θ1

θ2

isotherm

Q1

Q2

adiabatic

adiabatic

Carnot cycle

A

B

C

D

Takes in heat Q1 at temperature θ1Expels heat Q2 at θ2Does work W on surroundings

P

V

PV diagram

Symbolic diagram

paths reversible

W

θ2

θ1

Q2

Q1

hot

cold

E�ciencyη =W/Q1 =1-Q2/Q1

W

θ2

θ1

Q2

Q1

Refrigerator (reverse of Carnot on page 1

W

θ2

θ1

Q2’

Q1’

hot

cold

Q2

Q1

Irrev revI R

One engine drives the other

If I is irreversible can show no irreversible cycle more e�oicent than Carnot cycle.

If η(irrev) > η(rev), then Q1’ < Q1, Nonsense as implies net heat would �ow to hot reservior

Now make both cycles Carnot cycles. Must have Q1’=Q1.E�ciences must be universal function of θ1 and θ2 ONLY

THUS Q1/Q2 = f(θ1,θ2)

W

θ2

θ1

Q2

Q1

hot

cold

Now place two Carnot cycles in series

W’

θ3

Q3

colder

Q2

For two Carnot cycles in series (previous page)

𝑄𝑄1𝑄𝑄3

= 𝑄𝑄1𝑄𝑄2

𝑄𝑄2𝑄𝑄3

and thus 𝑓𝑓(𝜃𝜃1,𝜃𝜃3) = 𝑓𝑓(𝜃𝜃1,𝜃𝜃2)𝑓𝑓(𝜃𝜃2,𝜃𝜃3)

Must have 𝑓𝑓(𝜃𝜃1,𝜃𝜃2) = 𝜑𝜑(𝜃𝜃1)𝜑𝜑(𝜃𝜃2)

Hence 𝑄𝑄1𝑄𝑄2

= 𝜑𝜑 (𝜃𝜃1)𝜑𝜑(𝜃𝜃2_

DEFINE 𝑇𝑇 = 𝜑𝜑(𝜃𝜃) and thus 𝑄𝑄1𝑄𝑄2

= 𝑇𝑇1𝑇𝑇2

Now need to show T is the familiar Absolute T in Kelvin using ideal as laws.

𝑄𝑄1 = ∫ 𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑛𝑛𝑛𝑛𝐵𝐵𝐴𝐴 𝜃𝜃1𝑙𝑙𝑛𝑛

𝑉𝑉𝐴𝐴𝑉𝑉𝐵𝐵

𝑄𝑄2 = 𝑛𝑛𝑛𝑛𝜃𝜃2𝑙𝑙𝑛𝑛𝑉𝑉𝐶𝐶𝑉𝑉𝐷𝐷

𝑄𝑄1𝑄𝑄2

=𝑇𝑇1𝑇𝑇2

=𝜃𝜃1𝑙𝑙𝑛𝑛

𝑃𝑃𝐴𝐴𝑃𝑃𝐵𝐵

𝜃𝜃2 𝑙𝑙𝑛𝑛 𝑃𝑃𝐶𝐶𝑃𝑃𝐷𝐷

For adiabatic path BC, 𝑃𝑃𝐵𝐵𝑃𝑃𝐵𝐵𝛾𝛾 = 𝑃𝑃𝐶𝐶𝑃𝑃𝐶𝐶

𝛾𝛾 or 𝜃𝜃1𝑃𝑃𝐵𝐵𝛾𝛾−1 =

𝜃𝜃2𝑃𝑃𝐶𝐶𝛾𝛾−1 (used 𝑃𝑃𝐵𝐵𝑃𝑃𝐵𝐵 = 𝑛𝑛𝑛𝑛𝜃𝜃1)

Similarly, for path DA 𝜃𝜃1𝑃𝑃𝐴𝐴𝛾𝛾−1 = 𝜃𝜃2𝑃𝑃𝐷𝐷

𝛾𝛾−1. Hence 𝑙𝑙𝑛𝑛 𝑉𝑉𝐵𝐵𝑉𝑉𝐴𝐴

= 𝑙𝑙𝑛𝑛 𝑉𝑉𝐶𝐶𝑉𝑉𝐷𝐷

and 𝑇𝑇1𝑇𝑇2

= 𝜃𝜃1𝜃𝜃2

This implies T =constant θ. We choose constant =1, and 𝑻𝑻 = 𝜽𝜽.

Now adopt normal convention for sign of Q. Q positive for heat given to system from reservoir, thus for above reversible processes −𝑄𝑄2𝑄𝑄1

= 𝑇𝑇2𝑇𝑇1

or 𝑄𝑄1𝑇𝑇1

+ 𝑄𝑄2𝑇𝑇2

= 0, and for tiny reversible steps đ𝑄𝑄1𝑇𝑇1

+ đ𝑄𝑄2𝑇𝑇2

= 0

Or ∮ đ𝑄𝑄𝑇𝑇

= 0 . Hence ∮ đ𝑄𝑄𝑇𝑇

𝐵𝐵𝐴𝐴 is independent of path

The entropy is a function of state 𝑆𝑆(𝐵𝐵,𝐴𝐴) = ∮ đ𝑄𝑄𝑇𝑇

𝐵𝐵𝐴𝐴 . Only differences in S are well

defined. History leaves no imprint on system.

