As Leis da Termodinâmica

O que é termodinâmica?

Energia?

Trabalho? W = F⃗ . r⃗

W = |F⃗| |⃗r| cos (θ )

W = − Pext ΔV

W = ∑j

Pext , j ΔV

W = ∫V 0

V f

P dV

W = n R T ln(V f

V 0)

Gás ideal, sistema fechado, temperatura constante

Calor

Calor x Energia Térmica:

Calor: energia térmica em trânsito →ΔΔT

Energia térmica: movimento das partículas

Por que Cmolar

são tão diferentes?

?

Trocas de calor

ΔU = Q + W

Thermochemistry

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First Law of Thermodynamics• Energy is neither created nor destroyed.

• In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

Thermochemistry

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State Functions

Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

Thermochemistry

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State Functions• However, we do know that the internal energy

of a system is independent of the path by which the system achieved that state.– In the system below, the water could have reached

room temperature from either direction.

Thermochemistry

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State Functions• Therefore, internal energy is a state function.

• It depends only on the present state of the system, not on the path by which the system arrived at that state.

• And so, E depends only on Einitial and Efinal.

Thermochemistry

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State Functions

• However, q and w are not state functions.

• Whether the battery is shorted out or is discharged by running the fan, its E is the same.– But q and w are different

in the two cases.

Thermochemistry

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Hess’s Law

Hess’s law states that “[i]f a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

Thermochemistry

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Hess’s Law

Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

Thermochemistry

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Standard Enthalpies of Formation

Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure).

Thermochemistry

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Calculation of H

• Imagine this as occurringin three steps:

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

C3H8 (g) 3 C (graphite) + 4 H2 (g)

3 C (graphite) + 3 O2 (g) 3 CO2 (g)

4 H2 (g) + 2 O2 (g) 4 H2O (l)

Thermochemistry

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Calculation of H

• Imagine this as occurringin three steps:

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

C3H8 (g) 3 C (graphite) + 4 H2 (g)

3 C (graphite) + 3 O2 (g) 3 CO2 (g)

4 H2 (g) + 2 O2 (g) 4 H2O (l)

Thermochemistry

© 2009, Prentice-Hall, Inc.

Calculation of H

• Imagine this as occurringin three steps:

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

C3H8 (g) 3 C (graphite) + 4 H2 (g)

3 C (graphite) + 3 O2 (g) 3 CO2 (g)

4 H2 (g) + 2 O2 (g) 4 H2O (l)

Thermochemistry

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C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

C3H8 (g) 3 C (graphite) + 4 H2 (g)

3 C (graphite) + 3 O2 (g) 3 CO2 (g)

4 H2 (g) + 2 O2 (g) 4 H2O (l)

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

Calculation of H

• The sum of these equations is:

Thermochemistry

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Calculation of H

We can use Hess’s law in this way:

H = nHf°products – mHf° reactants

where n and m are the stoichiometric coefficients.

Thermochemistry

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H = [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)]= [(-1180.5 kJ) + (-1143.2 kJ)] – [(-103.85 kJ) + (0 kJ)]= (-2323.7 kJ) – (-103.85 kJ) = -2219.9 kJ

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

Calculation of H

U = f12

k T CV ,m =ΔUΔT

= f12

k

Entalpia

H = U + P V

CV ,m = ΔUΔT

=qV

Δ T

CP,m = Δ HΔT

=qP

ΔT

Δ H = ΔU + Δ (P V )

Δ H = ΔU + Δ(n R T )

Δ H = ΔU + R T Δ(ngás)

Δ H = qPSignificado físico

CV ,m =Δ UΔ T

CP,m =Δ HΔ T

C =calorΔ T

CP ,m =Δ HΔ T

=Δ U + n R ΔT

Δ T= CV ,m + R

Thermochemistry

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Energy in FoodsMost of the fuel in the food we eat comes from carbohydrates and fats.

Thermochemistry

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Energy in Fuels

The vast majority of the energy consumed in this country comes from fossil fuels.

A Segunda Lei da Termodinâmica

O que significa espontâneo ?

H2 (g)

+ O2 (g)

H2O

(l)

Entropia

Entropia e Desordem?

Afirmação:

A entropia de todo sistema isolado aumenta quando este sofre um processo espontâneo

Entropia x Desordem

Então, o que é entropia?

dS =dqrev

TΔ S = ∫ dqrev

TΔ S = ∫

1

2 dqrev

T

Δ S = n CV ,m ln( T final

T inicial) Gás ideal, processo reversível,

sistema fechado, volume constante

Δ S = n R ln( V final

V inicial)

Gás ideal, processo reversível, sistema fechado, temperatura constante

Δ S = n CV ,m ln( T final

T inicial)

Δ S = n R ln( V final

V inicial)

Δ S = n R ln( Pinicial

P final)

E se a pressão, não o volume, fossem fornecidos?

Calculate the change in entropy of 0.321 mol O2(g)

when its pressure is increased from 0.300 atm to 12.00 atm at constant temperature.

Δ S = n R ln( Pinicial

P final)

ΔS de Mudança de Estado Físico

Δ transição S =Δ transição H

T transição

Calculate the entropy of vaporization of acetone at 296 K with an external pressure of 1 bar. The molar heat capacity of liquid acetone is 127 JK-1mol-1, its boiling point is 329.4 K, and its enthalpy of vaporization is 29.1 kJmol-1.

S = k B ln (Ω)

Ω =N!

N1! N2! N3! ...

N j

N 0

α exp( −E j

kB T )

Qual é o fundamento microscópico para a entropia?

Procurando um critério de espontaneidade

qsurr = − qsist

T surr = constante ! ! ! ! !

Δ Ssurr =qsurr

T surr

=− qsist

T sist

= −qsist

T

Δ Ssurr =qsurr

T surr

=− qsist , P

T sist

= −Δ H sist

T

Δ Ssurr = −Δ H sist

T

E, para o Hg ?

Um processo é espontâneo se ele é

acompanhado por um aumento da entropia do

universo (sistema + vizinhanças)

Δ S total>0⇒ΔG sistema<0