THERMODINAMICS

Tóth Mónika2014

monika.a.toth@aok.pte.hu

Temperature

Temperature: is related to the average

energy of the motion of the particles of an

object or system.

SI unit of temperature: Kelvin (K) 0 oC=273,15 K

Different temperature scales. Thermometer with

Kelvin scale.

• Steel railroad tracks are laid when the T 0 o C. At standard section of rail is then 12 m long. What gap should be left between rail section so that is no compression when the T gets as high as 42 o C? (asteel = 11*10-6 1/ o C)

1. example

Gas state of the matter

1. The particles of the gases move randomly.

2. They can be highly compressed

3. The intermolecular forces are negligible. There isn’t internal friction between the particles.

4. The density and viscosity of gases much lower than solids and fluids

5. Most of the gases under standard condition behave as an ideal gas (almost).

They can be highly compressed!

Ideal gas

1.) The particles of the gases

move randomly.

2.) They collide with each other

and the wall of the container

completely ellastically (there is no

momentum and energy loss during

the collision).

3.) The intermolecular forces are

negligible.

4. Most of the gases under standard

condition behave as an ideal gas

(almost).

Avogadro’s law

Equal volumes of ideal or perfect gases, at the same temperature and pressure, contain the same number of

particles, or molecules.

Amedeo Avogadro

(1776 – 1856)

Combined and ideal gas law

Kmol

JR

3143,8

cVp

cT

V

cT

pGuy-Lussac II.:

Boyle-Marriote:

Guy-Lussac I.: cT

Vp

Tn

VpR

The ideal gas law can be derived from the combined gas law and

Avogadro’s law !

Universal gas constant (R) gives

the amount of energy required to

increase the temperature of

1 mol gas by one Kelvin.

Ideal gas law.

Combined gas law.

TRnVp

TkNVp k=1,381*10-23 J/K

GAS LOWS I. (EQUATION OF STATE)IZOTHERMAL PROCESS

1 1 2 2

constant

Boyle's law

constant

1constant

1( : )

T

pV

pV p V

pV

hyperbola yx

GAS LOW II.ISOBARIC PROCESS

1 2

1 2

constant

Gay - Lussac's I. law

constant

p

V

T

V V

T T

GAS LOWS III.ISOCHORIC PROCESS

1 2

1 2

constant

Gay - Lussac's II. law

constant

V

p

T

p p

T T

2. example

• We have a cylinder filled with gas with the volume of1 m3 and the pressure of the gas is the normalatmospheric pressure.

What will be the pressure of the gas after pushing inthe piston into the cylinder and so reducing thevolume to 0,3 m3?

(The temperature does not change.)

3. example

• We have a container closed with a piston. The container is filled with gas which tempearute is 20°C and it’s volume is 80 cm3.

• What will be the volume of the gas, after heating up it up to 60°C. (the piston can freely move, that is the pressure is constant)

4. example

Nitrogen gas is filled in a container. The volum of it 20 dm3. The temperature of gas is 289K Pa and thepressure is 3*105

Pa.

A, How many moles are in the container?

B, What is the mass of the gas?

(Mnitrogen=28*10-3kg/mol )

Thermodynamics is a Greek word which means flow of heat in physical and chemical reactions

Thermodynamics

THERMODYNAMIC SYSTEM

System: the material in the portion of space to be analyzedSurroundings/Environment: everything outside the systemBoundary: A separator, real or imaginary, between system and surroundings

System

Surroundings

Boundary

The thermodynamic state of a system is defined by specifying a set of measurable properties sufficient so that all remaining properties are determined.

THE PROPERTIES OF THE THERMODYNAMIC SYSTEM

macroscopic variables:

pressure (p) – momentum transferred to walls by molecular impacts

temperature (T) – molecular speeds (gas) or amplitudes of atomic vibrations (solids)

volume (V)

TERMODYNAMIC SYSTEMExchanges of work, heat, or matter between the system and the surroundings take place across this boundary.

Mass Energy

Mass

Energy

Mass (-)

Energy (-)

System

Thermodynamics systems

Open

Closed

Isolated

Mass and energy exchange!

Only energy exchange!

Neither mass nor energy exchange!

EXTENSIVE AND INTENSIVE QUANTITIES

The macroscopic quantities only have a well defiend values that can be determined at each certain state of the thermodynamic system (however the system is composed of sufficient number of microparticles).

Dividing the system into sub-system can be distinguished..

• Extensive quantities : value proportional to amount in system: m, V, E, Q (electric charge), N (particle count)

• Intensive quantities: value independent of the amount of material: p, T

HEAT: Q (Joule)• Energy transfer between the thermodynamic system and the

enviroment,

followed by heat production or phase transition

• Heat exchange:

- conduction: the heat flows through the particles of the body itself, through molecular vibration.

- convection: heat is transferred through the flow of a liquid or a gas.

- radiation: heat is transferred without heating the medium

• Heat is not a property of a system, but instead is always associated with a process

How can we calculate the amount of heat taken up by a system? (Heat capacity,

specific heat)Heat capacity (C): is the measure of heat energy required to increase the temperature of a system by 1 kelvin.

