Second and Third Laws of Thermodynamics

Second and Third Laws

of Thermodynamics

3.1 Entropy is a State Function

The mechanical equations of motion are invariant under time reversal.

The reverse process of a spontaneous process is nonspontaneous.

The first law provide no information regarding spontaneity of a process.

We need to devise a state function to identify a spontaneous process.

Consider that an ideal gas is heated reversibly.

rev rev rev

d

d d d d d d nRT d

U C T q w q p V q V

= = + = − = − V

d

d nRT d

q C T V

= + V

This is an inexact differential. ^V

0

T

C V

 ∂  =

 ∂ 

 

( / )

V

nRT V nR

T V

 ∂  =

 ∂ 

 

Try an integrating factor

1 T

d

d d

q C

nR

T V

T = T + V

This is an exact differential.

( / )

V

0

T

C T V

 ∂  =

 ∂ 

 

( / )

0

V

nR V T

 ∂  =

 ∂ 

 

dq

T

ENTROPY

3.2 The Second Law of Thermodynamics

d

d q

S = T

Entropy

dq

S T

∆ = ∫

dS = 0

∫

d q 0 T ≤

∫

The Clausius theorem leads to

d

d q

S ≥ T

The state function called the entropy S can be calculated from

The change in entropy in any process is given by , where the inequality applies to a spontaneous (irreversible) process and the equality applies to a reversible process.

d

d q

S = T d d q

S ≥ T

irrev

d d q S > T

1

2

irreversible

1

reversible

2 1

irrev rev

1 2

d d

q q 0

T + T <

∫ ∫

d

q d T = S

2 2

irrev

1 1

d q d 0

T − S <

∫ ∫

2 2

irrev

2 1

1 1

d d q

S S S S

∆ = − = ∫ > ∫ T

irrev

d d q

S > T

Entropy of a System

universe system surroundings

d S = d S + d S

A reversible process in which the system gains heat

dq

d

d

d

0

surr

S S q

= − T = d d

d

surr

q q

S = T = T

An irreversible process in which the system gains heat

dq

d

d

d

0

surr

S S q

= − T > d

d

surr

S q

> T

d

system

surr

S q

∆ ≥ ∫ T d

d

surr

S q

≥ T

Heat Transfer

T

T

dq d d 1 1

d d

a a

q q

S q

T

T T

T

 

= − =    −   

T

> T

T

= T

T

< T

∆ _{S > 0}

∆ _{S = 0}

∆ _{S < 0}

Spontaneous Equilibrium Impossible

The entropy of expansion of an ideal gas in an isolated system

Example 3.3

(a) Irreversible expansion of an ideal gas with no heat or work as Joule found no temperature change of a dilute gas allowed to expand in an isolated system.

(b) Reversible isothermal expansion of an ideal gas at 298 with

w = – RT ln 2

RT ln 2

1 1

ln 2 5.76 J K mol

S R

∆ = =

3.3Entropy Changes in Reversible Processes

phase change, liquid(T,P) →vapour(T,P)

rev

2 1

q H

S S S

T T

∆ = − = = ∆

dq

S T

∆ = ∫

heating at constant pressure, substance(T₁,P) →substance(T₂,P)

heating at constant volume, substance(T₁,V) →substance(T₂,V)

2 2

1 1

rev 2

1

d d ln

T T

V

T T V

q C T

S T C

T T T

∆ = ∫ = ∫ =

2 2

1 1

rev 2

1

d d ln

T T

P

T T P

q C T

S T C

T T T

∆ = ∫ = ∫ =

isothermal expansion, ideal gas(P₁, V₁, T) →idea gas(P₂,V₂, T)

2 2

1 1

2 2

1 1

d d 1

d d ln ln

V V

V V

V P

q w P

S V nR V nR nR

T T T V V P

∆ = ∫ = − ∫ = ∫ = ∫ = = −

The Molar Entropy of an Ideal Gas

Example 3.8

B(T₁,P₁) → B(T₂,P₁) → B(T₂,P₂) B(T₁,P₁) → B(T₂,P₁)

2

1

2 2

1 1

d ln ln

T

P P P

T

T T

S C T C nC

T T T

∆ = ∫ = =

B(T₂,P₁) → B(T₂,P₂) ²

1

ln P S nR

∆ = − P

5/ 2

2 2

1 1

ln T ln P

S nR

T P

     

 

∆ =    −   

   

 

 

Note

5

p

2

C = R

3.4 Entropy Changes in Irreversible Processes

H₂O(l, –10℃)

S H T

∆ = ∆

∆ S

273 liq 263

C d

S T

∆ = ∫ T

H₂O(s, –10℃) H₂O(s, 0℃) H₂O(1, 0℃)

263 ice 273

C d

S T

∆ = ∫ T

The entropy change in an irreversible process should be obtained by calculating ∆S along a reversible path between the initial state and the final state.