First law: 𝑃𝑃𝑑𝑑 = 𝑇𝑇𝑃𝑃𝑆𝑆 − 𝑃𝑃𝑃𝑃𝑃𝑃

Heat capacities. Constant volume: 𝐶𝐶𝑉𝑉 = 𝑇𝑇 �𝜕𝜕𝜕𝜕𝜕𝜕𝑇𝑇�𝑉𝑉

Constant pressure: 𝐶𝐶𝑃𝑃 = 𝑇𝑇 �𝜕𝜕𝜕𝜕𝜕𝜕𝑇𝑇�𝑃𝑃

If know 𝐶𝐶𝑃𝑃 as function of T

𝑆𝑆(𝑇𝑇𝐹𝐹) = 𝑆𝑆(𝑇𝑇𝐼𝐼) + ∫ 𝐶𝐶𝑃𝑃𝑑𝑑𝑇𝑇𝑇𝑇

𝐹𝐹𝐼𝐼 & if 𝐶𝐶𝑃𝑃 is constant 𝑆𝑆(𝑇𝑇𝐹𝐹) = 𝑆𝑆(𝑇𝑇𝐼𝐼) + 𝐶𝐶𝑃𝑃𝑙𝑙𝑛𝑛

𝑇𝑇𝐹𝐹𝑇𝑇𝐼𝐼

Need to know S(0).

New VERY useful function

HELMHOLTZ Free Energy

𝐹𝐹 = 𝑑𝑑 − 𝑇𝑇𝑆𝑆

𝑃𝑃𝐹𝐹 = 𝑃𝑃𝑑𝑑 − 𝑇𝑇𝑃𝑃𝑆𝑆 − 𝑆𝑆𝑃𝑃𝑇𝑇 Use 𝑃𝑃𝑑𝑑 = 𝑇𝑇𝑃𝑃𝑇𝑇 − 𝑃𝑃𝑃𝑃𝑃𝑃

𝑃𝑃𝐹𝐹 = −𝑆𝑆𝑃𝑃𝑇𝑇 – 𝑃𝑃𝑃𝑃𝑃𝑃

𝑆𝑆 = −�𝜕𝜕𝐹𝐹𝜕𝜕𝑇𝑇�𝑉𝑉

and 𝑃𝑃 = −�𝜕𝜕𝐹𝐹𝜕𝜕𝑉𝑉�𝑇𝑇

F function of state, deduce Maxwell relation �𝜕𝜕𝜕𝜕𝜕𝜕𝑉𝑉�𝑇𝑇

= �𝜕𝜕𝑃𝑃𝜕𝜕𝑇𝑇�𝑉𝑉

used 𝜕𝜕𝜕𝜕𝑉𝑉�𝜕𝜕𝐹𝐹𝜕𝜕𝑇𝑇� = 𝜕𝜕

𝜕𝜕𝑇𝑇�𝜕𝜕𝐹𝐹𝜕𝜕𝑉𝑉�.

Another free energy functional is the GIBBS Free Energy

𝐺𝐺 = 𝐻𝐻 − 𝑇𝑇𝑆𝑆 𝐻𝐻 is the enthalpy for which we had 𝑃𝑃𝐻𝐻 = 𝑃𝑃𝑄𝑄(𝑟𝑟𝑟𝑟𝑟𝑟) + 𝑃𝑃𝑃𝑃𝑃𝑃

Thus 𝑃𝑃𝐺𝐺 = 𝑇𝑇𝑃𝑃𝑆𝑆 + 𝑃𝑃𝑃𝑃𝑃𝑃 − 𝑇𝑇𝑃𝑃𝑆𝑆 − 𝑆𝑆𝑃𝑃𝑇𝑇 = −𝑆𝑆𝑃𝑃𝑇𝑇 + 𝑃𝑃𝑃𝑃𝑃𝑃 = �𝑑𝑑𝑑𝑑𝑑𝑑𝑇𝑇� 𝑃𝑃𝑇𝑇 + �𝑑𝑑𝑑𝑑

𝑑𝑑𝑃𝑃� 𝑃𝑃𝑃𝑃

Whence another Maxwell relation �𝜕𝜕𝑉𝑉𝜕𝜕𝑇𝑇�𝑃𝑃

= −�𝜕𝜕𝜕𝜕𝜕𝜕𝑃𝑃�𝑇𝑇

Law of increase of S

Consider two paths between A and B: one reversible, other irreversible

The efficiencies obey 𝜂𝜂𝑟𝑟𝑟𝑟𝑟𝑟 > 𝜂𝜂𝑖𝑖𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

or 1 − (−𝑄𝑄2)𝑄𝑄1

>1 − (−𝑄𝑄2′)𝑄𝑄1′

For Carnot (−𝑄𝑄2)𝑄𝑄1

= 𝑇𝑇2𝑇𝑇1

Thus (−𝑄𝑄2′)

𝑄𝑄1′> 𝑇𝑇2

𝑇𝑇1 or đ𝑄𝑄1

′

𝑇𝑇1+ đ𝑄𝑄2

′

𝑇𝑇2 < 0

�đ𝑄𝑄𝑖𝑖𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑇𝑇 < 0

∫ đ𝑄𝑄𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑇𝑇

𝑏𝑏𝑎𝑎 < ∫ đ𝑄𝑄𝑖𝑖𝑖𝑖𝑖𝑖

𝑇𝑇𝑏𝑏𝑎𝑎 = 𝑆𝑆(𝑏𝑏,𝑎𝑎)

Clausius inequality 𝛿𝛿𝑄𝑄𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑇𝑇

< 𝛿𝛿𝑆𝑆

Thermally isolated system 𝛿𝛿𝑄𝑄𝑖𝑖𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 0

THUS 𝛿𝛿𝑆𝑆 > 0 Law of increase of entropy

All changes in isolated system lead to increase in S,

OR keep s unchanged if process is reversible.