• ∆𝑈 = 𝑞 − 𝑤

• ∆𝑈, 𝑞 𝑎𝑛𝑑 𝑤 .

•

• ➔ ∆𝑈 = −𝑤 ➔ ∆𝑈 = 𝑞

• ∆𝑈

•

•

•

•

•

→

Heat Reservoir (HR) at high temperature T2

Heat Engine

(HE)

Heat Reservoir (HR) at low Temperature T1

q2 q1

w

∆𝑈 = 𝑞2 − 𝑞1 −𝑤

η =𝑤𝑜𝑟𝑘 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑

𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛𝑝𝑢𝑡

→ → → →

→

→

→

→

→

→

→ → → →

Δ ʹ 𝑞𝑖 = 𝑤𝑖 ⇒ 𝑞2− 𝑞1 = 𝑤

From the equation shows the efficiency, even for this idealized Carnot cycle (i.e., all processes assumed to be reversible),

is less than unity for finite values of q1 , the dissipated energy.

𝑇ℎ𝑒 𝐸𝑓𝑓𝑖𝑐𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑎𝑟𝑛𝑜𝑡 𝑐𝑦𝑐𝑙𝑒 𝑖 𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦, η =𝑤

𝑞2=𝑞2 − 𝑞1𝑞2

= 1 −𝑞1𝑞2

What is the consequence of all of the steps in the cyclic process having been conducted reversibly?

This greater efficiency could be obtained in either of two ways:

1. The same amount of thermal energy is withdrawn from the heat reservoir at t2 , i.e., 𝑞2(2)

= 𝑞2(1)

= 𝑞2 and more work ,

w(2) , is obtained from it than was obtained from the first engine; that is, w(2) > w(1) . Thus, the second engine rejects

less thermal energy to the cold reservoir at t1 than does the first engine; i.e., 𝑞1(2)

< 𝑞1(1)

.

2. The same work is obtained i.e., w(2) = w(1) = w , by withdrawing less thermal energy from the heat reservoir at t2, i.e., ,

𝑞2(2)

< 𝑞2(1)

=> 𝑞1(2)

< 𝑞1(1)

. Thus, less thermal energy, 𝑞1(2)

, is rejected into the heat reservoir at t1 by the second engine

η(2) > η(1) ⇒𝑤(2)

𝑞2(2)

>𝑤(1)

𝑞2(1)

=𝑞1(2)

𝑞2(2)

<𝑞1(1)

𝑞2(1)

Thought Experiment: Consider a second engine working with a different substance, operating between the same

temperatures t1 and t2 , and let this second engine be more efficient than the first one i.e. η(2) > η(1).

𝑋𝑛(𝑚)

X – quantitym superscript refers to engine #n subscript refers to heat reservoir #

HR2, T2

HE(2),T.S 2

HR1, T1

HE(1),T.S 1

𝑞2(1)

W(2) W(1)

𝑞2(2)

𝑞1(2)

𝑞1(1)

𝑋𝑛(𝑚)

X – quantitym superscript refers to engine #n subscript refers to heat reservoir #

HE – Heat EngineTS – Thermodynamic substanceHR – Heat ReservoirT – Temperaturew – work done/obtainedQ heat exchanged

Consider now that the second engine is run in the forward direction, and the first engine is run in the reverse direction;

that is, it acts as a heat pump .

Then, from (1) i.e., 𝑞2(2)

= 𝑞2(1)

for the second engine run in the forward direction, 𝑤(2) = 𝑞2(2)

− 𝑞1(2)

. For the first

engine run in the reverse direction, −𝑤(1)= −𝑞21+ 𝑞1

(1). The sum of the two processes is 𝑤(2)−𝑤(1)= −𝑞1

2+ 𝑞1

(1)

that is, an amount of work (w' – w ) has been obtained from a quantity of thermal energy (q1 q1) without any other

change occurring.

Although this conclusion does not contravene the First Law of Thermodynamics, it is contrary to human experience.

Such a process corresponds to perpetual motion of the second kind ; that is, heat is converted to work without leaving

a change in any other body. (Perpetual motion of the first kind is the creation of energy from nothing.)

w = 𝑞2(2)

− 𝑞1(1)

−w = −𝑞21+ 𝑞1

(1)𝑞2(2)

− 𝑞21= 𝑞1

1− 𝑞1

(2)