1

1 1

273 ( 6004 J mol )

263

(75.3 J K mol ) ln (36.8 J K mol ) ln 20.54 J K mol

263 273 K 273

S

− − − − − − −

∆ = + − + = −

1 1 1 1 1 1

(263 K) (75.3 J K mol )(10 K) 6004 J mol (36.8 J K mol )(10 K) 5619 J mol

H

∆ = − − = −

1

1 1

reservoir

(263 ) 5619 J mol

21.37 J K mol

263 263 K

H K

S

− − −

∆ = −∆ = =

1 1

universe reservoir 1 1

(21.73 20.54) J K mol 0.83 J K mol

S S

S ∆

∆ = ∆ + = − =

3.5 Entropy of Mixing Ideal Gases

P, T, V₂ n₁+n₂

P, T, V₁

P, T, V₂

P, T, V P, T, V

n₁ n₂

1 2

V = + V V

1 1

1 1 1 1 1

1 2

ln V ln n ln

S n R n R n R y

V n n

∆ = − = − = −

+

2 2

2 2 2 2 2

1 2

ln V ln n ln

S n R n R n R y

V n n

∆ = − = − = −

+

Step1

S S

S

n R

ln y

n R

ln y

∆ = ∆ + ∆ = − −

w = 0 and ∆U = 0 at constant T, then q = ∆U – w = 0

Step 2

S 0

∆ =

mix

S S

S

n R

ln y

n R

ln y

∆ = ∆ + ∆ = − −

Entropy of Mixing Two Ideal Gases

To produce one mole of an ideal mixture

The Gibbs Paradox

Example 3.11

1 2

mix 1 2 1 2

1 2

ln V ln V ln V ln V

S n R n R n R n R

V V V V

∆ = − − = +

3.6 Entropy and Statistical Probability

The equilibrium state of an isolated system is that in which entropy has its maximum value.

From a microscopic point of view, it would be the state with the maximum statistical probability.

What is the statistical probability that all of the molecules will be in the original chamber?

1 2

 N

  

Ω is the number of equally probable microscopic arrangement for a system S = k ln Ω

∆S = S

– S

= k ln(Ω

/Ω

)

The ratio Ω₂/Ω₁is equal to the ratio of the probabilities P₂/P₁. For one mole of gas molecules, P₂= 1 and

6.022 1023

1

1 P 2

 

=    

A

1 1

A

ln 1 / 1 ln 2 ln 2 (8.314 J K ) ln 2 = 5.76 J K 2

N

S k       kN R

∆ =     = = =

   

 

The Boltzmann constant k_B = 1.68 X10^–23J·K^–1 k_B N_A= R = 8.314 J·K^–1·mol^–1

S = k

ln Ω

3.7 Calorimetric Determination of Entropy

m b

m b

fus vap

0 0

m b

(s) (l) (g)

d d d

T P T P T P

T T T

H H

C C C

S S T T T

T T T T T

∆ ∆

− = ∫

+

+ ∫

+

+ ∫

 

Entropy of SO₂(g) at 298.15 K relative to its entropy at 0 K T/K Method of Calculation

0 – 15 Debye function 1.26

15 – 197.64 Graphical, solid 84.18

197.64 Fusion, 7402/197.64 37.45

197.64 – 263.08 Graphical, liquid 24.94

263.08 Vaporization, 24937/263.08 94.79

263.08 – 298.15 From C_Pof gas 5.23

247.85

1 1

/J K mol

S − −

∆ 

298.15 0

S

− S

=

Cryogenic Temperatures and Debye Function

 Liquid nitrogen b.p. 77K at 1 bar

 Liquid helium b.p. 4.2 K at 1 bar

 Liquid helium down to 0.3 K at low pressures

 Adiabatic demagnetization of a paramagnetic salt like gadolinium sulfate reaches 0.001 K

 Adiabatic demagnetization of nuclear spins reaches 10^–6K

The Debye T

Law

As T → 0 K,

C

(T ) → T

As T → 0 K,

C

(T) → aT + bT

The change in entropy approaches zero as the temperature was reduced.