Unit: J/K

Specific heat (c): is the measure of heat energy required to increase the temperature of 1 kg system by 1 Kelvin.

Unit: J/kg*K

The specific heat measured under isobar conditions (cp) is always higher than the specific heat under isometric conditions (cv)!

vp cc

Latent heat

Latent heat: the amount of heat which is absorbed (or realesed from) by the system during the phase transion.

Specific latent heat (L): the amount of heat which is absorbed (or realesed) by 1kg of system during the phase transion.

TCQ

K

JC

T

QC

With temperature chamge

Heat capacity

LmQ

kg

JL

m

QL ][

Without temperature change

Latent heat

Phase transitions of matter

Melting Evaporation

Freezing Condensation

Solid Fluid Gas

Solid: the position of atoms and molecules are fixed, only vibrational motion, low degree of freedom, highly ordered state of the matter.

Fluid: the position of atoms and molecules are not fixed, translational, rotational, vibrational motion, higher degree of freedom.

Gas: the position of atoms and molecules are not fixed, highest degree of freedom, most disordered state of matter.

Phase diagramm of water

Water at its triple point

(0,01 oC, 0,006 atm).

• How much heat is needed to melt a metal ball? The mass of it is 6g.

• Tmelting= 300 o C

• C= 0,84 J/g o C

• L= 63 J/g

5. example

Expansion work

INTERNAL ENERGY

U: Joules (J or kJ), calorie or kcal also use

1 cal = 4.184 J 1 kcal = 4.184 kJ

In thermodynamics, the internal energy (U) is the total energy contained by a thermodynamic system.

U= Eel+Evibr+Erot+Ekin+Eother

The internal energy is a state function of a system

It is an extensive quantity

THE INTERNAL ENERGY IS A STATE FUNCTION

• State function: its value depends only on the current state of the system and not on the path taken or process undergone to arrive at this state.

• Other state functions: enthalpy (H), free energy (F), free enthalpy (G), entropy (S)

Internal energy of the ideal gas• Thermodynamics often uses the concept of the ideal gas as a

working system

• ideal gas is a gas of particles considered as point objects

• Monoatomic particles do not rotate or vibrate

• internal energy changes in an ideal gas can be described by changes in its kinetic energy

• the internal energy of the perfect gas depends on its pressure, volume and temperature

• the internal energy is proportional to mass of the gas (number of moles) N and to its temperature T 3

2U N k T

• How much heat is required to vaporize1g of ice which is at -10 o C to create120 o C steam at normal pressure?

• Cice= 2,1 J/g o C; Cwater= 4,2 J/g o C; Csteam= 2 J/g o C

• Lice= 334 J/g; Lvap.= 2260 J/g

For an extra point

LAWS OF THERMODYNAMICS

ZERO LAW OF THERMODYNAMICS

• If two systems (A and B) are independently in equilibrium with a third one (C), then they are in equilibrium with each other as well.

• Between different points of a system in equilibrium, the intensive variables are equal (there are no thermodynamic currents).

∆U = Q + W (W = − p∆V)

• Law of conservation of energy, because energy can neither be created nor destroyed althogh it may be converted from one to other

• The total energy of the system remains constant.

• The change in the internal energy of the system is the sum of the supplied heat (Q) and the work (W) done on the system.

FIRST LAWS OF THERMODYNAMICS:

APPLICATIONS OF I. LAWS OF THERMODYNAMICS

1.) The gas expands, so it does work on the surroundings (volumetric work)

How does the internal energy of an ideal gas change in an isobaric process

VpW

WQU

2.) The temperature of the gas increases, so the internal energy of that increases as well

Q

W

TmcU p

Thermal efficiency: the ratio of the work done by the system and the heat taken up by the sytem.

0W

QU

1.) There is no change in the volume of the gas, so there is no volumetric work.

2.) The heat energy increases the internal energy

TcmQ v

How does the internal energy of an ideal gas change in an isochor process

How does the internal energy of an ideal gas change in an isotherm process

0 WQU

2

1

p

plnTRW

1.) The gas expands, so it does volumetric work on the surroundings.

Temperature remains constant, sothe internal energy doesn’t change!

1.) Heat energy is not given to thegas.

2.) The expansion of the gas decreases of the internal energy of the gas.

0Q

How does the internal energy of an ideal gas change in an adiabatic process

WU

ADIABATIC pocess: The cylinder with a gas(system) is in nonconducting material. No heat (Q) is allowed to enter or to leave system. An adiabaticprocess can be either reversible or irreversible.

EXOTERMAL process: a process releases heatinto the surroundings –Q

Endothermal process: a process absorbs heat(vaporization, because heat must be supplied to drive molecules of a liquid apart from one other) +Q

1. example

• We compress 0,1 mol air in a cylindervery slowly. The temperature does notchange, it is 20 o C. The volumedecreases to the half. How much work isdone by the gas?

2. example• The volume of 1 kg mass of water is 10-3 m3.