Conclusion of Thought Experiment: All reversible Carnot cycles operating between the same upper and lower temperatures

must have the same efficiency— namely, the maximum possible. This maximum efficiency is independent of the working

substance and is a function only of the working temperatures t1 and t2 . Thus,

Two Carnot cycles operating between t1 and t2 , and between, t2 and t3 are

equivalent to a single cycle operating between t1 and t3 . Thus,

𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = η =𝑞2 − 𝑞1𝑞2

= 𝑓′ 𝑡1, 𝑡2

𝑞1𝑞2

= 𝑓 𝑡1, 𝑡2

𝑞1𝑞2

= 𝑓 𝑡1, 𝑡2 ;𝑞2𝑞3

= 𝑓 𝑡2, 𝑡3 ;𝑞1𝑞3

= 𝑓 𝑡1, 𝑡3

𝑞1𝑞3

𝑋𝑞3𝑞2

=𝑓 𝑡1, 𝑡2𝑓 𝑡2, 𝑡3

= 𝑓 𝑡1, 𝑡3

⇒𝑞1𝑞2

= 𝑓 𝑡1, 𝑡2 =𝐹 𝑡1𝐹 𝑡2

𝑎𝑛𝑑𝑞2𝑞3

= 𝑓 𝑡2, 𝑡3 =𝐹 𝑡2𝐹 𝑡3

Kelvin took these functions to have the simplest possible form— namely, T1 and T2 . Thus,𝑞1

𝑞2=

𝑇1

𝑇2

SHOW THAT THE ABSOLUTE TEMPERTURE SCALE AND THE IDEAL GAS TEMPERATURE SCALE ARE IDENTICALHINT: CONSIDER IDEAL GAS AS THE THERMODYNAMIC SUBSTANCE IN THE CARNOT CYCLE

𝑞2−𝑞1

𝑞2=

𝑇2−𝑇1

𝑇2⇒

𝑞2

𝑇2−

𝑞1

𝑇1= 0

σ𝑞𝑖

𝑇𝑖= 0

ර𝛿𝑞𝑖𝑇

𝑟𝑒𝑣

= 0

ׯ𝛿𝑞𝑖

𝑇 𝑟𝑒𝑣= 0

𝑑𝑆′ =𝛿𝑞𝑟𝑒𝑣

𝑇

δ

ර𝑑𝑆′ = 0 = න𝐴

𝐵

𝑑𝑆′ +න𝐵

𝐴

𝑑𝑆′ = (𝑆𝐵−𝑆𝐴) + (𝑆𝐴 − 𝑆𝐵) = 0

𝑞2

𝑇2−

𝑞1

𝑇1= 0

Δ ʹ

ʹ

, 𝑑𝑆𝑖 ≥

σ𝑖 𝑑𝑆′ = 𝑑𝑆𝑖𝑟𝑟

′

ʹ

∆𝑈 = 𝑞 − 𝑤

𝑤𝑟𝑒𝑣 = 𝑉𝐴𝑉𝐵 𝑃𝑑𝑉 𝑉𝐴=

𝑉𝐵 𝑅𝑇

𝑉𝑑𝑉 = 𝑅𝑇𝑙𝑛

𝑉𝐵

𝑉𝐴

∆𝑆𝑔𝑎𝑠 = 𝑆𝐵 − 𝑆𝐴 =𝑞𝑟𝑒𝑣𝑇

=𝑤𝑟𝑒𝑣𝑇

= 𝑅𝑙𝑛𝑉𝐵𝑉𝐴

∆𝑆ℎ𝑒𝑎𝑡 𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟 = −𝑞𝑟𝑒𝑣𝑇

= −∆𝑆𝑔𝑎𝑠 = −𝑅𝑙𝑛𝑉𝐵𝑉𝐴

∆𝑆𝑡𝑜𝑡𝑎𝑙 = ∆𝑆𝑔𝑎𝑠 + ∆𝑆ℎ𝑒𝑎𝑡 𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟 = 0

•

•

•

Δ

Δ

∆𝑆𝑡𝑜𝑡𝑎𝑙 = ∆𝑆𝑔𝑎𝑠 + ∆𝑆𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟 = ∆𝑆𝑔𝑎𝑠 = 𝑅𝑙𝑛𝑉𝐵

𝑉𝐴

Δ

∆𝑆𝑔𝑎𝑠 = 𝑆𝐵 − 𝑆𝐴 = 𝑅𝑙𝑛𝑉𝐵

𝑉𝐴

0 ≤ 𝑤 ≤ 𝑤𝑚𝑎𝑥

0 ≤ 𝑞 ≤ 𝑞𝑟𝑒𝑣

⇒ 0 ≤ ∆𝑆𝑡𝑜𝑡𝑎𝑙 ≤𝑞𝑟𝑒𝑣𝑇

ΔΔ

Δ

∆𝑆 = 𝑆𝐵 − 𝑆𝐴 =𝑞𝑟𝑒𝑣

𝑇=

𝑞𝑖𝑟𝑟

𝑇+ ∆𝑆𝑖𝑟𝑟

the change in entropy can be determined only by the measurement of thermal energy transferred reversibly at the

temperature T , then entropy changes can be measured only for reversible processes, in which case the measured thermal

energy transferred is qrev and ΔSirr= 0.

For a change of state from A to B , the 1st LoT gives 𝑈𝐵 − 𝑈𝐴 = 𝑞 − 𝑤.

1st LoT gives no indication of the allowed magnitudes of q and w in the given process which vary depending on the degree of

irreversibility of the path taken between the state A and B.