S

S

Molar entropies of monoclinic and rhombic sulfur from absolute zero to the transition temperature 368.5 K

p

0 0^T

d

T

C

S S T

− = ∫ T

Assuming

S

= 0

368.5 368.5

tr

1 1

(monoclinic) (rhombic) 1.09 J K mol

S S S

− −

∆ = −

=

1 tr

H 401 J mol

∆ =

1

1 1

tr tr

tr

401 J mol

1.09 J K mol 368.5 K

S H T

− − −

∆ = ∆ = =

In 1905, Nernst concluded that as temperature approaches 0 K, ∆_rS for all reactions approaches zero.

0 r

lim 0

T

S

→

∆ =

3.8 The Third Law of Thermodynamics

In 1913, Max Planck stated.

The entropy of each pure element or substance in a perfect crystalline form is zero at absolute zero.

Corollary to Clausius’ statement of the second law.

It is impossible to reduce the temperature of a system to 0 K in a finite number of steps.

Residual entropies

N₂O

H₂O

NNO NNO NNO NNO NNO NNO NNO ONN NNO NNO ONN NNO

1 1

mix

1 1 1 1

ln ln 5.76 J K mol

2 2 2 2

S R  

∆ = −   +   =

1 1

residual

ln 3 3.37 J K mol

S R 2

∆ = =

3.9 Carnot Heat Engines

Heat reservoir at high temperature T₁

Heat reservoir at low temperature T₂

Surroundings Engine

working fluid

|q

|

|q

|

|w|

PV diagram for the working fluid in a Carnot engine. Heat |q₁| is absorbed in the isothermal expansion at T₁, and heat |q₂| is evolved in the isothermal compression at T₂. The other two steps are adiabatic.

A – B Reversible isothermal expansion at T₁. Absorbs heat |q₁| and does work |w₁|

B – C Reversible adiabatic expansion. Temperature drops to T₂and does work |w₂|

C – D Reversible isothermal compression at T₂. Discards heat |q₂| and absorbs work |w₃|

D – A Reversible adiabatic compression. Temperature rises to T₁and absorbs work |w₄|

Carnot Heat Engines

C – D Reversible isothermal compression at T₂. Discards heat |q₂| and absorbs work |w₃|

B – C Reversible adiabatic expansion. Temperature drops to T₂.

 A – B Reversible isothermal expansion at T₁. Absorbs heat |q₁| and does work |w₁|

1

B B

1 1 1 1

A A

d d d d d d

d d d

ln

q U P V C

T P V P V q w nRT V

V

V w q nRT dV nRT

V V

= + = + =

= =

 

= = =  

 

∫

1 1

1 B 2 C

T V

= T V

ln V

w q nRT  

= =  

Carnot Heat Engines

A – B isothermal expansion ∆U₁= q₁+ w₁

B – C adiabatic expansion ∆U₂= w₂

C – D isothermal compression ∆U₃= q₂+ w₃

D – A adiabatic compression ∆U₄= w₄

1 1

1 B 2 C

T V

= T V

1 1

1 A 2 D

T V

= T V

1 1 1 B 2 C

1 1

1 A 2 D

T V T V

T V T V

γ γ

γ γ

−

−

−

=

A D

V V

V = V

C

2 2

D

ln V q nRT

V

 

=  

 

B

1 1

A

ln V q nRT

V

 

=  

 

2 C 2

2

D

ln

ln nRT V

q V T

q V T

nRT V

 

 

 

= =

 

 

 

cy cy cy

cy 1 2

cy 1 2 3 4

cy cy 1 2

cy 1 2

0

| | | | | |

U q w

q q q

w w w w w

w q q q

w q q

∆ = + =

= +

= + + +

− = = +

= −

cy 1 2 2

1 1 1

| | 1 | |

w q q q

q q q

ε = ⁻ = ⁺ = −

2 2

1 1

q T q = T

2 1

1 T ε = − T

1 2

1 2

q q 0

T − T =