At this temperature is vaporized the water, the volume of the steam is 1,671 m3 . The pressure of the gas is the normalatmospheric pressure (p=1,013*105Pa.)

a. How much work is done?

a. By how much did the internal energy of the substance change? (Lvap.= 2260 kJ/kg)

ENTHALPY H (J)

The state function that allows us to keep track of energy changes at

constans pressure is called ENTHALPY

ΔH = ΔU + pV

P = const; V ≠ const

ΔH = Q + W + p ΔV

ΔH=Q

• The chemical reactions take place at constantpressure.

• Heat, that relase or required can be equated to thechange in enthalpy of the system

• We can measure the Q (heat) with calorimeter and can give the enthalpy change during the reaction

ENDOTHERMAL process: ΔH > 0 (absorb heat)

EXOTERMAL process: ΔH < 0 (release heat)

The 1. law of th. The 2. law of th.

Some things happen naturally some things do not.

What determines natural directions of change?

The 1. law tells us that, if a rections take place, the total energy of the universe (reaction system and its enviroment)

remains constant.

BUT it dosn’t address the questions why do some reactions have a tendency to occor whereas other don’t?

ENTROPY (S) is a state function, which can show the direction of te procceses.

The 2. law of the thermodynamics•Spontaneous change is a change that has tendency to occur without needing to the driven by an external influence. (Hot block of metal spontaneously cools to the temperature of its surroundings, the reverse process spontaneously dosn’t occur)

•The spontaneous changes do not need fast!

•LOW ENTRORY means LITTLE DISORDER,• HIGH ENTROPY means GREAT DISORDER.

•The natural progression of a system and its surroundings is from ORDER to DISORDER

•If two system is not in thermal equilibrium, then trough the entropy change their temperature can be equalised.

The 2. of the thermodynamics

•The entropy of an isolated system increases in the course of any spontaneous change.

•Quantitative deffinition of entropy: Under isothermal condition the change in entropy of system is equal to the ratio of heat exchange and the temperature.

•ENTROPY, S (J/K)

• extensive quantity, therfore the sum of the entropies of subsystems is equal the sum of the entropy of divided up systems.

revQS

T

• Microstate: microscopic parameters of all the particles of the system (e.g. position, velocity),

• Macrostate: distribution of macroscopic parameters (e.g.temperature, pressure, density, energy)

• The number of microstates that belong to the same macrostate is called thermodynamic probability:Ω

CARNOT CYCLE•How work the refrigerators and heat – engine•An important reversible cycle is the Carnot Cycle, described by SADI CARNOT in 1824•This cycle determines the limit of our ability to convert heat into work

CARNOT CYCLE

http://www.youtube.com/watch?v=kJlmRT4E6R0&list=TLty0riI8yl2bDTXf9198A

eZSWrRtvfwkE

•The net work is the area enclosed by path ABCD•Te net amount of heat QA-QB

•The results of the cycle is that heat has been converted into work by the system.•Any required amount of work can be obtained by simply repeating the cycle•Hence, the system acts like as HEAT - ENGINE

3. example

• An engine lifts up 25000kg water into 13 m of altitude with the burning of 10 kg of fuel (L=4000kJ/kg).

• How much are the efficiancy of the process?

FREE ENERGY , F

I. Law of TD: ΔU = Q + W

II. Law of TD: Q ≤ TΔS

ΔU ≤ TΔS + W

ΔU – TΔS ≤ W

ΔF = ΔU – TΔS ≤ W F = U - TS

U = F + TS

Free energy: out of the total energy, this amount can be use for effective work Bound energy: cannot be

used for effective work, it stays in the system as heat

F = F (V,T)

Helmholtz free energy

✔ Free energy = “Useful” work obtainable from a closed thermodynamic system at a constant temperature and volume.

✔ For such a system, the negative of the difference in the Helmholtz energy is equal to the maximum amount of work extractable from a thermodynamic process in which temperature and volume are held constant.

✔ Under these conditions (T = const, V = const), free energy is minimized at equilibrium.

FREE ENTHAPLY , G(Gibbs free energy)

G = H –TS

G = U + pV - TSH = G + TS

Gibbs free energy Bound energy

G = G

(p,T)

✔ Gibbs free energy = “Useful” work obtainable from a closed thermodynamic system at a constant temperature and pressure.

✔ Gibbs energy (also referred to as ∆G) is also the chemical potential that is minimized when a system reaches equilibrium at constant pressure and temperature.

✔ The change in Gibbs free energy associated with a chemical reaction is a useful indicator of whether the reaction will proceed spontaneously. Since the change in free energy is equal to the maximum useful work which can be accomplished by the reaction

ΔG = Wmax

✔ then a negative ΔG associated with a reaction indicates that it can happen spontaneously.

When can we reach equilibrium ?

1. In a closed system, if U = const and V = const and entropy is at maximum

1. If T = const and V = const and F free energy is at minimum

2. If T = const and p = const and Gibbs free energy is at minimum

CONNECTION BETWEEN POTENTIAL FUNCTIONS

H

U pV

pVFTS

TS G

For an extra point

• A Carnot heat-engine works at 480K (isothermal expansion) and 300K (isothermal compression), theoretically. In the reality 1,2kJ work is done by the heat-engine while 4,2kJ heat is added.

• How much is the efficiency theoretically and in the reality?