2nd LoT sets a definite limit on the maximum amount of work which can be obtained from the system during a given change of

state and, hence, sets a limit on the quantity of thermal energy which the system may absorb. For an infinitesimal change of

state,

𝑑𝑆𝑠𝑦𝑠𝑡𝑒𝑚′ =

𝛿𝑞

𝑇+ 𝑑𝑆𝑖𝑟𝑟

′

𝛿𝑞 = 𝑑𝑈𝑠𝑦𝑠𝑡𝑒𝑚′ + 𝛿𝑤

𝑑𝑆𝑠𝑦𝑠𝑡𝑒𝑚′ =

𝑑𝑈𝑠𝑦𝑠𝑡𝑒𝑚′ + 𝛿𝑤

𝑇+ 𝑑𝑆𝑖𝑟𝑟

′

𝛿𝑤 = 𝑇𝑑𝑆𝑠𝑦𝑠𝑡𝑒𝑚′ − 𝑑𝑈𝑠𝑦𝑠𝑡𝑒𝑚

′ − 𝑇𝑑𝑆𝑖𝑟𝑟′

⇒ 𝛿𝑤 ≤ 𝑇𝑑𝑆𝑠𝑦𝑠𝑡𝑒𝑚′ − 𝑑𝑈𝑠𝑦𝑠𝑡𝑒𝑚

′

This work, wmax , corresponds to the absorption of the maximum heat, qrev , and is the most work that can be performed

during the change of state. Since entropy is a state function, then, in undergoing any specific change of state from A to B,

The change in the entropy of the system is the same whether the process is conducted reversibly or irreversibly.

So, it is the heat effect which is different in the two cases; that is, if the process involves the absorption of thermal energy

and is conducted reversibly, then the thermal energy absorbed, qrev , is greater than the thermal energy which would have

been absorbed if the process had been conducted irreversibly.

If the temperature remains constant throughout the process (and equal to the temperature of the reservoir supplying heat to

the system), then the integration from state A to state B gives 𝑤 ≤ 𝑇 𝑆𝐵′ − 𝑆𝐴

′ − 𝑈𝐵′ − 𝑈𝐴

′ and since U and S are functions

of state, then w cannot be greater than a certain amount, wmax, the work which is obtained from the system when the process

is conducted reversibly; that is, 𝑤𝑚𝑎𝑥 = 𝑇 𝑆𝐵′ − 𝑆𝐴

′ − 𝑈𝐵′ − 𝑈𝐴

′

∆𝑠𝑖𝑟𝑟 = 𝑆𝐵 − 𝑆𝐴 = 𝑅𝑙𝑛𝑉𝐵

𝑉𝐴

Δ

0 ≤ ∆𝑆𝑖𝑟𝑟 ≤ 𝑅𝑙𝑛𝑉𝐵𝑉𝐴

Δ Sirr = 0 for a reversible isothermal expansion and Δ Sirr = R lnVB /VA for a free expansion.

The value of Δ Sirr is thus depends on the degree of irreversibility of the process

𝑑𝑈′ = 𝛿𝑞 − 𝛿𝑤

𝑑𝑆′ =𝛿𝑞

𝑇𝑜𝑟 𝛿𝑞 = 𝑇𝑑𝑆′

𝛿𝑤 = 𝑃𝑑𝑉′

𝑑𝑈′ = 𝑇𝑑𝑆′ − 𝑃𝑑𝑉′

𝑈′ = 𝑈′ 𝑆′, 𝑉′

𝑑𝑈′ =𝜕𝑈′

𝜕𝑆′𝑉

𝑑𝑆′ +𝜕𝑈′

𝜕𝑉′𝑆

𝑑𝑉′

, 𝑇 =𝜕𝑈′

𝜕𝑆′ 𝑉𝑎𝑛𝑑 𝑃 = −

𝜕𝑈′

𝜕𝑉′ 𝑆

𝑆′ = 𝑆′ 𝑈′, 𝑉′

𝑑𝑆′ =𝜕𝑆′

𝜕𝑈′𝑉′𝑑𝑈′ +

𝜕𝑆′

𝜕𝑉′𝑈′

𝑑𝑉′

𝑑𝑆′ =𝑑𝑈′

𝑇+𝑃𝑑𝑉′

𝑇

𝜕𝑆′

𝑑𝑈′𝑉′=1

𝑇𝑎𝑛𝑑

𝜕𝑆′

𝜕𝑉′𝑈′

=𝑃

𝑇

𝜕2𝑆′

𝑑𝑈′2𝑉′= −

1

𝑇2𝜕𝑇

𝜕𝑈′ < 0

Also, since P/T > 0, increasing the volume increases the entropy of a system at constant U . T

he plot of S versus V is similar to the one of S versus U : S increases with V and the plot has negative curvature.

There is no need to have a δq transfer of energy to have entropy increase! This type of entropy is related to the increase of

the space the system occupies . We will call this configurational entropy to distinguish it from so-called thermal entropy. The

further development of thermodynamics is a consequence of the fact that S and V (or S and P ) are an inconvenient pair of

independent variables. occupies the required volume.