∂T p

temperature

volume

Isothermal

κT

isothermal compressibility

1 ∂V

compressibility

κT = −

(5.20)

∂p

pressure

Adiabatic

1 ∂V

compressibility

κS = −

∂p

(5.21)

κS

adiabatic compressibility

Isothermal bulk

∂p

modulus

KT =

= −V

(5.22)

isothermal bulk modulus

κT

∂V

Adiabatic bulk

∂p

modulus

KS =

κS

= −V

∂V

(5.23)

adiabatic bulk modulus

aAlso called “cubic expansivity” or “volume expansivity.” The linear expansivity is αp = βp/3.

108 Thermodynamics

Expansion processes

Joule

∂T

T 2 ∂(p/T )

(5.24)

Joule coe cient

∂V

= − CV ∂T

pressure

temperature

expansion

∂p

(5.25)

internal energy

− CV

V −

∂T

heat capacity, V constant

µ =

∂T

∂(V /T )

(5.26)

Joule–Kelvin coe cient

Joule–Kelvin

∂p

∂T

volume

expansion

∂V

− V

enthalpy

(5.27)

heat capacity, p constant

∂T

aExpansion with no change in internal energy.

bExpansion with no change in enthalpy. Also known as a “Joule–Thomson expansion” or “throttling” process.

Thermodynamic potentialsa

			U	internal energy
	dU = T dS − pdV + µdN		T	temperature
Internal energy	dU = T dS − pdV + µdN	(5.28)	S	entropy
			µ	chemical potential
			N	number of particles
			H	enthalpy
	H = U + pV	(5.29)	H	enthalpy
Enthalpy	H = U + pV	(5.29)
Enthalpy	dH = T dS + V dp+ µdN	(5.30)	p	pressure
	dH = T dS + V dp+ µdN	(5.30)	V	volume
			V	volume

Helmholtz free	F = U − T S	(5.31)	F	Helmholtz free energy
energyb	dF = −S dT − pdV + µdN	(5.32)
	dF = −S dT − pdV + µdN	(5.32)
	G = U − T S + pV	(5.33)
Gibbs free energyc	= F + pV = H − T S	(5.34)	G	Gibbs free energy
	dG = −S dT + V dp+ µdN	(5.35)
Grand potential	Φ = F − µN	(5.36)	Φ	grand potential
	dΦ = −S dT − pdV − N dµ	(5.37)
Gibbs–Duhem	−S dT + V dp− N dµ = 0	(5.38)
relation	−S dT + V dp− N dµ = 0	(5.38)
			A	availability
	A = U − T0S + p0V	(5.39)	A	availability
Availability	A = U − T0S + p0V	(5.39)	T0	temperature of
	dA = (T − T0)dS − (p− p0)dV	(5.40)		surroundings
	dA = (T − T0)dS − (p− p0)dV	(5.40)	p0	pressure of surroundings

a dN=0 for a closed system.

bSometimes called the “work function.”

cSometimes called the “thermodynamic potential.”

5.2 Classical thermodynamics	109

Maxwell’s relations

∂T

∂p

∂2U

internal energy

Maxwell 1

∂V

− ∂S V

= ∂S∂V

(5.41)

volume

temperature

∂T

∂V

∂2H

enthalpy

Maxwell 2

= ∂p∂S

(5.42)

pressure

∂p S

∂S p

entropy

∂p

∂S

∂2F

Maxwell 3

(5.43)

Helmholtz free energy

∂T V

∂V T

∂T ∂V

∂V

∂S

∂ G

Maxwell 4

−

(5.44)

Gibbs free energy

∂T

∂p

∂p∂T

Gibbs–Helmholtz equations

U =

−

2 ∂(F/T )

(5.45)

Helmholtz free energy

internal energy

∂T

Gibbs free energy

2 ∂(F/V )

G =

−

(5.46)

enthalpy

∂V

temperature

2 ∂(G/T )

H =

−

(5.47)

pressure

∂T

volume

Phase transitions

Heat absorbed

L = T (S2 − S1)

(5.48)

(latent) heat absorbed (1 → 2)

temperature of phase change

entropy

S2 − S1

(5.49)

pressure

Clausius–Clapeyron

V2 − V1

volume

equation

(5.50)

1,2

phase states

T (V2 − V1)

Coexistence curveb

p(T )

exp

−L

(5.51)

molar gas constant

βp2 − βp1

(5.52)

βp

isobaric expansivity

Ehrenfest’s

κT 2 − κT 1

κT

isothermal compressibility

equationc

Cp2 − Cp1

(5.53)

heat capacity (p constant)

V T βp2 − βp1

number of phases in equilibrium

Gibbs’s phase rule

P + F = C + 2

(5.54)

number of degrees of freedom

number of components

aPhase boundary gradient for a ﬁrst-order transition. Equation (5.50) is sometimes called the “Clapeyron equation.” bFor V2 V1, e.g., if phase 1 is a liquid and phase 2 a vapour.

cFor a second-order phase transition.

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